Solving the integral $int_0^1int_0^1frac{1}{(1-alpha x^beta y^gamma)^2}dxdy$












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How to solve
$$int_0^1int_0^1frac{1}{(1-{alpha}x^{beta}y^{gamma})^2}dxdy$$
where $0leqalpha<1$, $betageq0$, and $gammageq0$? Does it have any closed-form?



I'm not getting any method to solve this integral. Any hint is appreciated.






integration definite-integrals







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    Expand it into series. Since $alpha < 1$, the denominator of the integrand is always positive, so the integrand is continuous, thus interchanging $sum$ and $int$ is permitted by Fubini.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Feb 9 '18 at 13:57












  • $begingroup$
    @GNUSupporter Thank you for your suggestion. Would you elaborate, please?
    $endgroup$
    – user519686
    Feb 9 '18 at 15:49








  • 2




    $begingroup$
    It has closed-form ,solution by hypergeometric function that CAS says.
    $endgroup$
    – Mariusz Iwaniuk
    Feb 9 '18 at 17:06






  • 1




    $begingroup$
    I mistakenly thought that $beta,gamma ge 0$. If not, this may not be calculable.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Feb 9 '18 at 17:09








  • 1




    $begingroup$
    @Lolita It would be helpful to know the conditions on $beta$ and $gamma$. Are they real or complex numbers? If real, are they positive?
    $endgroup$
    – David H
    Feb 9 '18 at 19:50
















11












$begingroup$


How to solve
$$int_0^1int_0^1frac{1}{(1-{alpha}x^{beta}y^{gamma})^2}dxdy$$
where $0leqalpha<1$, $betageq0$, and $gammageq0$? Does it have any closed-form?



I'm not getting any method to solve this integral. Any hint is appreciated.






integration definite-integrals







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Expand it into series. Since $alpha < 1$, the denominator of the integrand is always positive, so the integrand is continuous, thus interchanging $sum$ and $int$ is permitted by Fubini.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Feb 9 '18 at 13:57












  • $begingroup$
    @GNUSupporter Thank you for your suggestion. Would you elaborate, please?
    $endgroup$
    – user519686
    Feb 9 '18 at 15:49








  • 2




    $begingroup$
    It has closed-form ,solution by hypergeometric function that CAS says.
    $endgroup$
    – Mariusz Iwaniuk
    Feb 9 '18 at 17:06






  • 1




    $begingroup$
    I mistakenly thought that $beta,gamma ge 0$. If not, this may not be calculable.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Feb 9 '18 at 17:09








  • 1




    $begingroup$
    @Lolita It would be helpful to know the conditions on $beta$ and $gamma$. Are they real or complex numbers? If real, are they positive?
    $endgroup$
    – David H
    Feb 9 '18 at 19:50














11












11








11


5



$begingroup$


How to solve
$$int_0^1int_0^1frac{1}{(1-{alpha}x^{beta}y^{gamma})^2}dxdy$$
where $0leqalpha<1$, $betageq0$, and $gammageq0$? Does it have any closed-form?



I'm not getting any method to solve this integral. Any hint is appreciated.






integration definite-integrals







share|cite|improve this question











$endgroup$




How to solve
$$int_0^1int_0^1frac{1}{(1-{alpha}x^{beta}y^{gamma})^2}dxdy$$
where $0leqalpha<1$, $betageq0$, and $gammageq0$? Does it have any closed-form?



I'm not getting any method to solve this integral. Any hint is appreciated.






integration definite-integrals




integration definite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 5 at 12:32









Did

249k23226466




249k23226466










asked Feb 9 '18 at 13:38







user519686















  • 2




    $begingroup$
    Expand it into series. Since $alpha < 1$, the denominator of the integrand is always positive, so the integrand is continuous, thus interchanging $sum$ and $int$ is permitted by Fubini.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Feb 9 '18 at 13:57












  • $begingroup$
    @GNUSupporter Thank you for your suggestion. Would you elaborate, please?
    $endgroup$
    – user519686
    Feb 9 '18 at 15:49








  • 2




    $begingroup$
    It has closed-form ,solution by hypergeometric function that CAS says.
    $endgroup$
    – Mariusz Iwaniuk
    Feb 9 '18 at 17:06






  • 1




    $begingroup$
    I mistakenly thought that $beta,gamma ge 0$. If not, this may not be calculable.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Feb 9 '18 at 17:09








  • 1




    $begingroup$
    @Lolita It would be helpful to know the conditions on $beta$ and $gamma$. Are they real or complex numbers? If real, are they positive?
    $endgroup$
    – David H
    Feb 9 '18 at 19:50














  • 2




    $begingroup$
    Expand it into series. Since $alpha < 1$, the denominator of the integrand is always positive, so the integrand is continuous, thus interchanging $sum$ and $int$ is permitted by Fubini.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Feb 9 '18 at 13:57












  • $begingroup$
    @GNUSupporter Thank you for your suggestion. Would you elaborate, please?
    $endgroup$
    – user519686
    Feb 9 '18 at 15:49








  • 2




    $begingroup$
    It has closed-form ,solution by hypergeometric function that CAS says.
    $endgroup$
    – Mariusz Iwaniuk
    Feb 9 '18 at 17:06






  • 1




    $begingroup$
    I mistakenly thought that $beta,gamma ge 0$. If not, this may not be calculable.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Feb 9 '18 at 17:09








  • 1




    $begingroup$
    @Lolita It would be helpful to know the conditions on $beta$ and $gamma$. Are they real or complex numbers? If real, are they positive?
    $endgroup$
    – David H
    Feb 9 '18 at 19:50








2




2




$begingroup$
Expand it into series. Since $alpha < 1$, the denominator of the integrand is always positive, so the integrand is continuous, thus interchanging $sum$ and $int$ is permitted by Fubini.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Feb 9 '18 at 13:57






$begingroup$
Expand it into series. Since $alpha < 1$, the denominator of the integrand is always positive, so the integrand is continuous, thus interchanging $sum$ and $int$ is permitted by Fubini.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Feb 9 '18 at 13:57














$begingroup$
@GNUSupporter Thank you for your suggestion. Would you elaborate, please?
$endgroup$
– user519686
Feb 9 '18 at 15:49






$begingroup$
@GNUSupporter Thank you for your suggestion. Would you elaborate, please?
$endgroup$
– user519686
Feb 9 '18 at 15:49






2




2




$begingroup$
It has closed-form ,solution by hypergeometric function that CAS says.
$endgroup$
– Mariusz Iwaniuk
Feb 9 '18 at 17:06




$begingroup$
It has closed-form ,solution by hypergeometric function that CAS says.
$endgroup$
– Mariusz Iwaniuk
Feb 9 '18 at 17:06




1




1




$begingroup$
I mistakenly thought that $beta,gamma ge 0$. If not, this may not be calculable.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Feb 9 '18 at 17:09






$begingroup$
I mistakenly thought that $beta,gamma ge 0$. If not, this may not be calculable.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Feb 9 '18 at 17:09






1




1




$begingroup$
@Lolita It would be helpful to know the conditions on $beta$ and $gamma$. Are they real or complex numbers? If real, are they positive?
$endgroup$
– David H
Feb 9 '18 at 19:50




$begingroup$
@Lolita It would be helpful to know the conditions on $beta$ and $gamma$. Are they real or complex numbers? If real, are they positive?
$endgroup$
– David H
Feb 9 '18 at 19:50










2 Answers
2






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13





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Case 1: $beta neq gamma$ and $beta, gamma > 0$



As has been noted in the comments, the integral can be expressed in closed form in terms of the hypergeometric function $_2 F_1(a,b;c;z)$ provided $beta neq gamma$. To do this we will make use of the following integral representation for the hypergeometric function of
$$_2 F_1 (a,b;c;z) = frac{Gamma (c)}{Gamma (b) Gamma (c - b)} int_0^1 frac{t^{b - 1} (1 - t)^{c - b - 1}}{(1 - zt)^a} , dt. tag1$$



Now, let
$$I = int_0^1 int_0^1 frac{1}{(1 - alpha x^beta y^gamma)^2} , dx dy,$$
where $0 leqslant alpha < 1, beta > 0, gamma > 0$. Setting $z = x^beta y^gamma$, then
$$dx = frac{y^{-gamma/beta}}{beta z^{1 - 1/beta}} , dz,$$
and the integral becomes
$$I = frac{1}{beta} int_0^1 int_0^{y^gamma} frac{1}{(1 - alpha z)^2} frac{y^{-gamma/beta}}{z^{1 - 1/beta}} , dz dy.$$
Changing the order of integration gives
$$I = frac{1}{beta} int_0^1 int_{z^{1/gamma}}^1 frac{1}{(1 - alpha z)^2} frac{y^{-gamma/beta}}{z^{1 - 1/beta}} , dy dz.$$
The $y$-integration can now be readily performed. Provided $beta neq gamma$ the result is
begin{align*}
I &= frac{1}{beta - gamma} int_0^1 frac{z^{frac{1}{beta} - 1} - z^{frac{1}{gamma} - 1}}{(1 - alpha z)^2} , dz tag2
end{align*}



We now rewrite (2) in the form of (1) as follows
begin{align*}
I &= frac{1}{beta - gamma} int_0^1 frac{z^{frac{1}{beta} - 1} (1 - z)^{(frac{1}{beta} + 1) - frac{1}{beta} - 1}}{(1 - alpha z)^2} , dz - frac{1}{beta - gamma} int_0^1 frac{z^{frac{1}{gamma} - 1} (1 - z)^{(frac{1}{gamma} + 1) - frac{1}{gamma} - 1}}{(1 - alpha z)^2} , dz\
&= frac{1}{beta - gamma} frac{Gamma left (frac{1}{beta} right ) Gamma (1)}{Gamma left (1 + frac{1}{beta} right )} _2 F_1 left (2, frac{1}{beta}; 1 + frac{1}{beta}; alpha right ) - frac{1}{beta - gamma} frac{Gamma left (frac{1}{gamma} right ) Gamma (1)}{Gamma left (1 + frac{1}{gamma} right )} _2 F_1 left (2, frac{1}{gamma}; 1 + frac{1}{gamma}; alpha right ).
end{align*}



From the following property for the Gamma function, namely $Gamma (1 + z) = zGamma (z)$, the result can be simplified and leads to
$$int_0^1 int_0^1 frac{1}{(1 - alpha x^beta y^gamma)^2} , dx dy = frac{beta}{beta - gamma} _2 F_1 left (2, frac{1}{beta}; 1 + frac{1}{beta}; alpha right ) - frac{gamma}{beta - gamma} _2 F_1 left (2, frac{1}{gamma}; 1 + frac{1}{gamma}; alpha right ),$$
provided $beta neq gamma$.




Case 2: $beta = 0, gamma > 0$



The integral reduces to
$$I = int_0^1 int_0^1 frac{1}{(1 - alpha y^gamma)^2} , dx dy = int_0^1 frac{dy}{(1 - alpha y^gamma)^2} , dy.$$
Enforcing a substitution of $y mapsto y^{1/gamma}$ gives
begin{align*}
I &= frac{1}{gamma} int_0^1 frac{y^{frac{1}{gamma} - 1}}{(1 - alpha y)^2} , dy\
&= frac{1}{gamma} int_0^1 frac{y^{frac{1}{gamma} - 1}(1 - y)^{(frac{1}{gamma} + 1) - frac{1}{gamma} - 1}}{(1 - alpha y)^2} , dy\
&= frac{1}{gamma} frac{Gamma (1/gamma) Gamma (1)}{Gamma (1 + 1/gamma)} _2 F_1 left (2, frac{1}{gamma}; 1 + frac{1}{gamma}; alpha right )\
&= _2 F_1 left (2, frac{1}{gamma}; 1 + frac{1}{gamma}; alpha right ).
end{align*}




Case 3: $gamma = 0, beta > 0$



Similarly, by symmetry we have
$$I = _2 F_1 left (2, frac{1}{beta}; 1 + frac{1}{beta}; alpha right ).$$




Case 4: $beta = gamma = 0$



Here we trivially have
$$I = frac{1}{(1 - alpha)^2}.$$




Case 5: $beta = gamma$ and $beta, gamma > 0$



Thanks to @Paul Enta the solution to the integral for this particular case can now be given.



If $beta = gamma$, after performing the $y$-integration rather than the result given in (2) one will instead have
$$I = -frac{1}{beta^2} int_0^1 frac{z^{frac{1}{beta} - 1} ln z}{(1 - alpha z)^2} , dz. tag3$$



Now consider the integral
$$J(s) = -frac{1}{beta^2} int_0^1 frac{z^{s + frac{1}{beta} - 1}}{(1 - alpha z)^2} , dz.$$
Now observe that $J'(0) = I$ so in order to find $I$ we just need to find $J$ first before taking its derivative with respect to the parameter $s$ before setting it equal to zero.



Now
begin{align*}
J(s) &= -frac{1}{beta^2} int_0^1 frac{z^{s + frac{1}{beta} - 1} (1 - z)^{(s + frac{1}{beta} + 1) - s - frac{1}{beta} - 1}}{(1 - alpha z)^2} , dz\
&= -frac{1}{beta^2} frac{Gamma (s + 1/beta) Gamma (1)}{Gamma (s + 1/beta + 1)} _2F_1 left (2, s + frac{1}{beta}; s + 1 + frac{1}{beta}; alpha right )\
&= -frac{1}{beta^2} frac{1}{(s + frac{1}{beta})} _2F_1 left (2, s + frac{1}{beta}; s + 1 + frac{1}{beta}; alpha right ).
end{align*}
Thus
begin{align*}
J'(s) &= frac{1}{beta^2 (s + frac{1}{beta})^2} _2F_1 left (2,s + frac{1}{s}; s + 1 + frac{1}{s}; alpha right ) - frac{1}{beta^2 (s + frac{1}{beta})} frac{partial}{partial s} _2F_1 left (2,s + frac{1}{beta}; s + 1 + frac{1}{beta}; alpha right ).tag4
end{align*}
Now
begin{align*}
frac{partial}{partial s} _2F_1 left (2,s + frac{1}{beta}; s + 1 + frac{1}{beta}; alpha right ) &= frac{partial}{partial s} sum_{n = 0}^infty frac{(2)_n (s + frac{1}{beta})_n}{(s + 1 + frac{1}{beta})_n} frac{alpha^n}{n!}\
&= sum_{n = 0}^infty (2)_n frac{alpha^n}{n!} frac{partial}{partial s} left [frac{(s + frac{1}{beta})_n}{(s + 1 + frac{1}{beta})_n} right ]\
&= sum_{n = 0}^infty (2)_n frac{n}{(s + n + frac{1}{beta})^2} frac{alpha^n}{n!}.
end{align*}



We now make use of the following property for the Pochhammer symbol $(x)_n$ (the raising factorial) of
$$(x)_n = frac{Gamma (x + n)}{Gamma (x)}.$$



So
begin{align*}
frac{partial}{partial s} _2F_1 left (2,s + frac{1}{beta}; s + 1 + frac{1}{beta}; alpha right ) &= sum_{n = 0}^infty frac{n(n +1) alpha^n}{(s + n + frac{1}{beta})^2} = sum_{n = 1}^infty frac{n(n +1) alpha^n}{(s + n + frac{1}{beta})^2},
end{align*}
and we have
$$J'(0) = _2F_1 left (2,frac{1}{beta};1 + frac{1}{beta}; alpha right ) - frac{1}{beta} sum_{n = 1}^infty frac{n (n + 1) alpha^n}{(n + frac{1}{beta})^2}. tag5$$



It now remains to write the infinite sum appearing in (5) in closed form in terms of a hypergeometric function. Setting
$$S = sum_{n = 1}^infty frac{n(n + 1) alpha^n}{(n + frac{1}{beta})^2}.$$
Shifting the index $n mapsto n + 1$ gives
$$S = alpha sum_{n = 0}^infty frac{(n + 1)(n + 2) alpha^n}{(n + frac{1}{beta} + 1)^2}.tag6 $$
Now, noting that
$$left (1 + frac{1}{beta} right )_n = frac{Gamma left (n + frac{1}{beta} + 1 right )}{Gamma left (1 + frac{1}{beta} right )} = frac{left (frac{1}{beta} + n right ) Gamma left (frac{1}{beta} + n right )}{frac{1}{beta} Gamma left (frac{1}{beta} right )}.$$
and
$$left (2 + frac{1}{beta} right )_n = frac{Gamma left (n + frac{1}{beta} + 2 right )}{Gamma left (2 + frac{1}{beta} right )} = frac{left (frac{1}{beta} + n + 1 right ) left (frac{1}{beta} + n right ) Gamma left (frac{1}{beta} + n right )}{left (1 + frac{1}{beta} right ) frac{1}{beta} Gamma left (frac{1}{beta} right )},$$
we see that
$$frac{left (1 + frac{1}{beta} right )_n}{left (2 + frac{1}{beta} right )_n} = frac{1 + frac{1}{beta}}{1 + frac{1}{beta} + n}.tag7$$
Also
$$(3)_n = frac{Gamma (3 + n)}{Gamma (3)} = frac{(2 + n)!}{2!} = frac{(n + 1)(n + 2) n!}{2}.tag8$$



Using (7) and (8), the sum in (6) can be written as
begin{align*}
S &= frac{2 alpha}{(1 + frac{1}{beta})^2} sum_{n = 0}^infty frac{(3)_n left (1 + frac{1}{beta} right )_n left (1 + frac{1}{beta} right )_n}{left (2 + frac{1}{beta} right )_n left (2 + frac{1}{beta} right )_n} frac{alpha^n}{n!}\
&= frac{2 alpha beta^2}{(1 + beta)^2} _3F_2 left (3, 1 + frac{1}{beta}, 1 + frac{1}{beta}; 2 + frac{1}{beta}, 2 + frac{1}{beta}; alpha right ).
end{align*}



So finally, for the case $beta = gamma$ such that $beta, gamma > 0$ we have
$$I = _2F_1 left (2, frac{1}{beta}; 1 + frac{1}{beta}; alpha right ) - frac{2 alpha beta}{(1 + beta)^2} _3F_2 left (3,1 + frac{1}{beta}, 1 + frac{1}{beta};2 + frac{1}{beta}, 2 + frac{1}{beta}; alpha right ).$$






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  • 1




    $begingroup$
    This last integral can be evaluated by derivation under the integral: $$ I= -frac{1}{beta^2}left. frac{d}{ds} int_0^1 frac{z^{s+frac{1}{beta} - 1} ln z}{(1 - alpha z)^2} , dzright|_{s=0}$$ which can be calculated as the derivative of an hypergeometric function. The result is $$ I= _2 F_1 left (2, frac{1}{beta}; 1 + frac{1}{beta}; alpha right )-frac{2alphabeta}{left( 1+beta right)^2} _3 F_2 left (3, frac{1}{beta}, frac{1}{beta}; 2 + frac{1}{beta}, 2+frac{1}{beta}; alpha right ) $$
    $endgroup$
    – Paul Enta
    Feb 16 '18 at 18:28






  • 1




    $begingroup$
    Alternatively, from the solution obtained for $betaneqgamma$, it can also be expressed as $$ lim_{betatogamma}frac{beta _2 F_1 left (2, frac{1}{beta}; 1 + frac{1}{beta}; alpha right )-gamma _2 F_1 left (2, frac{1}{gamma}; 1 + frac{1}{gamma}; alpha right )}{beta-gamma}=frac{d}{dbeta}left[beta _2 F_1 left (2, frac{1}{beta}; 1 + frac{1}{beta}; alpha right )right]$$
    $endgroup$
    – Paul Enta
    Feb 16 '18 at 18:28












  • $begingroup$
    @Paul Enta - Thanks for your suggestion on how to evaluate the integral when $beta = gamma$. It takes quite an effort to evaluate, so thanks again. Just one remark. I have updated my solution where you will find you have made a small error in the argument for the $_3F_2$ term.
    $endgroup$
    – omegadot
    Feb 17 '18 at 3:54












  • $begingroup$
    You're welcome. Sorry for the typo. To construct the hypergeometric function, you may find it easier to work with Gamma functions from the beginning, to build the ratio of two successive terms of the series and to identify the constants just like here.
    $endgroup$
    – Paul Enta
    Feb 17 '18 at 9:35










  • $begingroup$
    Beautiful answer! Unfortunately, I have reached my daily vote limit and I have to wait a couple of hours before I can vote $(+1)$.
    $endgroup$
    – user477343
    Feb 19 '18 at 14:59



















4












$begingroup$

One feels slightly sorry to post this in view of the truly Herculean task that a previously posted answer represents, but it happens that, at least for every $alpha$ in $[0,1)$ and every nonnegative $beta$ and $gamma$,




$$int_0^1!!int_0^1frac{dxdy}{(1-{alpha}x^{beta}y^{gamma})^2}=sum_{n=0}^inftyfrac{n+1}{(nbeta+1)(ngamma+1)}alpha^n$$




Two remarks:




  • This identity actually holds for every complex parameters such that $|alpha|<1$, $Rebeta>0$ and $Regamma>0$.


  • Various equivalent presentations of the RHS exist, as series and/or as special functions evaluated at $alpha$, but these alternative formulas/denominations seem to bring no further mathematical understanding of the situation hence we omit them.



The proof of the identity above is a one-liner: integrate term by term the series expansion $$frac1{(1-{alpha}x^{beta}y^{gamma})^2}=sum_{n=0}^infty(n+1)alpha^nx^{nbeta}y^{ngamma}$$






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  • 3




    $begingroup$
    Sure, but as the OP explicitly asked for a closed-form solution, and by closed-form solution I assumed a solution in terms of any of the known functions, special or otherwise, of mathematical physics, this is what I attempted to provide. I usually don't think of infinite sums as closed in form but of course all functions of mathematical physics can be represented by infinite sums.
    $endgroup$
    – omegadot
    Jan 5 at 12:42






  • 3




    $begingroup$
    @omegadot The difference is specious. As hinted at in my answer, one could easily express the sum of the series above using one or several from the vast list of special functions with a name, but this brings zero extra understanding. (In addition, and since the goal of your comment here seems actually to justify your answer, I might mention that I fail to understand why all the complications and separate cases in it are needed at all.)
    $endgroup$
    – Did
    Jan 5 at 14:37














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2 Answers
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2 Answers
2






active

oldest

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active

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active

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13





+100







$begingroup$

Case 1: $beta neq gamma$ and $beta, gamma > 0$



As has been noted in the comments, the integral can be expressed in closed form in terms of the hypergeometric function $_2 F_1(a,b;c;z)$ provided $beta neq gamma$. To do this we will make use of the following integral representation for the hypergeometric function of
$$_2 F_1 (a,b;c;z) = frac{Gamma (c)}{Gamma (b) Gamma (c - b)} int_0^1 frac{t^{b - 1} (1 - t)^{c - b - 1}}{(1 - zt)^a} , dt. tag1$$



Now, let
$$I = int_0^1 int_0^1 frac{1}{(1 - alpha x^beta y^gamma)^2} , dx dy,$$
where $0 leqslant alpha < 1, beta > 0, gamma > 0$. Setting $z = x^beta y^gamma$, then
$$dx = frac{y^{-gamma/beta}}{beta z^{1 - 1/beta}} , dz,$$
and the integral becomes
$$I = frac{1}{beta} int_0^1 int_0^{y^gamma} frac{1}{(1 - alpha z)^2} frac{y^{-gamma/beta}}{z^{1 - 1/beta}} , dz dy.$$
Changing the order of integration gives
$$I = frac{1}{beta} int_0^1 int_{z^{1/gamma}}^1 frac{1}{(1 - alpha z)^2} frac{y^{-gamma/beta}}{z^{1 - 1/beta}} , dy dz.$$
The $y$-integration can now be readily performed. Provided $beta neq gamma$ the result is
begin{align*}
I &= frac{1}{beta - gamma} int_0^1 frac{z^{frac{1}{beta} - 1} - z^{frac{1}{gamma} - 1}}{(1 - alpha z)^2} , dz tag2
end{align*}



We now rewrite (2) in the form of (1) as follows
begin{align*}
I &= frac{1}{beta - gamma} int_0^1 frac{z^{frac{1}{beta} - 1} (1 - z)^{(frac{1}{beta} + 1) - frac{1}{beta} - 1}}{(1 - alpha z)^2} , dz - frac{1}{beta - gamma} int_0^1 frac{z^{frac{1}{gamma} - 1} (1 - z)^{(frac{1}{gamma} + 1) - frac{1}{gamma} - 1}}{(1 - alpha z)^2} , dz\
&= frac{1}{beta - gamma} frac{Gamma left (frac{1}{beta} right ) Gamma (1)}{Gamma left (1 + frac{1}{beta} right )} _2 F_1 left (2, frac{1}{beta}; 1 + frac{1}{beta}; alpha right ) - frac{1}{beta - gamma} frac{Gamma left (frac{1}{gamma} right ) Gamma (1)}{Gamma left (1 + frac{1}{gamma} right )} _2 F_1 left (2, frac{1}{gamma}; 1 + frac{1}{gamma}; alpha right ).
end{align*}



From the following property for the Gamma function, namely $Gamma (1 + z) = zGamma (z)$, the result can be simplified and leads to
$$int_0^1 int_0^1 frac{1}{(1 - alpha x^beta y^gamma)^2} , dx dy = frac{beta}{beta - gamma} _2 F_1 left (2, frac{1}{beta}; 1 + frac{1}{beta}; alpha right ) - frac{gamma}{beta - gamma} _2 F_1 left (2, frac{1}{gamma}; 1 + frac{1}{gamma}; alpha right ),$$
provided $beta neq gamma$.




Case 2: $beta = 0, gamma > 0$



The integral reduces to
$$I = int_0^1 int_0^1 frac{1}{(1 - alpha y^gamma)^2} , dx dy = int_0^1 frac{dy}{(1 - alpha y^gamma)^2} , dy.$$
Enforcing a substitution of $y mapsto y^{1/gamma}$ gives
begin{align*}
I &= frac{1}{gamma} int_0^1 frac{y^{frac{1}{gamma} - 1}}{(1 - alpha y)^2} , dy\
&= frac{1}{gamma} int_0^1 frac{y^{frac{1}{gamma} - 1}(1 - y)^{(frac{1}{gamma} + 1) - frac{1}{gamma} - 1}}{(1 - alpha y)^2} , dy\
&= frac{1}{gamma} frac{Gamma (1/gamma) Gamma (1)}{Gamma (1 + 1/gamma)} _2 F_1 left (2, frac{1}{gamma}; 1 + frac{1}{gamma}; alpha right )\
&= _2 F_1 left (2, frac{1}{gamma}; 1 + frac{1}{gamma}; alpha right ).
end{align*}




Case 3: $gamma = 0, beta > 0$



Similarly, by symmetry we have
$$I = _2 F_1 left (2, frac{1}{beta}; 1 + frac{1}{beta}; alpha right ).$$




Case 4: $beta = gamma = 0$



Here we trivially have
$$I = frac{1}{(1 - alpha)^2}.$$




Case 5: $beta = gamma$ and $beta, gamma > 0$



Thanks to @Paul Enta the solution to the integral for this particular case can now be given.



If $beta = gamma$, after performing the $y$-integration rather than the result given in (2) one will instead have
$$I = -frac{1}{beta^2} int_0^1 frac{z^{frac{1}{beta} - 1} ln z}{(1 - alpha z)^2} , dz. tag3$$



Now consider the integral
$$J(s) = -frac{1}{beta^2} int_0^1 frac{z^{s + frac{1}{beta} - 1}}{(1 - alpha z)^2} , dz.$$
Now observe that $J'(0) = I$ so in order to find $I$ we just need to find $J$ first before taking its derivative with respect to the parameter $s$ before setting it equal to zero.



Now
begin{align*}
J(s) &= -frac{1}{beta^2} int_0^1 frac{z^{s + frac{1}{beta} - 1} (1 - z)^{(s + frac{1}{beta} + 1) - s - frac{1}{beta} - 1}}{(1 - alpha z)^2} , dz\
&= -frac{1}{beta^2} frac{Gamma (s + 1/beta) Gamma (1)}{Gamma (s + 1/beta + 1)} _2F_1 left (2, s + frac{1}{beta}; s + 1 + frac{1}{beta}; alpha right )\
&= -frac{1}{beta^2} frac{1}{(s + frac{1}{beta})} _2F_1 left (2, s + frac{1}{beta}; s + 1 + frac{1}{beta}; alpha right ).
end{align*}
Thus
begin{align*}
J'(s) &= frac{1}{beta^2 (s + frac{1}{beta})^2} _2F_1 left (2,s + frac{1}{s}; s + 1 + frac{1}{s}; alpha right ) - frac{1}{beta^2 (s + frac{1}{beta})} frac{partial}{partial s} _2F_1 left (2,s + frac{1}{beta}; s + 1 + frac{1}{beta}; alpha right ).tag4
end{align*}
Now
begin{align*}
frac{partial}{partial s} _2F_1 left (2,s + frac{1}{beta}; s + 1 + frac{1}{beta}; alpha right ) &= frac{partial}{partial s} sum_{n = 0}^infty frac{(2)_n (s + frac{1}{beta})_n}{(s + 1 + frac{1}{beta})_n} frac{alpha^n}{n!}\
&= sum_{n = 0}^infty (2)_n frac{alpha^n}{n!} frac{partial}{partial s} left [frac{(s + frac{1}{beta})_n}{(s + 1 + frac{1}{beta})_n} right ]\
&= sum_{n = 0}^infty (2)_n frac{n}{(s + n + frac{1}{beta})^2} frac{alpha^n}{n!}.
end{align*}



We now make use of the following property for the Pochhammer symbol $(x)_n$ (the raising factorial) of
$$(x)_n = frac{Gamma (x + n)}{Gamma (x)}.$$



So
begin{align*}
frac{partial}{partial s} _2F_1 left (2,s + frac{1}{beta}; s + 1 + frac{1}{beta}; alpha right ) &= sum_{n = 0}^infty frac{n(n +1) alpha^n}{(s + n + frac{1}{beta})^2} = sum_{n = 1}^infty frac{n(n +1) alpha^n}{(s + n + frac{1}{beta})^2},
end{align*}
and we have
$$J'(0) = _2F_1 left (2,frac{1}{beta};1 + frac{1}{beta}; alpha right ) - frac{1}{beta} sum_{n = 1}^infty frac{n (n + 1) alpha^n}{(n + frac{1}{beta})^2}. tag5$$



It now remains to write the infinite sum appearing in (5) in closed form in terms of a hypergeometric function. Setting
$$S = sum_{n = 1}^infty frac{n(n + 1) alpha^n}{(n + frac{1}{beta})^2}.$$
Shifting the index $n mapsto n + 1$ gives
$$S = alpha sum_{n = 0}^infty frac{(n + 1)(n + 2) alpha^n}{(n + frac{1}{beta} + 1)^2}.tag6 $$
Now, noting that
$$left (1 + frac{1}{beta} right )_n = frac{Gamma left (n + frac{1}{beta} + 1 right )}{Gamma left (1 + frac{1}{beta} right )} = frac{left (frac{1}{beta} + n right ) Gamma left (frac{1}{beta} + n right )}{frac{1}{beta} Gamma left (frac{1}{beta} right )}.$$
and
$$left (2 + frac{1}{beta} right )_n = frac{Gamma left (n + frac{1}{beta} + 2 right )}{Gamma left (2 + frac{1}{beta} right )} = frac{left (frac{1}{beta} + n + 1 right ) left (frac{1}{beta} + n right ) Gamma left (frac{1}{beta} + n right )}{left (1 + frac{1}{beta} right ) frac{1}{beta} Gamma left (frac{1}{beta} right )},$$
we see that
$$frac{left (1 + frac{1}{beta} right )_n}{left (2 + frac{1}{beta} right )_n} = frac{1 + frac{1}{beta}}{1 + frac{1}{beta} + n}.tag7$$
Also
$$(3)_n = frac{Gamma (3 + n)}{Gamma (3)} = frac{(2 + n)!}{2!} = frac{(n + 1)(n + 2) n!}{2}.tag8$$



Using (7) and (8), the sum in (6) can be written as
begin{align*}
S &= frac{2 alpha}{(1 + frac{1}{beta})^2} sum_{n = 0}^infty frac{(3)_n left (1 + frac{1}{beta} right )_n left (1 + frac{1}{beta} right )_n}{left (2 + frac{1}{beta} right )_n left (2 + frac{1}{beta} right )_n} frac{alpha^n}{n!}\
&= frac{2 alpha beta^2}{(1 + beta)^2} _3F_2 left (3, 1 + frac{1}{beta}, 1 + frac{1}{beta}; 2 + frac{1}{beta}, 2 + frac{1}{beta}; alpha right ).
end{align*}



So finally, for the case $beta = gamma$ such that $beta, gamma > 0$ we have
$$I = _2F_1 left (2, frac{1}{beta}; 1 + frac{1}{beta}; alpha right ) - frac{2 alpha beta}{(1 + beta)^2} _3F_2 left (3,1 + frac{1}{beta}, 1 + frac{1}{beta};2 + frac{1}{beta}, 2 + frac{1}{beta}; alpha right ).$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    This last integral can be evaluated by derivation under the integral: $$ I= -frac{1}{beta^2}left. frac{d}{ds} int_0^1 frac{z^{s+frac{1}{beta} - 1} ln z}{(1 - alpha z)^2} , dzright|_{s=0}$$ which can be calculated as the derivative of an hypergeometric function. The result is $$ I= _2 F_1 left (2, frac{1}{beta}; 1 + frac{1}{beta}; alpha right )-frac{2alphabeta}{left( 1+beta right)^2} _3 F_2 left (3, frac{1}{beta}, frac{1}{beta}; 2 + frac{1}{beta}, 2+frac{1}{beta}; alpha right ) $$
    $endgroup$
    – Paul Enta
    Feb 16 '18 at 18:28






  • 1




    $begingroup$
    Alternatively, from the solution obtained for $betaneqgamma$, it can also be expressed as $$ lim_{betatogamma}frac{beta _2 F_1 left (2, frac{1}{beta}; 1 + frac{1}{beta}; alpha right )-gamma _2 F_1 left (2, frac{1}{gamma}; 1 + frac{1}{gamma}; alpha right )}{beta-gamma}=frac{d}{dbeta}left[beta _2 F_1 left (2, frac{1}{beta}; 1 + frac{1}{beta}; alpha right )right]$$
    $endgroup$
    – Paul Enta
    Feb 16 '18 at 18:28












  • $begingroup$
    @Paul Enta - Thanks for your suggestion on how to evaluate the integral when $beta = gamma$. It takes quite an effort to evaluate, so thanks again. Just one remark. I have updated my solution where you will find you have made a small error in the argument for the $_3F_2$ term.
    $endgroup$
    – omegadot
    Feb 17 '18 at 3:54












  • $begingroup$
    You're welcome. Sorry for the typo. To construct the hypergeometric function, you may find it easier to work with Gamma functions from the beginning, to build the ratio of two successive terms of the series and to identify the constants just like here.
    $endgroup$
    – Paul Enta
    Feb 17 '18 at 9:35










  • $begingroup$
    Beautiful answer! Unfortunately, I have reached my daily vote limit and I have to wait a couple of hours before I can vote $(+1)$.
    $endgroup$
    – user477343
    Feb 19 '18 at 14:59
















13





+100







$begingroup$

Case 1: $beta neq gamma$ and $beta, gamma > 0$



As has been noted in the comments, the integral can be expressed in closed form in terms of the hypergeometric function $_2 F_1(a,b;c;z)$ provided $beta neq gamma$. To do this we will make use of the following integral representation for the hypergeometric function of
$$_2 F_1 (a,b;c;z) = frac{Gamma (c)}{Gamma (b) Gamma (c - b)} int_0^1 frac{t^{b - 1} (1 - t)^{c - b - 1}}{(1 - zt)^a} , dt. tag1$$



Now, let
$$I = int_0^1 int_0^1 frac{1}{(1 - alpha x^beta y^gamma)^2} , dx dy,$$
where $0 leqslant alpha < 1, beta > 0, gamma > 0$. Setting $z = x^beta y^gamma$, then
$$dx = frac{y^{-gamma/beta}}{beta z^{1 - 1/beta}} , dz,$$
and the integral becomes
$$I = frac{1}{beta} int_0^1 int_0^{y^gamma} frac{1}{(1 - alpha z)^2} frac{y^{-gamma/beta}}{z^{1 - 1/beta}} , dz dy.$$
Changing the order of integration gives
$$I = frac{1}{beta} int_0^1 int_{z^{1/gamma}}^1 frac{1}{(1 - alpha z)^2} frac{y^{-gamma/beta}}{z^{1 - 1/beta}} , dy dz.$$
The $y$-integration can now be readily performed. Provided $beta neq gamma$ the result is
begin{align*}
I &= frac{1}{beta - gamma} int_0^1 frac{z^{frac{1}{beta} - 1} - z^{frac{1}{gamma} - 1}}{(1 - alpha z)^2} , dz tag2
end{align*}



We now rewrite (2) in the form of (1) as follows
begin{align*}
I &= frac{1}{beta - gamma} int_0^1 frac{z^{frac{1}{beta} - 1} (1 - z)^{(frac{1}{beta} + 1) - frac{1}{beta} - 1}}{(1 - alpha z)^2} , dz - frac{1}{beta - gamma} int_0^1 frac{z^{frac{1}{gamma} - 1} (1 - z)^{(frac{1}{gamma} + 1) - frac{1}{gamma} - 1}}{(1 - alpha z)^2} , dz\
&= frac{1}{beta - gamma} frac{Gamma left (frac{1}{beta} right ) Gamma (1)}{Gamma left (1 + frac{1}{beta} right )} _2 F_1 left (2, frac{1}{beta}; 1 + frac{1}{beta}; alpha right ) - frac{1}{beta - gamma} frac{Gamma left (frac{1}{gamma} right ) Gamma (1)}{Gamma left (1 + frac{1}{gamma} right )} _2 F_1 left (2, frac{1}{gamma}; 1 + frac{1}{gamma}; alpha right ).
end{align*}



From the following property for the Gamma function, namely $Gamma (1 + z) = zGamma (z)$, the result can be simplified and leads to
$$int_0^1 int_0^1 frac{1}{(1 - alpha x^beta y^gamma)^2} , dx dy = frac{beta}{beta - gamma} _2 F_1 left (2, frac{1}{beta}; 1 + frac{1}{beta}; alpha right ) - frac{gamma}{beta - gamma} _2 F_1 left (2, frac{1}{gamma}; 1 + frac{1}{gamma}; alpha right ),$$
provided $beta neq gamma$.




Case 2: $beta = 0, gamma > 0$



The integral reduces to
$$I = int_0^1 int_0^1 frac{1}{(1 - alpha y^gamma)^2} , dx dy = int_0^1 frac{dy}{(1 - alpha y^gamma)^2} , dy.$$
Enforcing a substitution of $y mapsto y^{1/gamma}$ gives
begin{align*}
I &= frac{1}{gamma} int_0^1 frac{y^{frac{1}{gamma} - 1}}{(1 - alpha y)^2} , dy\
&= frac{1}{gamma} int_0^1 frac{y^{frac{1}{gamma} - 1}(1 - y)^{(frac{1}{gamma} + 1) - frac{1}{gamma} - 1}}{(1 - alpha y)^2} , dy\
&= frac{1}{gamma} frac{Gamma (1/gamma) Gamma (1)}{Gamma (1 + 1/gamma)} _2 F_1 left (2, frac{1}{gamma}; 1 + frac{1}{gamma}; alpha right )\
&= _2 F_1 left (2, frac{1}{gamma}; 1 + frac{1}{gamma}; alpha right ).
end{align*}




Case 3: $gamma = 0, beta > 0$



Similarly, by symmetry we have
$$I = _2 F_1 left (2, frac{1}{beta}; 1 + frac{1}{beta}; alpha right ).$$




Case 4: $beta = gamma = 0$



Here we trivially have
$$I = frac{1}{(1 - alpha)^2}.$$




Case 5: $beta = gamma$ and $beta, gamma > 0$



Thanks to @Paul Enta the solution to the integral for this particular case can now be given.



If $beta = gamma$, after performing the $y$-integration rather than the result given in (2) one will instead have
$$I = -frac{1}{beta^2} int_0^1 frac{z^{frac{1}{beta} - 1} ln z}{(1 - alpha z)^2} , dz. tag3$$



Now consider the integral
$$J(s) = -frac{1}{beta^2} int_0^1 frac{z^{s + frac{1}{beta} - 1}}{(1 - alpha z)^2} , dz.$$
Now observe that $J'(0) = I$ so in order to find $I$ we just need to find $J$ first before taking its derivative with respect to the parameter $s$ before setting it equal to zero.



Now
begin{align*}
J(s) &= -frac{1}{beta^2} int_0^1 frac{z^{s + frac{1}{beta} - 1} (1 - z)^{(s + frac{1}{beta} + 1) - s - frac{1}{beta} - 1}}{(1 - alpha z)^2} , dz\
&= -frac{1}{beta^2} frac{Gamma (s + 1/beta) Gamma (1)}{Gamma (s + 1/beta + 1)} _2F_1 left (2, s + frac{1}{beta}; s + 1 + frac{1}{beta}; alpha right )\
&= -frac{1}{beta^2} frac{1}{(s + frac{1}{beta})} _2F_1 left (2, s + frac{1}{beta}; s + 1 + frac{1}{beta}; alpha right ).
end{align*}
Thus
begin{align*}
J'(s) &= frac{1}{beta^2 (s + frac{1}{beta})^2} _2F_1 left (2,s + frac{1}{s}; s + 1 + frac{1}{s}; alpha right ) - frac{1}{beta^2 (s + frac{1}{beta})} frac{partial}{partial s} _2F_1 left (2,s + frac{1}{beta}; s + 1 + frac{1}{beta}; alpha right ).tag4
end{align*}
Now
begin{align*}
frac{partial}{partial s} _2F_1 left (2,s + frac{1}{beta}; s + 1 + frac{1}{beta}; alpha right ) &= frac{partial}{partial s} sum_{n = 0}^infty frac{(2)_n (s + frac{1}{beta})_n}{(s + 1 + frac{1}{beta})_n} frac{alpha^n}{n!}\
&= sum_{n = 0}^infty (2)_n frac{alpha^n}{n!} frac{partial}{partial s} left [frac{(s + frac{1}{beta})_n}{(s + 1 + frac{1}{beta})_n} right ]\
&= sum_{n = 0}^infty (2)_n frac{n}{(s + n + frac{1}{beta})^2} frac{alpha^n}{n!}.
end{align*}



We now make use of the following property for the Pochhammer symbol $(x)_n$ (the raising factorial) of
$$(x)_n = frac{Gamma (x + n)}{Gamma (x)}.$$



So
begin{align*}
frac{partial}{partial s} _2F_1 left (2,s + frac{1}{beta}; s + 1 + frac{1}{beta}; alpha right ) &= sum_{n = 0}^infty frac{n(n +1) alpha^n}{(s + n + frac{1}{beta})^2} = sum_{n = 1}^infty frac{n(n +1) alpha^n}{(s + n + frac{1}{beta})^2},
end{align*}
and we have
$$J'(0) = _2F_1 left (2,frac{1}{beta};1 + frac{1}{beta}; alpha right ) - frac{1}{beta} sum_{n = 1}^infty frac{n (n + 1) alpha^n}{(n + frac{1}{beta})^2}. tag5$$



It now remains to write the infinite sum appearing in (5) in closed form in terms of a hypergeometric function. Setting
$$S = sum_{n = 1}^infty frac{n(n + 1) alpha^n}{(n + frac{1}{beta})^2}.$$
Shifting the index $n mapsto n + 1$ gives
$$S = alpha sum_{n = 0}^infty frac{(n + 1)(n + 2) alpha^n}{(n + frac{1}{beta} + 1)^2}.tag6 $$
Now, noting that
$$left (1 + frac{1}{beta} right )_n = frac{Gamma left (n + frac{1}{beta} + 1 right )}{Gamma left (1 + frac{1}{beta} right )} = frac{left (frac{1}{beta} + n right ) Gamma left (frac{1}{beta} + n right )}{frac{1}{beta} Gamma left (frac{1}{beta} right )}.$$
and
$$left (2 + frac{1}{beta} right )_n = frac{Gamma left (n + frac{1}{beta} + 2 right )}{Gamma left (2 + frac{1}{beta} right )} = frac{left (frac{1}{beta} + n + 1 right ) left (frac{1}{beta} + n right ) Gamma left (frac{1}{beta} + n right )}{left (1 + frac{1}{beta} right ) frac{1}{beta} Gamma left (frac{1}{beta} right )},$$
we see that
$$frac{left (1 + frac{1}{beta} right )_n}{left (2 + frac{1}{beta} right )_n} = frac{1 + frac{1}{beta}}{1 + frac{1}{beta} + n}.tag7$$
Also
$$(3)_n = frac{Gamma (3 + n)}{Gamma (3)} = frac{(2 + n)!}{2!} = frac{(n + 1)(n + 2) n!}{2}.tag8$$



Using (7) and (8), the sum in (6) can be written as
begin{align*}
S &= frac{2 alpha}{(1 + frac{1}{beta})^2} sum_{n = 0}^infty frac{(3)_n left (1 + frac{1}{beta} right )_n left (1 + frac{1}{beta} right )_n}{left (2 + frac{1}{beta} right )_n left (2 + frac{1}{beta} right )_n} frac{alpha^n}{n!}\
&= frac{2 alpha beta^2}{(1 + beta)^2} _3F_2 left (3, 1 + frac{1}{beta}, 1 + frac{1}{beta}; 2 + frac{1}{beta}, 2 + frac{1}{beta}; alpha right ).
end{align*}



So finally, for the case $beta = gamma$ such that $beta, gamma > 0$ we have
$$I = _2F_1 left (2, frac{1}{beta}; 1 + frac{1}{beta}; alpha right ) - frac{2 alpha beta}{(1 + beta)^2} _3F_2 left (3,1 + frac{1}{beta}, 1 + frac{1}{beta};2 + frac{1}{beta}, 2 + frac{1}{beta}; alpha right ).$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    This last integral can be evaluated by derivation under the integral: $$ I= -frac{1}{beta^2}left. frac{d}{ds} int_0^1 frac{z^{s+frac{1}{beta} - 1} ln z}{(1 - alpha z)^2} , dzright|_{s=0}$$ which can be calculated as the derivative of an hypergeometric function. The result is $$ I= _2 F_1 left (2, frac{1}{beta}; 1 + frac{1}{beta}; alpha right )-frac{2alphabeta}{left( 1+beta right)^2} _3 F_2 left (3, frac{1}{beta}, frac{1}{beta}; 2 + frac{1}{beta}, 2+frac{1}{beta}; alpha right ) $$
    $endgroup$
    – Paul Enta
    Feb 16 '18 at 18:28






  • 1




    $begingroup$
    Alternatively, from the solution obtained for $betaneqgamma$, it can also be expressed as $$ lim_{betatogamma}frac{beta _2 F_1 left (2, frac{1}{beta}; 1 + frac{1}{beta}; alpha right )-gamma _2 F_1 left (2, frac{1}{gamma}; 1 + frac{1}{gamma}; alpha right )}{beta-gamma}=frac{d}{dbeta}left[beta _2 F_1 left (2, frac{1}{beta}; 1 + frac{1}{beta}; alpha right )right]$$
    $endgroup$
    – Paul Enta
    Feb 16 '18 at 18:28












  • $begingroup$
    @Paul Enta - Thanks for your suggestion on how to evaluate the integral when $beta = gamma$. It takes quite an effort to evaluate, so thanks again. Just one remark. I have updated my solution where you will find you have made a small error in the argument for the $_3F_2$ term.
    $endgroup$
    – omegadot
    Feb 17 '18 at 3:54












  • $begingroup$
    You're welcome. Sorry for the typo. To construct the hypergeometric function, you may find it easier to work with Gamma functions from the beginning, to build the ratio of two successive terms of the series and to identify the constants just like here.
    $endgroup$
    – Paul Enta
    Feb 17 '18 at 9:35










  • $begingroup$
    Beautiful answer! Unfortunately, I have reached my daily vote limit and I have to wait a couple of hours before I can vote $(+1)$.
    $endgroup$
    – user477343
    Feb 19 '18 at 14:59














13





+100







13





+100



13




+100



$begingroup$

Case 1: $beta neq gamma$ and $beta, gamma > 0$



As has been noted in the comments, the integral can be expressed in closed form in terms of the hypergeometric function $_2 F_1(a,b;c;z)$ provided $beta neq gamma$. To do this we will make use of the following integral representation for the hypergeometric function of
$$_2 F_1 (a,b;c;z) = frac{Gamma (c)}{Gamma (b) Gamma (c - b)} int_0^1 frac{t^{b - 1} (1 - t)^{c - b - 1}}{(1 - zt)^a} , dt. tag1$$



Now, let
$$I = int_0^1 int_0^1 frac{1}{(1 - alpha x^beta y^gamma)^2} , dx dy,$$
where $0 leqslant alpha < 1, beta > 0, gamma > 0$. Setting $z = x^beta y^gamma$, then
$$dx = frac{y^{-gamma/beta}}{beta z^{1 - 1/beta}} , dz,$$
and the integral becomes
$$I = frac{1}{beta} int_0^1 int_0^{y^gamma} frac{1}{(1 - alpha z)^2} frac{y^{-gamma/beta}}{z^{1 - 1/beta}} , dz dy.$$
Changing the order of integration gives
$$I = frac{1}{beta} int_0^1 int_{z^{1/gamma}}^1 frac{1}{(1 - alpha z)^2} frac{y^{-gamma/beta}}{z^{1 - 1/beta}} , dy dz.$$
The $y$-integration can now be readily performed. Provided $beta neq gamma$ the result is
begin{align*}
I &= frac{1}{beta - gamma} int_0^1 frac{z^{frac{1}{beta} - 1} - z^{frac{1}{gamma} - 1}}{(1 - alpha z)^2} , dz tag2
end{align*}



We now rewrite (2) in the form of (1) as follows
begin{align*}
I &= frac{1}{beta - gamma} int_0^1 frac{z^{frac{1}{beta} - 1} (1 - z)^{(frac{1}{beta} + 1) - frac{1}{beta} - 1}}{(1 - alpha z)^2} , dz - frac{1}{beta - gamma} int_0^1 frac{z^{frac{1}{gamma} - 1} (1 - z)^{(frac{1}{gamma} + 1) - frac{1}{gamma} - 1}}{(1 - alpha z)^2} , dz\
&= frac{1}{beta - gamma} frac{Gamma left (frac{1}{beta} right ) Gamma (1)}{Gamma left (1 + frac{1}{beta} right )} _2 F_1 left (2, frac{1}{beta}; 1 + frac{1}{beta}; alpha right ) - frac{1}{beta - gamma} frac{Gamma left (frac{1}{gamma} right ) Gamma (1)}{Gamma left (1 + frac{1}{gamma} right )} _2 F_1 left (2, frac{1}{gamma}; 1 + frac{1}{gamma}; alpha right ).
end{align*}



From the following property for the Gamma function, namely $Gamma (1 + z) = zGamma (z)$, the result can be simplified and leads to
$$int_0^1 int_0^1 frac{1}{(1 - alpha x^beta y^gamma)^2} , dx dy = frac{beta}{beta - gamma} _2 F_1 left (2, frac{1}{beta}; 1 + frac{1}{beta}; alpha right ) - frac{gamma}{beta - gamma} _2 F_1 left (2, frac{1}{gamma}; 1 + frac{1}{gamma}; alpha right ),$$
provided $beta neq gamma$.




Case 2: $beta = 0, gamma > 0$



The integral reduces to
$$I = int_0^1 int_0^1 frac{1}{(1 - alpha y^gamma)^2} , dx dy = int_0^1 frac{dy}{(1 - alpha y^gamma)^2} , dy.$$
Enforcing a substitution of $y mapsto y^{1/gamma}$ gives
begin{align*}
I &= frac{1}{gamma} int_0^1 frac{y^{frac{1}{gamma} - 1}}{(1 - alpha y)^2} , dy\
&= frac{1}{gamma} int_0^1 frac{y^{frac{1}{gamma} - 1}(1 - y)^{(frac{1}{gamma} + 1) - frac{1}{gamma} - 1}}{(1 - alpha y)^2} , dy\
&= frac{1}{gamma} frac{Gamma (1/gamma) Gamma (1)}{Gamma (1 + 1/gamma)} _2 F_1 left (2, frac{1}{gamma}; 1 + frac{1}{gamma}; alpha right )\
&= _2 F_1 left (2, frac{1}{gamma}; 1 + frac{1}{gamma}; alpha right ).
end{align*}




Case 3: $gamma = 0, beta > 0$



Similarly, by symmetry we have
$$I = _2 F_1 left (2, frac{1}{beta}; 1 + frac{1}{beta}; alpha right ).$$




Case 4: $beta = gamma = 0$



Here we trivially have
$$I = frac{1}{(1 - alpha)^2}.$$




Case 5: $beta = gamma$ and $beta, gamma > 0$



Thanks to @Paul Enta the solution to the integral for this particular case can now be given.



If $beta = gamma$, after performing the $y$-integration rather than the result given in (2) one will instead have
$$I = -frac{1}{beta^2} int_0^1 frac{z^{frac{1}{beta} - 1} ln z}{(1 - alpha z)^2} , dz. tag3$$



Now consider the integral
$$J(s) = -frac{1}{beta^2} int_0^1 frac{z^{s + frac{1}{beta} - 1}}{(1 - alpha z)^2} , dz.$$
Now observe that $J'(0) = I$ so in order to find $I$ we just need to find $J$ first before taking its derivative with respect to the parameter $s$ before setting it equal to zero.



Now
begin{align*}
J(s) &= -frac{1}{beta^2} int_0^1 frac{z^{s + frac{1}{beta} - 1} (1 - z)^{(s + frac{1}{beta} + 1) - s - frac{1}{beta} - 1}}{(1 - alpha z)^2} , dz\
&= -frac{1}{beta^2} frac{Gamma (s + 1/beta) Gamma (1)}{Gamma (s + 1/beta + 1)} _2F_1 left (2, s + frac{1}{beta}; s + 1 + frac{1}{beta}; alpha right )\
&= -frac{1}{beta^2} frac{1}{(s + frac{1}{beta})} _2F_1 left (2, s + frac{1}{beta}; s + 1 + frac{1}{beta}; alpha right ).
end{align*}
Thus
begin{align*}
J'(s) &= frac{1}{beta^2 (s + frac{1}{beta})^2} _2F_1 left (2,s + frac{1}{s}; s + 1 + frac{1}{s}; alpha right ) - frac{1}{beta^2 (s + frac{1}{beta})} frac{partial}{partial s} _2F_1 left (2,s + frac{1}{beta}; s + 1 + frac{1}{beta}; alpha right ).tag4
end{align*}
Now
begin{align*}
frac{partial}{partial s} _2F_1 left (2,s + frac{1}{beta}; s + 1 + frac{1}{beta}; alpha right ) &= frac{partial}{partial s} sum_{n = 0}^infty frac{(2)_n (s + frac{1}{beta})_n}{(s + 1 + frac{1}{beta})_n} frac{alpha^n}{n!}\
&= sum_{n = 0}^infty (2)_n frac{alpha^n}{n!} frac{partial}{partial s} left [frac{(s + frac{1}{beta})_n}{(s + 1 + frac{1}{beta})_n} right ]\
&= sum_{n = 0}^infty (2)_n frac{n}{(s + n + frac{1}{beta})^2} frac{alpha^n}{n!}.
end{align*}



We now make use of the following property for the Pochhammer symbol $(x)_n$ (the raising factorial) of
$$(x)_n = frac{Gamma (x + n)}{Gamma (x)}.$$



So
begin{align*}
frac{partial}{partial s} _2F_1 left (2,s + frac{1}{beta}; s + 1 + frac{1}{beta}; alpha right ) &= sum_{n = 0}^infty frac{n(n +1) alpha^n}{(s + n + frac{1}{beta})^2} = sum_{n = 1}^infty frac{n(n +1) alpha^n}{(s + n + frac{1}{beta})^2},
end{align*}
and we have
$$J'(0) = _2F_1 left (2,frac{1}{beta};1 + frac{1}{beta}; alpha right ) - frac{1}{beta} sum_{n = 1}^infty frac{n (n + 1) alpha^n}{(n + frac{1}{beta})^2}. tag5$$



It now remains to write the infinite sum appearing in (5) in closed form in terms of a hypergeometric function. Setting
$$S = sum_{n = 1}^infty frac{n(n + 1) alpha^n}{(n + frac{1}{beta})^2}.$$
Shifting the index $n mapsto n + 1$ gives
$$S = alpha sum_{n = 0}^infty frac{(n + 1)(n + 2) alpha^n}{(n + frac{1}{beta} + 1)^2}.tag6 $$
Now, noting that
$$left (1 + frac{1}{beta} right )_n = frac{Gamma left (n + frac{1}{beta} + 1 right )}{Gamma left (1 + frac{1}{beta} right )} = frac{left (frac{1}{beta} + n right ) Gamma left (frac{1}{beta} + n right )}{frac{1}{beta} Gamma left (frac{1}{beta} right )}.$$
and
$$left (2 + frac{1}{beta} right )_n = frac{Gamma left (n + frac{1}{beta} + 2 right )}{Gamma left (2 + frac{1}{beta} right )} = frac{left (frac{1}{beta} + n + 1 right ) left (frac{1}{beta} + n right ) Gamma left (frac{1}{beta} + n right )}{left (1 + frac{1}{beta} right ) frac{1}{beta} Gamma left (frac{1}{beta} right )},$$
we see that
$$frac{left (1 + frac{1}{beta} right )_n}{left (2 + frac{1}{beta} right )_n} = frac{1 + frac{1}{beta}}{1 + frac{1}{beta} + n}.tag7$$
Also
$$(3)_n = frac{Gamma (3 + n)}{Gamma (3)} = frac{(2 + n)!}{2!} = frac{(n + 1)(n + 2) n!}{2}.tag8$$



Using (7) and (8), the sum in (6) can be written as
begin{align*}
S &= frac{2 alpha}{(1 + frac{1}{beta})^2} sum_{n = 0}^infty frac{(3)_n left (1 + frac{1}{beta} right )_n left (1 + frac{1}{beta} right )_n}{left (2 + frac{1}{beta} right )_n left (2 + frac{1}{beta} right )_n} frac{alpha^n}{n!}\
&= frac{2 alpha beta^2}{(1 + beta)^2} _3F_2 left (3, 1 + frac{1}{beta}, 1 + frac{1}{beta}; 2 + frac{1}{beta}, 2 + frac{1}{beta}; alpha right ).
end{align*}



So finally, for the case $beta = gamma$ such that $beta, gamma > 0$ we have
$$I = _2F_1 left (2, frac{1}{beta}; 1 + frac{1}{beta}; alpha right ) - frac{2 alpha beta}{(1 + beta)^2} _3F_2 left (3,1 + frac{1}{beta}, 1 + frac{1}{beta};2 + frac{1}{beta}, 2 + frac{1}{beta}; alpha right ).$$






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$endgroup$



Case 1: $beta neq gamma$ and $beta, gamma > 0$



As has been noted in the comments, the integral can be expressed in closed form in terms of the hypergeometric function $_2 F_1(a,b;c;z)$ provided $beta neq gamma$. To do this we will make use of the following integral representation for the hypergeometric function of
$$_2 F_1 (a,b;c;z) = frac{Gamma (c)}{Gamma (b) Gamma (c - b)} int_0^1 frac{t^{b - 1} (1 - t)^{c - b - 1}}{(1 - zt)^a} , dt. tag1$$



Now, let
$$I = int_0^1 int_0^1 frac{1}{(1 - alpha x^beta y^gamma)^2} , dx dy,$$
where $0 leqslant alpha < 1, beta > 0, gamma > 0$. Setting $z = x^beta y^gamma$, then
$$dx = frac{y^{-gamma/beta}}{beta z^{1 - 1/beta}} , dz,$$
and the integral becomes
$$I = frac{1}{beta} int_0^1 int_0^{y^gamma} frac{1}{(1 - alpha z)^2} frac{y^{-gamma/beta}}{z^{1 - 1/beta}} , dz dy.$$
Changing the order of integration gives
$$I = frac{1}{beta} int_0^1 int_{z^{1/gamma}}^1 frac{1}{(1 - alpha z)^2} frac{y^{-gamma/beta}}{z^{1 - 1/beta}} , dy dz.$$
The $y$-integration can now be readily performed. Provided $beta neq gamma$ the result is
begin{align*}
I &= frac{1}{beta - gamma} int_0^1 frac{z^{frac{1}{beta} - 1} - z^{frac{1}{gamma} - 1}}{(1 - alpha z)^2} , dz tag2
end{align*}



We now rewrite (2) in the form of (1) as follows
begin{align*}
I &= frac{1}{beta - gamma} int_0^1 frac{z^{frac{1}{beta} - 1} (1 - z)^{(frac{1}{beta} + 1) - frac{1}{beta} - 1}}{(1 - alpha z)^2} , dz - frac{1}{beta - gamma} int_0^1 frac{z^{frac{1}{gamma} - 1} (1 - z)^{(frac{1}{gamma} + 1) - frac{1}{gamma} - 1}}{(1 - alpha z)^2} , dz\
&= frac{1}{beta - gamma} frac{Gamma left (frac{1}{beta} right ) Gamma (1)}{Gamma left (1 + frac{1}{beta} right )} _2 F_1 left (2, frac{1}{beta}; 1 + frac{1}{beta}; alpha right ) - frac{1}{beta - gamma} frac{Gamma left (frac{1}{gamma} right ) Gamma (1)}{Gamma left (1 + frac{1}{gamma} right )} _2 F_1 left (2, frac{1}{gamma}; 1 + frac{1}{gamma}; alpha right ).
end{align*}



From the following property for the Gamma function, namely $Gamma (1 + z) = zGamma (z)$, the result can be simplified and leads to
$$int_0^1 int_0^1 frac{1}{(1 - alpha x^beta y^gamma)^2} , dx dy = frac{beta}{beta - gamma} _2 F_1 left (2, frac{1}{beta}; 1 + frac{1}{beta}; alpha right ) - frac{gamma}{beta - gamma} _2 F_1 left (2, frac{1}{gamma}; 1 + frac{1}{gamma}; alpha right ),$$
provided $beta neq gamma$.




Case 2: $beta = 0, gamma > 0$



The integral reduces to
$$I = int_0^1 int_0^1 frac{1}{(1 - alpha y^gamma)^2} , dx dy = int_0^1 frac{dy}{(1 - alpha y^gamma)^2} , dy.$$
Enforcing a substitution of $y mapsto y^{1/gamma}$ gives
begin{align*}
I &= frac{1}{gamma} int_0^1 frac{y^{frac{1}{gamma} - 1}}{(1 - alpha y)^2} , dy\
&= frac{1}{gamma} int_0^1 frac{y^{frac{1}{gamma} - 1}(1 - y)^{(frac{1}{gamma} + 1) - frac{1}{gamma} - 1}}{(1 - alpha y)^2} , dy\
&= frac{1}{gamma} frac{Gamma (1/gamma) Gamma (1)}{Gamma (1 + 1/gamma)} _2 F_1 left (2, frac{1}{gamma}; 1 + frac{1}{gamma}; alpha right )\
&= _2 F_1 left (2, frac{1}{gamma}; 1 + frac{1}{gamma}; alpha right ).
end{align*}




Case 3: $gamma = 0, beta > 0$



Similarly, by symmetry we have
$$I = _2 F_1 left (2, frac{1}{beta}; 1 + frac{1}{beta}; alpha right ).$$




Case 4: $beta = gamma = 0$



Here we trivially have
$$I = frac{1}{(1 - alpha)^2}.$$




Case 5: $beta = gamma$ and $beta, gamma > 0$



Thanks to @Paul Enta the solution to the integral for this particular case can now be given.



If $beta = gamma$, after performing the $y$-integration rather than the result given in (2) one will instead have
$$I = -frac{1}{beta^2} int_0^1 frac{z^{frac{1}{beta} - 1} ln z}{(1 - alpha z)^2} , dz. tag3$$



Now consider the integral
$$J(s) = -frac{1}{beta^2} int_0^1 frac{z^{s + frac{1}{beta} - 1}}{(1 - alpha z)^2} , dz.$$
Now observe that $J'(0) = I$ so in order to find $I$ we just need to find $J$ first before taking its derivative with respect to the parameter $s$ before setting it equal to zero.



Now
begin{align*}
J(s) &= -frac{1}{beta^2} int_0^1 frac{z^{s + frac{1}{beta} - 1} (1 - z)^{(s + frac{1}{beta} + 1) - s - frac{1}{beta} - 1}}{(1 - alpha z)^2} , dz\
&= -frac{1}{beta^2} frac{Gamma (s + 1/beta) Gamma (1)}{Gamma (s + 1/beta + 1)} _2F_1 left (2, s + frac{1}{beta}; s + 1 + frac{1}{beta}; alpha right )\
&= -frac{1}{beta^2} frac{1}{(s + frac{1}{beta})} _2F_1 left (2, s + frac{1}{beta}; s + 1 + frac{1}{beta}; alpha right ).
end{align*}
Thus
begin{align*}
J'(s) &= frac{1}{beta^2 (s + frac{1}{beta})^2} _2F_1 left (2,s + frac{1}{s}; s + 1 + frac{1}{s}; alpha right ) - frac{1}{beta^2 (s + frac{1}{beta})} frac{partial}{partial s} _2F_1 left (2,s + frac{1}{beta}; s + 1 + frac{1}{beta}; alpha right ).tag4
end{align*}
Now
begin{align*}
frac{partial}{partial s} _2F_1 left (2,s + frac{1}{beta}; s + 1 + frac{1}{beta}; alpha right ) &= frac{partial}{partial s} sum_{n = 0}^infty frac{(2)_n (s + frac{1}{beta})_n}{(s + 1 + frac{1}{beta})_n} frac{alpha^n}{n!}\
&= sum_{n = 0}^infty (2)_n frac{alpha^n}{n!} frac{partial}{partial s} left [frac{(s + frac{1}{beta})_n}{(s + 1 + frac{1}{beta})_n} right ]\
&= sum_{n = 0}^infty (2)_n frac{n}{(s + n + frac{1}{beta})^2} frac{alpha^n}{n!}.
end{align*}



We now make use of the following property for the Pochhammer symbol $(x)_n$ (the raising factorial) of
$$(x)_n = frac{Gamma (x + n)}{Gamma (x)}.$$



So
begin{align*}
frac{partial}{partial s} _2F_1 left (2,s + frac{1}{beta}; s + 1 + frac{1}{beta}; alpha right ) &= sum_{n = 0}^infty frac{n(n +1) alpha^n}{(s + n + frac{1}{beta})^2} = sum_{n = 1}^infty frac{n(n +1) alpha^n}{(s + n + frac{1}{beta})^2},
end{align*}
and we have
$$J'(0) = _2F_1 left (2,frac{1}{beta};1 + frac{1}{beta}; alpha right ) - frac{1}{beta} sum_{n = 1}^infty frac{n (n + 1) alpha^n}{(n + frac{1}{beta})^2}. tag5$$



It now remains to write the infinite sum appearing in (5) in closed form in terms of a hypergeometric function. Setting
$$S = sum_{n = 1}^infty frac{n(n + 1) alpha^n}{(n + frac{1}{beta})^2}.$$
Shifting the index $n mapsto n + 1$ gives
$$S = alpha sum_{n = 0}^infty frac{(n + 1)(n + 2) alpha^n}{(n + frac{1}{beta} + 1)^2}.tag6 $$
Now, noting that
$$left (1 + frac{1}{beta} right )_n = frac{Gamma left (n + frac{1}{beta} + 1 right )}{Gamma left (1 + frac{1}{beta} right )} = frac{left (frac{1}{beta} + n right ) Gamma left (frac{1}{beta} + n right )}{frac{1}{beta} Gamma left (frac{1}{beta} right )}.$$
and
$$left (2 + frac{1}{beta} right )_n = frac{Gamma left (n + frac{1}{beta} + 2 right )}{Gamma left (2 + frac{1}{beta} right )} = frac{left (frac{1}{beta} + n + 1 right ) left (frac{1}{beta} + n right ) Gamma left (frac{1}{beta} + n right )}{left (1 + frac{1}{beta} right ) frac{1}{beta} Gamma left (frac{1}{beta} right )},$$
we see that
$$frac{left (1 + frac{1}{beta} right )_n}{left (2 + frac{1}{beta} right )_n} = frac{1 + frac{1}{beta}}{1 + frac{1}{beta} + n}.tag7$$
Also
$$(3)_n = frac{Gamma (3 + n)}{Gamma (3)} = frac{(2 + n)!}{2!} = frac{(n + 1)(n + 2) n!}{2}.tag8$$



Using (7) and (8), the sum in (6) can be written as
begin{align*}
S &= frac{2 alpha}{(1 + frac{1}{beta})^2} sum_{n = 0}^infty frac{(3)_n left (1 + frac{1}{beta} right )_n left (1 + frac{1}{beta} right )_n}{left (2 + frac{1}{beta} right )_n left (2 + frac{1}{beta} right )_n} frac{alpha^n}{n!}\
&= frac{2 alpha beta^2}{(1 + beta)^2} _3F_2 left (3, 1 + frac{1}{beta}, 1 + frac{1}{beta}; 2 + frac{1}{beta}, 2 + frac{1}{beta}; alpha right ).
end{align*}



So finally, for the case $beta = gamma$ such that $beta, gamma > 0$ we have
$$I = _2F_1 left (2, frac{1}{beta}; 1 + frac{1}{beta}; alpha right ) - frac{2 alpha beta}{(1 + beta)^2} _3F_2 left (3,1 + frac{1}{beta}, 1 + frac{1}{beta};2 + frac{1}{beta}, 2 + frac{1}{beta}; alpha right ).$$







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share|cite|improve this answer








edited Feb 18 '18 at 3:59

























answered Feb 14 '18 at 10:16









omegadotomegadot

6,2592829




6,2592829








  • 1




    $begingroup$
    This last integral can be evaluated by derivation under the integral: $$ I= -frac{1}{beta^2}left. frac{d}{ds} int_0^1 frac{z^{s+frac{1}{beta} - 1} ln z}{(1 - alpha z)^2} , dzright|_{s=0}$$ which can be calculated as the derivative of an hypergeometric function. The result is $$ I= _2 F_1 left (2, frac{1}{beta}; 1 + frac{1}{beta}; alpha right )-frac{2alphabeta}{left( 1+beta right)^2} _3 F_2 left (3, frac{1}{beta}, frac{1}{beta}; 2 + frac{1}{beta}, 2+frac{1}{beta}; alpha right ) $$
    $endgroup$
    – Paul Enta
    Feb 16 '18 at 18:28






  • 1




    $begingroup$
    Alternatively, from the solution obtained for $betaneqgamma$, it can also be expressed as $$ lim_{betatogamma}frac{beta _2 F_1 left (2, frac{1}{beta}; 1 + frac{1}{beta}; alpha right )-gamma _2 F_1 left (2, frac{1}{gamma}; 1 + frac{1}{gamma}; alpha right )}{beta-gamma}=frac{d}{dbeta}left[beta _2 F_1 left (2, frac{1}{beta}; 1 + frac{1}{beta}; alpha right )right]$$
    $endgroup$
    – Paul Enta
    Feb 16 '18 at 18:28












  • $begingroup$
    @Paul Enta - Thanks for your suggestion on how to evaluate the integral when $beta = gamma$. It takes quite an effort to evaluate, so thanks again. Just one remark. I have updated my solution where you will find you have made a small error in the argument for the $_3F_2$ term.
    $endgroup$
    – omegadot
    Feb 17 '18 at 3:54












  • $begingroup$
    You're welcome. Sorry for the typo. To construct the hypergeometric function, you may find it easier to work with Gamma functions from the beginning, to build the ratio of two successive terms of the series and to identify the constants just like here.
    $endgroup$
    – Paul Enta
    Feb 17 '18 at 9:35










  • $begingroup$
    Beautiful answer! Unfortunately, I have reached my daily vote limit and I have to wait a couple of hours before I can vote $(+1)$.
    $endgroup$
    – user477343
    Feb 19 '18 at 14:59














  • 1




    $begingroup$
    This last integral can be evaluated by derivation under the integral: $$ I= -frac{1}{beta^2}left. frac{d}{ds} int_0^1 frac{z^{s+frac{1}{beta} - 1} ln z}{(1 - alpha z)^2} , dzright|_{s=0}$$ which can be calculated as the derivative of an hypergeometric function. The result is $$ I= _2 F_1 left (2, frac{1}{beta}; 1 + frac{1}{beta}; alpha right )-frac{2alphabeta}{left( 1+beta right)^2} _3 F_2 left (3, frac{1}{beta}, frac{1}{beta}; 2 + frac{1}{beta}, 2+frac{1}{beta}; alpha right ) $$
    $endgroup$
    – Paul Enta
    Feb 16 '18 at 18:28






  • 1




    $begingroup$
    Alternatively, from the solution obtained for $betaneqgamma$, it can also be expressed as $$ lim_{betatogamma}frac{beta _2 F_1 left (2, frac{1}{beta}; 1 + frac{1}{beta}; alpha right )-gamma _2 F_1 left (2, frac{1}{gamma}; 1 + frac{1}{gamma}; alpha right )}{beta-gamma}=frac{d}{dbeta}left[beta _2 F_1 left (2, frac{1}{beta}; 1 + frac{1}{beta}; alpha right )right]$$
    $endgroup$
    – Paul Enta
    Feb 16 '18 at 18:28












  • $begingroup$
    @Paul Enta - Thanks for your suggestion on how to evaluate the integral when $beta = gamma$. It takes quite an effort to evaluate, so thanks again. Just one remark. I have updated my solution where you will find you have made a small error in the argument for the $_3F_2$ term.
    $endgroup$
    – omegadot
    Feb 17 '18 at 3:54












  • $begingroup$
    You're welcome. Sorry for the typo. To construct the hypergeometric function, you may find it easier to work with Gamma functions from the beginning, to build the ratio of two successive terms of the series and to identify the constants just like here.
    $endgroup$
    – Paul Enta
    Feb 17 '18 at 9:35










  • $begingroup$
    Beautiful answer! Unfortunately, I have reached my daily vote limit and I have to wait a couple of hours before I can vote $(+1)$.
    $endgroup$
    – user477343
    Feb 19 '18 at 14:59








1




1




$begingroup$
This last integral can be evaluated by derivation under the integral: $$ I= -frac{1}{beta^2}left. frac{d}{ds} int_0^1 frac{z^{s+frac{1}{beta} - 1} ln z}{(1 - alpha z)^2} , dzright|_{s=0}$$ which can be calculated as the derivative of an hypergeometric function. The result is $$ I= _2 F_1 left (2, frac{1}{beta}; 1 + frac{1}{beta}; alpha right )-frac{2alphabeta}{left( 1+beta right)^2} _3 F_2 left (3, frac{1}{beta}, frac{1}{beta}; 2 + frac{1}{beta}, 2+frac{1}{beta}; alpha right ) $$
$endgroup$
– Paul Enta
Feb 16 '18 at 18:28




$begingroup$
This last integral can be evaluated by derivation under the integral: $$ I= -frac{1}{beta^2}left. frac{d}{ds} int_0^1 frac{z^{s+frac{1}{beta} - 1} ln z}{(1 - alpha z)^2} , dzright|_{s=0}$$ which can be calculated as the derivative of an hypergeometric function. The result is $$ I= _2 F_1 left (2, frac{1}{beta}; 1 + frac{1}{beta}; alpha right )-frac{2alphabeta}{left( 1+beta right)^2} _3 F_2 left (3, frac{1}{beta}, frac{1}{beta}; 2 + frac{1}{beta}, 2+frac{1}{beta}; alpha right ) $$
$endgroup$
– Paul Enta
Feb 16 '18 at 18:28




1




1




$begingroup$
Alternatively, from the solution obtained for $betaneqgamma$, it can also be expressed as $$ lim_{betatogamma}frac{beta _2 F_1 left (2, frac{1}{beta}; 1 + frac{1}{beta}; alpha right )-gamma _2 F_1 left (2, frac{1}{gamma}; 1 + frac{1}{gamma}; alpha right )}{beta-gamma}=frac{d}{dbeta}left[beta _2 F_1 left (2, frac{1}{beta}; 1 + frac{1}{beta}; alpha right )right]$$
$endgroup$
– Paul Enta
Feb 16 '18 at 18:28






$begingroup$
Alternatively, from the solution obtained for $betaneqgamma$, it can also be expressed as $$ lim_{betatogamma}frac{beta _2 F_1 left (2, frac{1}{beta}; 1 + frac{1}{beta}; alpha right )-gamma _2 F_1 left (2, frac{1}{gamma}; 1 + frac{1}{gamma}; alpha right )}{beta-gamma}=frac{d}{dbeta}left[beta _2 F_1 left (2, frac{1}{beta}; 1 + frac{1}{beta}; alpha right )right]$$
$endgroup$
– Paul Enta
Feb 16 '18 at 18:28














$begingroup$
@Paul Enta - Thanks for your suggestion on how to evaluate the integral when $beta = gamma$. It takes quite an effort to evaluate, so thanks again. Just one remark. I have updated my solution where you will find you have made a small error in the argument for the $_3F_2$ term.
$endgroup$
– omegadot
Feb 17 '18 at 3:54






$begingroup$
@Paul Enta - Thanks for your suggestion on how to evaluate the integral when $beta = gamma$. It takes quite an effort to evaluate, so thanks again. Just one remark. I have updated my solution where you will find you have made a small error in the argument for the $_3F_2$ term.
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– omegadot
Feb 17 '18 at 3:54














$begingroup$
You're welcome. Sorry for the typo. To construct the hypergeometric function, you may find it easier to work with Gamma functions from the beginning, to build the ratio of two successive terms of the series and to identify the constants just like here.
$endgroup$
– Paul Enta
Feb 17 '18 at 9:35




$begingroup$
You're welcome. Sorry for the typo. To construct the hypergeometric function, you may find it easier to work with Gamma functions from the beginning, to build the ratio of two successive terms of the series and to identify the constants just like here.
$endgroup$
– Paul Enta
Feb 17 '18 at 9:35












$begingroup$
Beautiful answer! Unfortunately, I have reached my daily vote limit and I have to wait a couple of hours before I can vote $(+1)$.
$endgroup$
– user477343
Feb 19 '18 at 14:59




$begingroup$
Beautiful answer! Unfortunately, I have reached my daily vote limit and I have to wait a couple of hours before I can vote $(+1)$.
$endgroup$
– user477343
Feb 19 '18 at 14:59











4












$begingroup$

One feels slightly sorry to post this in view of the truly Herculean task that a previously posted answer represents, but it happens that, at least for every $alpha$ in $[0,1)$ and every nonnegative $beta$ and $gamma$,




$$int_0^1!!int_0^1frac{dxdy}{(1-{alpha}x^{beta}y^{gamma})^2}=sum_{n=0}^inftyfrac{n+1}{(nbeta+1)(ngamma+1)}alpha^n$$




Two remarks:




  • This identity actually holds for every complex parameters such that $|alpha|<1$, $Rebeta>0$ and $Regamma>0$.


  • Various equivalent presentations of the RHS exist, as series and/or as special functions evaluated at $alpha$, but these alternative formulas/denominations seem to bring no further mathematical understanding of the situation hence we omit them.



The proof of the identity above is a one-liner: integrate term by term the series expansion $$frac1{(1-{alpha}x^{beta}y^{gamma})^2}=sum_{n=0}^infty(n+1)alpha^nx^{nbeta}y^{ngamma}$$






share|cite|improve this answer











$endgroup$









  • 3




    $begingroup$
    Sure, but as the OP explicitly asked for a closed-form solution, and by closed-form solution I assumed a solution in terms of any of the known functions, special or otherwise, of mathematical physics, this is what I attempted to provide. I usually don't think of infinite sums as closed in form but of course all functions of mathematical physics can be represented by infinite sums.
    $endgroup$
    – omegadot
    Jan 5 at 12:42






  • 3




    $begingroup$
    @omegadot The difference is specious. As hinted at in my answer, one could easily express the sum of the series above using one or several from the vast list of special functions with a name, but this brings zero extra understanding. (In addition, and since the goal of your comment here seems actually to justify your answer, I might mention that I fail to understand why all the complications and separate cases in it are needed at all.)
    $endgroup$
    – Did
    Jan 5 at 14:37


















4












$begingroup$

One feels slightly sorry to post this in view of the truly Herculean task that a previously posted answer represents, but it happens that, at least for every $alpha$ in $[0,1)$ and every nonnegative $beta$ and $gamma$,




$$int_0^1!!int_0^1frac{dxdy}{(1-{alpha}x^{beta}y^{gamma})^2}=sum_{n=0}^inftyfrac{n+1}{(nbeta+1)(ngamma+1)}alpha^n$$




Two remarks:




  • This identity actually holds for every complex parameters such that $|alpha|<1$, $Rebeta>0$ and $Regamma>0$.


  • Various equivalent presentations of the RHS exist, as series and/or as special functions evaluated at $alpha$, but these alternative formulas/denominations seem to bring no further mathematical understanding of the situation hence we omit them.



The proof of the identity above is a one-liner: integrate term by term the series expansion $$frac1{(1-{alpha}x^{beta}y^{gamma})^2}=sum_{n=0}^infty(n+1)alpha^nx^{nbeta}y^{ngamma}$$






share|cite|improve this answer











$endgroup$









  • 3




    $begingroup$
    Sure, but as the OP explicitly asked for a closed-form solution, and by closed-form solution I assumed a solution in terms of any of the known functions, special or otherwise, of mathematical physics, this is what I attempted to provide. I usually don't think of infinite sums as closed in form but of course all functions of mathematical physics can be represented by infinite sums.
    $endgroup$
    – omegadot
    Jan 5 at 12:42






  • 3




    $begingroup$
    @omegadot The difference is specious. As hinted at in my answer, one could easily express the sum of the series above using one or several from the vast list of special functions with a name, but this brings zero extra understanding. (In addition, and since the goal of your comment here seems actually to justify your answer, I might mention that I fail to understand why all the complications and separate cases in it are needed at all.)
    $endgroup$
    – Did
    Jan 5 at 14:37
















4












4








4





$begingroup$

One feels slightly sorry to post this in view of the truly Herculean task that a previously posted answer represents, but it happens that, at least for every $alpha$ in $[0,1)$ and every nonnegative $beta$ and $gamma$,




$$int_0^1!!int_0^1frac{dxdy}{(1-{alpha}x^{beta}y^{gamma})^2}=sum_{n=0}^inftyfrac{n+1}{(nbeta+1)(ngamma+1)}alpha^n$$




Two remarks:




  • This identity actually holds for every complex parameters such that $|alpha|<1$, $Rebeta>0$ and $Regamma>0$.


  • Various equivalent presentations of the RHS exist, as series and/or as special functions evaluated at $alpha$, but these alternative formulas/denominations seem to bring no further mathematical understanding of the situation hence we omit them.



The proof of the identity above is a one-liner: integrate term by term the series expansion $$frac1{(1-{alpha}x^{beta}y^{gamma})^2}=sum_{n=0}^infty(n+1)alpha^nx^{nbeta}y^{ngamma}$$






share|cite|improve this answer











$endgroup$



One feels slightly sorry to post this in view of the truly Herculean task that a previously posted answer represents, but it happens that, at least for every $alpha$ in $[0,1)$ and every nonnegative $beta$ and $gamma$,




$$int_0^1!!int_0^1frac{dxdy}{(1-{alpha}x^{beta}y^{gamma})^2}=sum_{n=0}^inftyfrac{n+1}{(nbeta+1)(ngamma+1)}alpha^n$$




Two remarks:




  • This identity actually holds for every complex parameters such that $|alpha|<1$, $Rebeta>0$ and $Regamma>0$.


  • Various equivalent presentations of the RHS exist, as series and/or as special functions evaluated at $alpha$, but these alternative formulas/denominations seem to bring no further mathematical understanding of the situation hence we omit them.



The proof of the identity above is a one-liner: integrate term by term the series expansion $$frac1{(1-{alpha}x^{beta}y^{gamma})^2}=sum_{n=0}^infty(n+1)alpha^nx^{nbeta}y^{ngamma}$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 12 at 9:49

























answered Jan 5 at 12:17









DidDid

249k23226466




249k23226466








  • 3




    $begingroup$
    Sure, but as the OP explicitly asked for a closed-form solution, and by closed-form solution I assumed a solution in terms of any of the known functions, special or otherwise, of mathematical physics, this is what I attempted to provide. I usually don't think of infinite sums as closed in form but of course all functions of mathematical physics can be represented by infinite sums.
    $endgroup$
    – omegadot
    Jan 5 at 12:42






  • 3




    $begingroup$
    @omegadot The difference is specious. As hinted at in my answer, one could easily express the sum of the series above using one or several from the vast list of special functions with a name, but this brings zero extra understanding. (In addition, and since the goal of your comment here seems actually to justify your answer, I might mention that I fail to understand why all the complications and separate cases in it are needed at all.)
    $endgroup$
    – Did
    Jan 5 at 14:37
















  • 3




    $begingroup$
    Sure, but as the OP explicitly asked for a closed-form solution, and by closed-form solution I assumed a solution in terms of any of the known functions, special or otherwise, of mathematical physics, this is what I attempted to provide. I usually don't think of infinite sums as closed in form but of course all functions of mathematical physics can be represented by infinite sums.
    $endgroup$
    – omegadot
    Jan 5 at 12:42






  • 3




    $begingroup$
    @omegadot The difference is specious. As hinted at in my answer, one could easily express the sum of the series above using one or several from the vast list of special functions with a name, but this brings zero extra understanding. (In addition, and since the goal of your comment here seems actually to justify your answer, I might mention that I fail to understand why all the complications and separate cases in it are needed at all.)
    $endgroup$
    – Did
    Jan 5 at 14:37










3




3




$begingroup$
Sure, but as the OP explicitly asked for a closed-form solution, and by closed-form solution I assumed a solution in terms of any of the known functions, special or otherwise, of mathematical physics, this is what I attempted to provide. I usually don't think of infinite sums as closed in form but of course all functions of mathematical physics can be represented by infinite sums.
$endgroup$
– omegadot
Jan 5 at 12:42




$begingroup$
Sure, but as the OP explicitly asked for a closed-form solution, and by closed-form solution I assumed a solution in terms of any of the known functions, special or otherwise, of mathematical physics, this is what I attempted to provide. I usually don't think of infinite sums as closed in form but of course all functions of mathematical physics can be represented by infinite sums.
$endgroup$
– omegadot
Jan 5 at 12:42




3




3




$begingroup$
@omegadot The difference is specious. As hinted at in my answer, one could easily express the sum of the series above using one or several from the vast list of special functions with a name, but this brings zero extra understanding. (In addition, and since the goal of your comment here seems actually to justify your answer, I might mention that I fail to understand why all the complications and separate cases in it are needed at all.)
$endgroup$
– Did
Jan 5 at 14:37






$begingroup$
@omegadot The difference is specious. As hinted at in my answer, one could easily express the sum of the series above using one or several from the vast list of special functions with a name, but this brings zero extra understanding. (In addition, and since the goal of your comment here seems actually to justify your answer, I might mention that I fail to understand why all the complications and separate cases in it are needed at all.)
$endgroup$
– Did
Jan 5 at 14:37




















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