integration question $int_{-1}^{1}left(frac{1-x}{1+x}right)^afrac{dx}{(x-b)^2}$












3












$begingroup$


How to integrate this integral $displaystyleint_{-1}^{1}bigg(frac{1-x}{1+x}bigg)^afrac{dx}{(x-b)^2}$ where $0<a<1$ and $b>1$



Answer. I put this into online integral calculator, but it said it cann't do this integration.






calculus integration definite-integrals







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    There are a lot of integrals that the online integral calculator can't handle. Try wolfram alpha
    $endgroup$
    – clathratus
    Dec 21 '18 at 5:40










  • $begingroup$
    @clathratus, I used wolfram alpha also, it said 'Standard computation time exceeded'
    $endgroup$
    – Dhamnekar Winod
    Dec 21 '18 at 11:54
















3












$begingroup$


How to integrate this integral $displaystyleint_{-1}^{1}bigg(frac{1-x}{1+x}bigg)^afrac{dx}{(x-b)^2}$ where $0<a<1$ and $b>1$



Answer. I put this into online integral calculator, but it said it cann't do this integration.






calculus integration definite-integrals







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    There are a lot of integrals that the online integral calculator can't handle. Try wolfram alpha
    $endgroup$
    – clathratus
    Dec 21 '18 at 5:40










  • $begingroup$
    @clathratus, I used wolfram alpha also, it said 'Standard computation time exceeded'
    $endgroup$
    – Dhamnekar Winod
    Dec 21 '18 at 11:54














3












3








3


0



$begingroup$


How to integrate this integral $displaystyleint_{-1}^{1}bigg(frac{1-x}{1+x}bigg)^afrac{dx}{(x-b)^2}$ where $0<a<1$ and $b>1$



Answer. I put this into online integral calculator, but it said it cann't do this integration.






calculus integration definite-integrals







share|cite|improve this question











$endgroup$




How to integrate this integral $displaystyleint_{-1}^{1}bigg(frac{1-x}{1+x}bigg)^afrac{dx}{(x-b)^2}$ where $0<a<1$ and $b>1$



Answer. I put this into online integral calculator, but it said it cann't do this integration.






calculus integration definite-integrals




calculus integration definite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 21 '18 at 5:56









Kemono Chen

3,1891844




3,1891844










asked Dec 21 '18 at 5:33









Dhamnekar WinodDhamnekar Winod

427514




427514








  • 2




    $begingroup$
    There are a lot of integrals that the online integral calculator can't handle. Try wolfram alpha
    $endgroup$
    – clathratus
    Dec 21 '18 at 5:40










  • $begingroup$
    @clathratus, I used wolfram alpha also, it said 'Standard computation time exceeded'
    $endgroup$
    – Dhamnekar Winod
    Dec 21 '18 at 11:54














  • 2




    $begingroup$
    There are a lot of integrals that the online integral calculator can't handle. Try wolfram alpha
    $endgroup$
    – clathratus
    Dec 21 '18 at 5:40










  • $begingroup$
    @clathratus, I used wolfram alpha also, it said 'Standard computation time exceeded'
    $endgroup$
    – Dhamnekar Winod
    Dec 21 '18 at 11:54








2




2




$begingroup$
There are a lot of integrals that the online integral calculator can't handle. Try wolfram alpha
$endgroup$
– clathratus
Dec 21 '18 at 5:40




$begingroup$
There are a lot of integrals that the online integral calculator can't handle. Try wolfram alpha
$endgroup$
– clathratus
Dec 21 '18 at 5:40












$begingroup$
@clathratus, I used wolfram alpha also, it said 'Standard computation time exceeded'
$endgroup$
– Dhamnekar Winod
Dec 21 '18 at 11:54




$begingroup$
@clathratus, I used wolfram alpha also, it said 'Standard computation time exceeded'
$endgroup$
– Dhamnekar Winod
Dec 21 '18 at 11:54










1 Answer
1






active

oldest

votes


















9












$begingroup$

First, substitute $t = (1-x)/(1+x)$.



$$begin{aligned} I = int_{-1}^{1}left(frac{1-x}{1+x}right)^{a}frac{mathrm{d}x}{(x-b)^{2}} &= frac{1}{2}int_{0}^{infty}t^{a}left(frac{x+1}{x-b}right)^{2}mathrm{d}t \
&= 2int_{0}^{infty}frac{t^{a},mathrm{d}t}{(1-t-b(1+t))^{2}} \
&= 2int_{0}^{infty}frac{t^{a},mathrm{d}t}{((1-b) - (1+b)t)^{2}} \
&= 2int_{0}^{infty}frac{t^{a},mathrm{d}t}{((b+1)t + (b-1))^{2}}end{aligned}$$



Let $b_{pm} = bpm 1$. Then after $u = b_{+}t/b_{-}$,



$$begin{aligned} I = 2int_{0}^{infty}frac{t^{a},mathrm{d}t}{(b_{+}t + b_{-})^{2}} &= frac{2}{b_{-}^{2}}left(frac{b_{-}}{b_{+}}right)^{a+1}int_{0}^{infty}frac{u^{a},mathrm{d}u}{(u+1)^{2}}.end{aligned}$$



Using the beta function



$$ frac{Gamma(x)Gamma(y)}{Gamma(x+y)} = int_{0}^{infty}frac{z^{y-1},mathrm{d}z}{(z+1)^{x+y}},$$



we identify $x=1-a$ and $y=1+a$. Then



$$ I = frac{2}{b_{-}^{2}}left(frac{b_{-}}{b_{+}}right)^{a+1}frac{Gamma(1-a)Gamma(1+a)}{Gamma(2)}.$$



Using the reflection identity $Gamma(z)Gamma(1-z) = pi/sinpi z$ and recursion $Gamma(1+a) = aGamma(a)$, we have



$$ frac{Gamma(1-a)Gamma(1+a)}{Gamma(2)} = Gamma(1-a)aGamma(a) = frac{pi a}{sinpi a}.$$



The answer is then



$$ I = frac{2}{(b-1)^{2}}left(frac{b-1}{b+1}right)^{a}frac{b-1}{b+1}frac{pi a}{sinpi a} = boxed{frac{2}{b^{2}-1}left(frac{b-1}{b+1}right)^{a}frac{pi a}{sinpi a}.}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    why did you multiply the right hand side integral by $frac{1}{b^2-1}$?
    $endgroup$
    – Dhamnekar Winod
    Dec 21 '18 at 13:57










  • $begingroup$
    That comes from simplifying the $b$ terms: $(b-1)(b+1)$. Notice the exponent on $b_{-}/b_{+}$.
    $endgroup$
    – Ininterrompue
    Dec 21 '18 at 16:17










  • $begingroup$
    ,but $du=frac{b+1}{b-1}dt$ and $u^a=bigg(frac{(b+1)t}{b-1}bigg)^a$. So the term $bigg(frac{b-1}{b+1}bigg)^{a+1}$outside the integral cancels the term Inside the integral.
    $endgroup$
    – Dhamnekar Winod
    Dec 21 '18 at 16:51












  • $begingroup$
    So first, you pull $1/b_{-}^{2}$ out of the integral, then you do u-sub. The term $t^{a}$ pulls $(b_{-}/b_{+})^{a}$ out of the integral, while $mathrm{d}t$ pulls another factor of $b_{-}/b_{+}$. There is no cancellation; $(b_{-}/b_{+})^{a+1}$ is the result of the substitution.
    $endgroup$
    – Ininterrompue
    Dec 21 '18 at 18:45













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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









9












$begingroup$

First, substitute $t = (1-x)/(1+x)$.



$$begin{aligned} I = int_{-1}^{1}left(frac{1-x}{1+x}right)^{a}frac{mathrm{d}x}{(x-b)^{2}} &= frac{1}{2}int_{0}^{infty}t^{a}left(frac{x+1}{x-b}right)^{2}mathrm{d}t \
&= 2int_{0}^{infty}frac{t^{a},mathrm{d}t}{(1-t-b(1+t))^{2}} \
&= 2int_{0}^{infty}frac{t^{a},mathrm{d}t}{((1-b) - (1+b)t)^{2}} \
&= 2int_{0}^{infty}frac{t^{a},mathrm{d}t}{((b+1)t + (b-1))^{2}}end{aligned}$$



Let $b_{pm} = bpm 1$. Then after $u = b_{+}t/b_{-}$,



$$begin{aligned} I = 2int_{0}^{infty}frac{t^{a},mathrm{d}t}{(b_{+}t + b_{-})^{2}} &= frac{2}{b_{-}^{2}}left(frac{b_{-}}{b_{+}}right)^{a+1}int_{0}^{infty}frac{u^{a},mathrm{d}u}{(u+1)^{2}}.end{aligned}$$



Using the beta function



$$ frac{Gamma(x)Gamma(y)}{Gamma(x+y)} = int_{0}^{infty}frac{z^{y-1},mathrm{d}z}{(z+1)^{x+y}},$$



we identify $x=1-a$ and $y=1+a$. Then



$$ I = frac{2}{b_{-}^{2}}left(frac{b_{-}}{b_{+}}right)^{a+1}frac{Gamma(1-a)Gamma(1+a)}{Gamma(2)}.$$



Using the reflection identity $Gamma(z)Gamma(1-z) = pi/sinpi z$ and recursion $Gamma(1+a) = aGamma(a)$, we have



$$ frac{Gamma(1-a)Gamma(1+a)}{Gamma(2)} = Gamma(1-a)aGamma(a) = frac{pi a}{sinpi a}.$$



The answer is then



$$ I = frac{2}{(b-1)^{2}}left(frac{b-1}{b+1}right)^{a}frac{b-1}{b+1}frac{pi a}{sinpi a} = boxed{frac{2}{b^{2}-1}left(frac{b-1}{b+1}right)^{a}frac{pi a}{sinpi a}.}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    why did you multiply the right hand side integral by $frac{1}{b^2-1}$?
    $endgroup$
    – Dhamnekar Winod
    Dec 21 '18 at 13:57










  • $begingroup$
    That comes from simplifying the $b$ terms: $(b-1)(b+1)$. Notice the exponent on $b_{-}/b_{+}$.
    $endgroup$
    – Ininterrompue
    Dec 21 '18 at 16:17










  • $begingroup$
    ,but $du=frac{b+1}{b-1}dt$ and $u^a=bigg(frac{(b+1)t}{b-1}bigg)^a$. So the term $bigg(frac{b-1}{b+1}bigg)^{a+1}$outside the integral cancels the term Inside the integral.
    $endgroup$
    – Dhamnekar Winod
    Dec 21 '18 at 16:51












  • $begingroup$
    So first, you pull $1/b_{-}^{2}$ out of the integral, then you do u-sub. The term $t^{a}$ pulls $(b_{-}/b_{+})^{a}$ out of the integral, while $mathrm{d}t$ pulls another factor of $b_{-}/b_{+}$. There is no cancellation; $(b_{-}/b_{+})^{a+1}$ is the result of the substitution.
    $endgroup$
    – Ininterrompue
    Dec 21 '18 at 18:45


















9












$begingroup$

First, substitute $t = (1-x)/(1+x)$.



$$begin{aligned} I = int_{-1}^{1}left(frac{1-x}{1+x}right)^{a}frac{mathrm{d}x}{(x-b)^{2}} &= frac{1}{2}int_{0}^{infty}t^{a}left(frac{x+1}{x-b}right)^{2}mathrm{d}t \
&= 2int_{0}^{infty}frac{t^{a},mathrm{d}t}{(1-t-b(1+t))^{2}} \
&= 2int_{0}^{infty}frac{t^{a},mathrm{d}t}{((1-b) - (1+b)t)^{2}} \
&= 2int_{0}^{infty}frac{t^{a},mathrm{d}t}{((b+1)t + (b-1))^{2}}end{aligned}$$



Let $b_{pm} = bpm 1$. Then after $u = b_{+}t/b_{-}$,



$$begin{aligned} I = 2int_{0}^{infty}frac{t^{a},mathrm{d}t}{(b_{+}t + b_{-})^{2}} &= frac{2}{b_{-}^{2}}left(frac{b_{-}}{b_{+}}right)^{a+1}int_{0}^{infty}frac{u^{a},mathrm{d}u}{(u+1)^{2}}.end{aligned}$$



Using the beta function



$$ frac{Gamma(x)Gamma(y)}{Gamma(x+y)} = int_{0}^{infty}frac{z^{y-1},mathrm{d}z}{(z+1)^{x+y}},$$



we identify $x=1-a$ and $y=1+a$. Then



$$ I = frac{2}{b_{-}^{2}}left(frac{b_{-}}{b_{+}}right)^{a+1}frac{Gamma(1-a)Gamma(1+a)}{Gamma(2)}.$$



Using the reflection identity $Gamma(z)Gamma(1-z) = pi/sinpi z$ and recursion $Gamma(1+a) = aGamma(a)$, we have



$$ frac{Gamma(1-a)Gamma(1+a)}{Gamma(2)} = Gamma(1-a)aGamma(a) = frac{pi a}{sinpi a}.$$



The answer is then



$$ I = frac{2}{(b-1)^{2}}left(frac{b-1}{b+1}right)^{a}frac{b-1}{b+1}frac{pi a}{sinpi a} = boxed{frac{2}{b^{2}-1}left(frac{b-1}{b+1}right)^{a}frac{pi a}{sinpi a}.}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    why did you multiply the right hand side integral by $frac{1}{b^2-1}$?
    $endgroup$
    – Dhamnekar Winod
    Dec 21 '18 at 13:57










  • $begingroup$
    That comes from simplifying the $b$ terms: $(b-1)(b+1)$. Notice the exponent on $b_{-}/b_{+}$.
    $endgroup$
    – Ininterrompue
    Dec 21 '18 at 16:17










  • $begingroup$
    ,but $du=frac{b+1}{b-1}dt$ and $u^a=bigg(frac{(b+1)t}{b-1}bigg)^a$. So the term $bigg(frac{b-1}{b+1}bigg)^{a+1}$outside the integral cancels the term Inside the integral.
    $endgroup$
    – Dhamnekar Winod
    Dec 21 '18 at 16:51












  • $begingroup$
    So first, you pull $1/b_{-}^{2}$ out of the integral, then you do u-sub. The term $t^{a}$ pulls $(b_{-}/b_{+})^{a}$ out of the integral, while $mathrm{d}t$ pulls another factor of $b_{-}/b_{+}$. There is no cancellation; $(b_{-}/b_{+})^{a+1}$ is the result of the substitution.
    $endgroup$
    – Ininterrompue
    Dec 21 '18 at 18:45
















9












9








9





$begingroup$

First, substitute $t = (1-x)/(1+x)$.



$$begin{aligned} I = int_{-1}^{1}left(frac{1-x}{1+x}right)^{a}frac{mathrm{d}x}{(x-b)^{2}} &= frac{1}{2}int_{0}^{infty}t^{a}left(frac{x+1}{x-b}right)^{2}mathrm{d}t \
&= 2int_{0}^{infty}frac{t^{a},mathrm{d}t}{(1-t-b(1+t))^{2}} \
&= 2int_{0}^{infty}frac{t^{a},mathrm{d}t}{((1-b) - (1+b)t)^{2}} \
&= 2int_{0}^{infty}frac{t^{a},mathrm{d}t}{((b+1)t + (b-1))^{2}}end{aligned}$$



Let $b_{pm} = bpm 1$. Then after $u = b_{+}t/b_{-}$,



$$begin{aligned} I = 2int_{0}^{infty}frac{t^{a},mathrm{d}t}{(b_{+}t + b_{-})^{2}} &= frac{2}{b_{-}^{2}}left(frac{b_{-}}{b_{+}}right)^{a+1}int_{0}^{infty}frac{u^{a},mathrm{d}u}{(u+1)^{2}}.end{aligned}$$



Using the beta function



$$ frac{Gamma(x)Gamma(y)}{Gamma(x+y)} = int_{0}^{infty}frac{z^{y-1},mathrm{d}z}{(z+1)^{x+y}},$$



we identify $x=1-a$ and $y=1+a$. Then



$$ I = frac{2}{b_{-}^{2}}left(frac{b_{-}}{b_{+}}right)^{a+1}frac{Gamma(1-a)Gamma(1+a)}{Gamma(2)}.$$



Using the reflection identity $Gamma(z)Gamma(1-z) = pi/sinpi z$ and recursion $Gamma(1+a) = aGamma(a)$, we have



$$ frac{Gamma(1-a)Gamma(1+a)}{Gamma(2)} = Gamma(1-a)aGamma(a) = frac{pi a}{sinpi a}.$$



The answer is then



$$ I = frac{2}{(b-1)^{2}}left(frac{b-1}{b+1}right)^{a}frac{b-1}{b+1}frac{pi a}{sinpi a} = boxed{frac{2}{b^{2}-1}left(frac{b-1}{b+1}right)^{a}frac{pi a}{sinpi a}.}$$






share|cite|improve this answer











$endgroup$



First, substitute $t = (1-x)/(1+x)$.



$$begin{aligned} I = int_{-1}^{1}left(frac{1-x}{1+x}right)^{a}frac{mathrm{d}x}{(x-b)^{2}} &= frac{1}{2}int_{0}^{infty}t^{a}left(frac{x+1}{x-b}right)^{2}mathrm{d}t \
&= 2int_{0}^{infty}frac{t^{a},mathrm{d}t}{(1-t-b(1+t))^{2}} \
&= 2int_{0}^{infty}frac{t^{a},mathrm{d}t}{((1-b) - (1+b)t)^{2}} \
&= 2int_{0}^{infty}frac{t^{a},mathrm{d}t}{((b+1)t + (b-1))^{2}}end{aligned}$$



Let $b_{pm} = bpm 1$. Then after $u = b_{+}t/b_{-}$,



$$begin{aligned} I = 2int_{0}^{infty}frac{t^{a},mathrm{d}t}{(b_{+}t + b_{-})^{2}} &= frac{2}{b_{-}^{2}}left(frac{b_{-}}{b_{+}}right)^{a+1}int_{0}^{infty}frac{u^{a},mathrm{d}u}{(u+1)^{2}}.end{aligned}$$



Using the beta function



$$ frac{Gamma(x)Gamma(y)}{Gamma(x+y)} = int_{0}^{infty}frac{z^{y-1},mathrm{d}z}{(z+1)^{x+y}},$$



we identify $x=1-a$ and $y=1+a$. Then



$$ I = frac{2}{b_{-}^{2}}left(frac{b_{-}}{b_{+}}right)^{a+1}frac{Gamma(1-a)Gamma(1+a)}{Gamma(2)}.$$



Using the reflection identity $Gamma(z)Gamma(1-z) = pi/sinpi z$ and recursion $Gamma(1+a) = aGamma(a)$, we have



$$ frac{Gamma(1-a)Gamma(1+a)}{Gamma(2)} = Gamma(1-a)aGamma(a) = frac{pi a}{sinpi a}.$$



The answer is then



$$ I = frac{2}{(b-1)^{2}}left(frac{b-1}{b+1}right)^{a}frac{b-1}{b+1}frac{pi a}{sinpi a} = boxed{frac{2}{b^{2}-1}left(frac{b-1}{b+1}right)^{a}frac{pi a}{sinpi a}.}$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 21 '18 at 7:04

























answered Dec 21 '18 at 6:52









IninterrompueIninterrompue

67519




67519












  • $begingroup$
    why did you multiply the right hand side integral by $frac{1}{b^2-1}$?
    $endgroup$
    – Dhamnekar Winod
    Dec 21 '18 at 13:57










  • $begingroup$
    That comes from simplifying the $b$ terms: $(b-1)(b+1)$. Notice the exponent on $b_{-}/b_{+}$.
    $endgroup$
    – Ininterrompue
    Dec 21 '18 at 16:17










  • $begingroup$
    ,but $du=frac{b+1}{b-1}dt$ and $u^a=bigg(frac{(b+1)t}{b-1}bigg)^a$. So the term $bigg(frac{b-1}{b+1}bigg)^{a+1}$outside the integral cancels the term Inside the integral.
    $endgroup$
    – Dhamnekar Winod
    Dec 21 '18 at 16:51












  • $begingroup$
    So first, you pull $1/b_{-}^{2}$ out of the integral, then you do u-sub. The term $t^{a}$ pulls $(b_{-}/b_{+})^{a}$ out of the integral, while $mathrm{d}t$ pulls another factor of $b_{-}/b_{+}$. There is no cancellation; $(b_{-}/b_{+})^{a+1}$ is the result of the substitution.
    $endgroup$
    – Ininterrompue
    Dec 21 '18 at 18:45




















  • $begingroup$
    why did you multiply the right hand side integral by $frac{1}{b^2-1}$?
    $endgroup$
    – Dhamnekar Winod
    Dec 21 '18 at 13:57










  • $begingroup$
    That comes from simplifying the $b$ terms: $(b-1)(b+1)$. Notice the exponent on $b_{-}/b_{+}$.
    $endgroup$
    – Ininterrompue
    Dec 21 '18 at 16:17










  • $begingroup$
    ,but $du=frac{b+1}{b-1}dt$ and $u^a=bigg(frac{(b+1)t}{b-1}bigg)^a$. So the term $bigg(frac{b-1}{b+1}bigg)^{a+1}$outside the integral cancels the term Inside the integral.
    $endgroup$
    – Dhamnekar Winod
    Dec 21 '18 at 16:51












  • $begingroup$
    So first, you pull $1/b_{-}^{2}$ out of the integral, then you do u-sub. The term $t^{a}$ pulls $(b_{-}/b_{+})^{a}$ out of the integral, while $mathrm{d}t$ pulls another factor of $b_{-}/b_{+}$. There is no cancellation; $(b_{-}/b_{+})^{a+1}$ is the result of the substitution.
    $endgroup$
    – Ininterrompue
    Dec 21 '18 at 18:45


















$begingroup$
why did you multiply the right hand side integral by $frac{1}{b^2-1}$?
$endgroup$
– Dhamnekar Winod
Dec 21 '18 at 13:57




$begingroup$
why did you multiply the right hand side integral by $frac{1}{b^2-1}$?
$endgroup$
– Dhamnekar Winod
Dec 21 '18 at 13:57












$begingroup$
That comes from simplifying the $b$ terms: $(b-1)(b+1)$. Notice the exponent on $b_{-}/b_{+}$.
$endgroup$
– Ininterrompue
Dec 21 '18 at 16:17




$begingroup$
That comes from simplifying the $b$ terms: $(b-1)(b+1)$. Notice the exponent on $b_{-}/b_{+}$.
$endgroup$
– Ininterrompue
Dec 21 '18 at 16:17












$begingroup$
,but $du=frac{b+1}{b-1}dt$ and $u^a=bigg(frac{(b+1)t}{b-1}bigg)^a$. So the term $bigg(frac{b-1}{b+1}bigg)^{a+1}$outside the integral cancels the term Inside the integral.
$endgroup$
– Dhamnekar Winod
Dec 21 '18 at 16:51






$begingroup$
,but $du=frac{b+1}{b-1}dt$ and $u^a=bigg(frac{(b+1)t}{b-1}bigg)^a$. So the term $bigg(frac{b-1}{b+1}bigg)^{a+1}$outside the integral cancels the term Inside the integral.
$endgroup$
– Dhamnekar Winod
Dec 21 '18 at 16:51














$begingroup$
So first, you pull $1/b_{-}^{2}$ out of the integral, then you do u-sub. The term $t^{a}$ pulls $(b_{-}/b_{+})^{a}$ out of the integral, while $mathrm{d}t$ pulls another factor of $b_{-}/b_{+}$. There is no cancellation; $(b_{-}/b_{+})^{a+1}$ is the result of the substitution.
$endgroup$
– Ininterrompue
Dec 21 '18 at 18:45






$begingroup$
So first, you pull $1/b_{-}^{2}$ out of the integral, then you do u-sub. The term $t^{a}$ pulls $(b_{-}/b_{+})^{a}$ out of the integral, while $mathrm{d}t$ pulls another factor of $b_{-}/b_{+}$. There is no cancellation; $(b_{-}/b_{+})^{a+1}$ is the result of the substitution.
$endgroup$
– Ininterrompue
Dec 21 '18 at 18:45




















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