Solving used Real Based Methods: $int_0^x frac{t^k}{left(t^n + aright)^m}:dt$
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In working on integrals for the past couple of months, I've come across different cases of the following integral:
begin{equation}
Ileft(x,a,k,n,mright) = int_0^x frac{t^k}{left(t^n + aright)^m}:dt
end{equation}
Where $x,ain mathbb{R}^{+}$.
Here the method that I've taken is rather simple and I was curious as to other 'Real' Based methods could be employed with this integral? I also believe that with the conditions I've set on the parameters that it is convergent. If I'm able to expand those conditions, could you please advise.
Interested in special cases too!
The method I took:
First I wanted to bring the 'a' out the front:
begin{equation}
I(x,a,k,n,m) = int_0^x frac{t^k}{left(aleft[left(a^{-frac{1}{n}}tright)^n + 1right]right)^m}:dt = frac{1}{a^m} int_0^x frac{t^k}{left(left(a^{-frac{1}{n}}tright)^n + 1right)^m}:dt
end{equation}
Here let $u = a^{-frac{1}{n}}t$ Thus,
begin{equation}
I(x,a,k,n,m) = frac{1}{a^m} int_0^{a^{-frac{1}{n}}x} frac{left(a^{frac{1}{n}}uright)^k}{left(u^n + 1right)^m}a^{frac{1}{n}}:du = a^{frac{k + 1}{n} - m}int_0^{a^{-frac{1}{n}}x} frac{u^k}{left(u^n + 1right)^m}:du = a^{frac{k + 1}{n} - m}I(a^{-frac{1}{n}}x,1,k,n,m)
end{equation}
From here I will use $I$ in place of $I(x,a,k,n,m)$ for ease of typing. The next step is to make the substitution $w = u^n$ to yield:
begin{equation}
I = a^{frac{k + 1}{n} - m}int_0^{ax^n} frac{w^frac{k}{n}}{left(w + 1right)^m}frac{:dw}{nw^{frac{n - 1}{n}}} = frac{1}{n}a^{frac{k + 1}{n} - m}int_0^{ax^n} frac{w^{frac{k + 1}{n} - 1}}{left(w + 1right)^m}:dw
end{equation}
Here make the substitution $z = frac{1}{1 + w}$ to yield:
begin{align}
I &= frac{1}{n}a^{frac{k + 1}{n} - m}int_1^{frac{1}{1 + ax^n}} z^m left(frac{1 - z}{z}right)^{frac{k + 1}{n} - 1}left(-frac{1}{z^2}right) :dz = frac{1}{n}a^{frac{k + 1}{n} - m}int_{frac{1}{1 + ax^n}}^1 z^{m - frac{k + 1}{n} - 1}left(1 - zright)^{frac{k + 1}{n} - 1}:dz \
&= frac{1}{n}a^{frac{k + 1}{n} - m} left[int_0^1 z^{m - frac{k + 1}{n} - 1}left(1 - zright)^{frac{k + 1}{n} - 1}:dz - int_0^{frac{1}{1 + ax^n}} z^{m - frac{k + 1}{n} - 1}left(1 - zright)^{frac{k + 1}{n} - 1}:dz right] \
&= frac{1}{n}a^{frac{k + 1}{n} - m} left[Bleft(m - frac{k + 1}{n}, frac{k + 1}{n}right) - Bleft( frac{1}{1 + ax^n}; m - frac{k + 1}{n}, frac{k + 1}{n} right)right]
end{align}
Where $B(a,b)$ is the Beta Function and $B(x; a,b)$ is the Incomplete Beta Function.
And so, we arrive at:
begin{equation}
int_0^x frac{t^k}{left(t^n + aright)^m}:dt = frac{1}{n}a^{frac{k + 1}{n} - m} left[Bleft(m - frac{k + 1}{n}, frac{k + 1}{n}right) - Bleft(frac{1}{1 + ax^n}; m - frac{k + 1}{n}, frac{k + 1}{n} right)right]
end{equation}
Here we observe that for convergence:
begin{equation}
m - frac{k + 1}{n} gt 0,quad frac{k + 1}{n} gt 0,quad n neq 0
end{equation}
Note: when $x rightarrow infty$ we have:
begin{equation}
int_0^infty frac{t^k}{left(t^n + aright)^m}:dt = frac{1}{n}a^{frac{k + 1}{n} - m} Bleft(m - frac{k + 1}{n}, frac{k + 1}{n}right)
end{equation}
Update: Today I realised that we can use this result for another integral:
begin{equation}
int_0^infty frac{ln(t)}{left(t^n + 1right)^m}:dt
end{equation}
This is achieved through a simple use of Feynman's Trick. Here we consider the case when $x rightarrow infty$ and $a = 1$. We see that
begin{align}
frac{d}{dk}left[ int_0^infty frac{t^k}{left(t^n + 1right)^m}:dt right]&= frac{d}{dk}left[frac{1}{n}Bleft(m - frac{k + 1}{n}, frac{k + 1}{n} right)right] \
int_0^infty frac{t^k ln(t)}{left(t^n + 1right)^m}:dt &= frac{1}{n^2}Bleft(m - frac{k + 1}{n}, frac{k + 1}{n} right)left[psi^{(0)}left(frac{k + 1}{n}right) - psi^{(0)}left(m - frac{k + 1}{n}right) right]
end{align}
Thus,
begin{equation}
lim_{k rightarrow 0} int_0^infty frac{t^k ln(t)}{left(t^n + 1right)^m}:dt = lim_{k rightarrow 0}frac{1}{n^2}Bleft(m - frac{k + 1}{n}, frac{k + 1}{n} right)left[psi^{(0)}left(frac{k + 1}{n}right) - psi^{(0)}left(m - frac{k + 1}{n}right) right]
end{equation}
And finally:
begin{equation}
int_0^infty frac{ ln(t)}{left(t^n + 1right)^m}:dt = frac{1}{n^2}Bleft(m - frac{1}{n}, frac{1}{n} right)left[psi^{(0)}left(frac{1}{n}right) - psi^{(0)}left(m - frac{1}{n}right) right]
end{equation}
Note: In the case where $m = 1$ we arrive:
begin{align}
int_0^infty frac{ ln(t)}{left(t^n + 1right)^1}:dt &= frac{1}{n^2}Bleft(1 - frac{1}{n}, frac{1}{n} right)left[psi^{(0)}left(frac{1}{n}right) - psi^{(0)}left(1 - frac{1}{n}right) right] \
&= frac{1}{n^2} Gammaleft(frac{1}{n} right)Gammaleft(1 - frac{1}{n} right) cdot -picotleft(frac{pi}{n}right) \
&= frac{1}{n^2} frac{pi}{sinleft(frac{pi}{n}right)}cdot -picotleft(frac{pi}{n}right)
end{align}
Thus:
begin{equation}
int_0^infty frac{ ln(t)}{t^n + 1}:dt = -frac{pi^2}{n^2} operatorname{cosec}left(frac{pi}{n} right)cotleft(frac{pi}{n}right)
end{equation}
real-analysis integration definite-integrals beta-function
edited Mar 13 at 3:46
NickD
1,1921512
asked Dec 30 '18 at 23:58
DavidGDavidG
1
$endgroup$
|
show 3 more comments
$begingroup$
In working on integrals for the past couple of months, I've come across different cases of the following integral:
begin{equation}
Ileft(x,a,k,n,mright) = int_0^x frac{t^k}{left(t^n + aright)^m}:dt
end{equation}
Where $x,ain mathbb{R}^{+}$.
Here the method that I've taken is rather simple and I was curious as to other 'Real' Based methods could be employed with this integral? I also believe that with the conditions I've set on the parameters that it is convergent. If I'm able to expand those conditions, could you please advise.
Interested in special cases too!
The method I took:
First I wanted to bring the 'a' out the front:
begin{equation}
I(x,a,k,n,m) = int_0^x frac{t^k}{left(aleft[left(a^{-frac{1}{n}}tright)^n + 1right]right)^m}:dt = frac{1}{a^m} int_0^x frac{t^k}{left(left(a^{-frac{1}{n}}tright)^n + 1right)^m}:dt
end{equation}
Here let $u = a^{-frac{1}{n}}t$ Thus,
begin{equation}
I(x,a,k,n,m) = frac{1}{a^m} int_0^{a^{-frac{1}{n}}x} frac{left(a^{frac{1}{n}}uright)^k}{left(u^n + 1right)^m}a^{frac{1}{n}}:du = a^{frac{k + 1}{n} - m}int_0^{a^{-frac{1}{n}}x} frac{u^k}{left(u^n + 1right)^m}:du = a^{frac{k + 1}{n} - m}I(a^{-frac{1}{n}}x,1,k,n,m)
end{equation}
From here I will use $I$ in place of $I(x,a,k,n,m)$ for ease of typing. The next step is to make the substitution $w = u^n$ to yield:
begin{equation}
I = a^{frac{k + 1}{n} - m}int_0^{ax^n} frac{w^frac{k}{n}}{left(w + 1right)^m}frac{:dw}{nw^{frac{n - 1}{n}}} = frac{1}{n}a^{frac{k + 1}{n} - m}int_0^{ax^n} frac{w^{frac{k + 1}{n} - 1}}{left(w + 1right)^m}:dw
end{equation}
Here make the substitution $z = frac{1}{1 + w}$ to yield:
begin{align}
I &= frac{1}{n}a^{frac{k + 1}{n} - m}int_1^{frac{1}{1 + ax^n}} z^m left(frac{1 - z}{z}right)^{frac{k + 1}{n} - 1}left(-frac{1}{z^2}right) :dz = frac{1}{n}a^{frac{k + 1}{n} - m}int_{frac{1}{1 + ax^n}}^1 z^{m - frac{k + 1}{n} - 1}left(1 - zright)^{frac{k + 1}{n} - 1}:dz \
&= frac{1}{n}a^{frac{k + 1}{n} - m} left[int_0^1 z^{m - frac{k + 1}{n} - 1}left(1 - zright)^{frac{k + 1}{n} - 1}:dz - int_0^{frac{1}{1 + ax^n}} z^{m - frac{k + 1}{n} - 1}left(1 - zright)^{frac{k + 1}{n} - 1}:dz right] \
&= frac{1}{n}a^{frac{k + 1}{n} - m} left[Bleft(m - frac{k + 1}{n}, frac{k + 1}{n}right) - Bleft( frac{1}{1 + ax^n}; m - frac{k + 1}{n}, frac{k + 1}{n} right)right]
end{align}
Where $B(a,b)$ is the Beta Function and $B(x; a,b)$ is the Incomplete Beta Function.
And so, we arrive at:
begin{equation}
int_0^x frac{t^k}{left(t^n + aright)^m}:dt = frac{1}{n}a^{frac{k + 1}{n} - m} left[Bleft(m - frac{k + 1}{n}, frac{k + 1}{n}right) - Bleft(frac{1}{1 + ax^n}; m - frac{k + 1}{n}, frac{k + 1}{n} right)right]
end{equation}
Here we observe that for convergence:
begin{equation}
m - frac{k + 1}{n} gt 0,quad frac{k + 1}{n} gt 0,quad n neq 0
end{equation}
Note: when $x rightarrow infty$ we have:
begin{equation}
int_0^infty frac{t^k}{left(t^n + aright)^m}:dt = frac{1}{n}a^{frac{k + 1}{n} - m} Bleft(m - frac{k + 1}{n}, frac{k + 1}{n}right)
end{equation}
Update: Today I realised that we can use this result for another integral:
begin{equation}
int_0^infty frac{ln(t)}{left(t^n + 1right)^m}:dt
end{equation}
This is achieved through a simple use of Feynman's Trick. Here we consider the case when $x rightarrow infty$ and $a = 1$. We see that
begin{align}
frac{d}{dk}left[ int_0^infty frac{t^k}{left(t^n + 1right)^m}:dt right]&= frac{d}{dk}left[frac{1}{n}Bleft(m - frac{k + 1}{n}, frac{k + 1}{n} right)right] \
int_0^infty frac{t^k ln(t)}{left(t^n + 1right)^m}:dt &= frac{1}{n^2}Bleft(m - frac{k + 1}{n}, frac{k + 1}{n} right)left[psi^{(0)}left(frac{k + 1}{n}right) - psi^{(0)}left(m - frac{k + 1}{n}right) right]
end{align}
Thus,
begin{equation}
lim_{k rightarrow 0} int_0^infty frac{t^k ln(t)}{left(t^n + 1right)^m}:dt = lim_{k rightarrow 0}frac{1}{n^2}Bleft(m - frac{k + 1}{n}, frac{k + 1}{n} right)left[psi^{(0)}left(frac{k + 1}{n}right) - psi^{(0)}left(m - frac{k + 1}{n}right) right]
end{equation}
And finally:
begin{equation}
int_0^infty frac{ ln(t)}{left(t^n + 1right)^m}:dt = frac{1}{n^2}Bleft(m - frac{1}{n}, frac{1}{n} right)left[psi^{(0)}left(frac{1}{n}right) - psi^{(0)}left(m - frac{1}{n}right) right]
end{equation}
Note: In the case where $m = 1$ we arrive:
begin{align}
int_0^infty frac{ ln(t)}{left(t^n + 1right)^1}:dt &= frac{1}{n^2}Bleft(1 - frac{1}{n}, frac{1}{n} right)left[psi^{(0)}left(frac{1}{n}right) - psi^{(0)}left(1 - frac{1}{n}right) right] \
&= frac{1}{n^2} Gammaleft(frac{1}{n} right)Gammaleft(1 - frac{1}{n} right) cdot -picotleft(frac{pi}{n}right) \
&= frac{1}{n^2} frac{pi}{sinleft(frac{pi}{n}right)}cdot -picotleft(frac{pi}{n}right)
end{align}
Thus:
begin{equation}
int_0^infty frac{ ln(t)}{t^n + 1}:dt = -frac{pi^2}{n^2} operatorname{cosec}left(frac{pi}{n} right)cotleft(frac{pi}{n}right)
end{equation}
real-analysis integration definite-integrals beta-function
edited Mar 13 at 3:46
NickD
1,1921512
asked Dec 30 '18 at 23:58
DavidGDavidG
1
$endgroup$
1
$begingroup$
damn bro that's a crazy integral
$endgroup$
– clathratus
Dec 31 '18 at 6:07
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It's more that I see it so often I wanted to finally get it out in a closed form so I could be done with solving it 10 different ways haha. Please if you have time look in detail. I do believe it's correct, but I need some wise eyes to have a look over.
$endgroup$
– DavidG
Dec 31 '18 at 6:15
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Would this work for $n=2$, $a=frac{1+b}{1-b}$ for some $|b|<1$, and some integer $mgeq1$, and $k=2K$ for integer $0leq Kleq m-1$? I need to know for $$F(m;b)=int_0^pi frac{mathrm dx}{(1+bcos x)^m}$$
$endgroup$
– clathratus
Jan 17 at 18:03
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@clathratus - Given that the integrand is discontinuous at $x = frac{pi}{2}$ - do you expect there to be an elementary primitive?. Regardless, if you changed the upper bound to $frac{pi}{2}$ then the half-tangent substitution should yield an elementary form.
$endgroup$
– DavidG
Jan 18 at 9:36
1
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Check out this: math.stackexchange.com/q/3077345/583016 It's also on our document. I have made a desmos page and it seems as if it works.
$endgroup$
– clathratus
Jan 18 at 17:27
|
show 3 more comments
$begingroup$
In working on integrals for the past couple of months, I've come across different cases of the following integral:
begin{equation}
Ileft(x,a,k,n,mright) = int_0^x frac{t^k}{left(t^n + aright)^m}:dt
end{equation}
Where $x,ain mathbb{R}^{+}$.
Here the method that I've taken is rather simple and I was curious as to other 'Real' Based methods could be employed with this integral? I also believe that with the conditions I've set on the parameters that it is convergent. If I'm able to expand those conditions, could you please advise.
Interested in special cases too!
The method I took:
First I wanted to bring the 'a' out the front:
begin{equation}
I(x,a,k,n,m) = int_0^x frac{t^k}{left(aleft[left(a^{-frac{1}{n}}tright)^n + 1right]right)^m}:dt = frac{1}{a^m} int_0^x frac{t^k}{left(left(a^{-frac{1}{n}}tright)^n + 1right)^m}:dt
end{equation}
Here let $u = a^{-frac{1}{n}}t$ Thus,
begin{equation}
I(x,a,k,n,m) = frac{1}{a^m} int_0^{a^{-frac{1}{n}}x} frac{left(a^{frac{1}{n}}uright)^k}{left(u^n + 1right)^m}a^{frac{1}{n}}:du = a^{frac{k + 1}{n} - m}int_0^{a^{-frac{1}{n}}x} frac{u^k}{left(u^n + 1right)^m}:du = a^{frac{k + 1}{n} - m}I(a^{-frac{1}{n}}x,1,k,n,m)
end{equation}
From here I will use $I$ in place of $I(x,a,k,n,m)$ for ease of typing. The next step is to make the substitution $w = u^n$ to yield:
begin{equation}
I = a^{frac{k + 1}{n} - m}int_0^{ax^n} frac{w^frac{k}{n}}{left(w + 1right)^m}frac{:dw}{nw^{frac{n - 1}{n}}} = frac{1}{n}a^{frac{k + 1}{n} - m}int_0^{ax^n} frac{w^{frac{k + 1}{n} - 1}}{left(w + 1right)^m}:dw
end{equation}
Here make the substitution $z = frac{1}{1 + w}$ to yield:
begin{align}
I &= frac{1}{n}a^{frac{k + 1}{n} - m}int_1^{frac{1}{1 + ax^n}} z^m left(frac{1 - z}{z}right)^{frac{k + 1}{n} - 1}left(-frac{1}{z^2}right) :dz = frac{1}{n}a^{frac{k + 1}{n} - m}int_{frac{1}{1 + ax^n}}^1 z^{m - frac{k + 1}{n} - 1}left(1 - zright)^{frac{k + 1}{n} - 1}:dz \
&= frac{1}{n}a^{frac{k + 1}{n} - m} left[int_0^1 z^{m - frac{k + 1}{n} - 1}left(1 - zright)^{frac{k + 1}{n} - 1}:dz - int_0^{frac{1}{1 + ax^n}} z^{m - frac{k + 1}{n} - 1}left(1 - zright)^{frac{k + 1}{n} - 1}:dz right] \
&= frac{1}{n}a^{frac{k + 1}{n} - m} left[Bleft(m - frac{k + 1}{n}, frac{k + 1}{n}right) - Bleft( frac{1}{1 + ax^n}; m - frac{k + 1}{n}, frac{k + 1}{n} right)right]
end{align}
Where $B(a,b)$ is the Beta Function and $B(x; a,b)$ is the Incomplete Beta Function.
And so, we arrive at:
begin{equation}
int_0^x frac{t^k}{left(t^n + aright)^m}:dt = frac{1}{n}a^{frac{k + 1}{n} - m} left[Bleft(m - frac{k + 1}{n}, frac{k + 1}{n}right) - Bleft(frac{1}{1 + ax^n}; m - frac{k + 1}{n}, frac{k + 1}{n} right)right]
end{equation}
Here we observe that for convergence:
begin{equation}
m - frac{k + 1}{n} gt 0,quad frac{k + 1}{n} gt 0,quad n neq 0
end{equation}
Note: when $x rightarrow infty$ we have:
begin{equation}
int_0^infty frac{t^k}{left(t^n + aright)^m}:dt = frac{1}{n}a^{frac{k + 1}{n} - m} Bleft(m - frac{k + 1}{n}, frac{k + 1}{n}right)
end{equation}
Update: Today I realised that we can use this result for another integral:
begin{equation}
int_0^infty frac{ln(t)}{left(t^n + 1right)^m}:dt
end{equation}
This is achieved through a simple use of Feynman's Trick. Here we consider the case when $x rightarrow infty$ and $a = 1$. We see that
begin{align}
frac{d}{dk}left[ int_0^infty frac{t^k}{left(t^n + 1right)^m}:dt right]&= frac{d}{dk}left[frac{1}{n}Bleft(m - frac{k + 1}{n}, frac{k + 1}{n} right)right] \
int_0^infty frac{t^k ln(t)}{left(t^n + 1right)^m}:dt &= frac{1}{n^2}Bleft(m - frac{k + 1}{n}, frac{k + 1}{n} right)left[psi^{(0)}left(frac{k + 1}{n}right) - psi^{(0)}left(m - frac{k + 1}{n}right) right]
end{align}
Thus,
begin{equation}
lim_{k rightarrow 0} int_0^infty frac{t^k ln(t)}{left(t^n + 1right)^m}:dt = lim_{k rightarrow 0}frac{1}{n^2}Bleft(m - frac{k + 1}{n}, frac{k + 1}{n} right)left[psi^{(0)}left(frac{k + 1}{n}right) - psi^{(0)}left(m - frac{k + 1}{n}right) right]
end{equation}
And finally:
begin{equation}
int_0^infty frac{ ln(t)}{left(t^n + 1right)^m}:dt = frac{1}{n^2}Bleft(m - frac{1}{n}, frac{1}{n} right)left[psi^{(0)}left(frac{1}{n}right) - psi^{(0)}left(m - frac{1}{n}right) right]
end{equation}
Note: In the case where $m = 1$ we arrive:
begin{align}
int_0^infty frac{ ln(t)}{left(t^n + 1right)^1}:dt &= frac{1}{n^2}Bleft(1 - frac{1}{n}, frac{1}{n} right)left[psi^{(0)}left(frac{1}{n}right) - psi^{(0)}left(1 - frac{1}{n}right) right] \
&= frac{1}{n^2} Gammaleft(frac{1}{n} right)Gammaleft(1 - frac{1}{n} right) cdot -picotleft(frac{pi}{n}right) \
&= frac{1}{n^2} frac{pi}{sinleft(frac{pi}{n}right)}cdot -picotleft(frac{pi}{n}right)
end{align}
Thus:
begin{equation}
int_0^infty frac{ ln(t)}{t^n + 1}:dt = -frac{pi^2}{n^2} operatorname{cosec}left(frac{pi}{n} right)cotleft(frac{pi}{n}right)
end{equation}
real-analysis integration definite-integrals beta-function
edited Mar 13 at 3:46
NickD
1,1921512
asked Dec 30 '18 at 23:58
DavidGDavidG
1
$endgroup$
In working on integrals for the past couple of months, I've come across different cases of the following integral:
begin{equation}
Ileft(x,a,k,n,mright) = int_0^x frac{t^k}{left(t^n + aright)^m}:dt
end{equation}
Where $x,ain mathbb{R}^{+}$.
Here the method that I've taken is rather simple and I was curious as to other 'Real' Based methods could be employed with this integral? I also believe that with the conditions I've set on the parameters that it is convergent. If I'm able to expand those conditions, could you please advise.
Interested in special cases too!
The method I took:
First I wanted to bring the 'a' out the front:
begin{equation}
I(x,a,k,n,m) = int_0^x frac{t^k}{left(aleft[left(a^{-frac{1}{n}}tright)^n + 1right]right)^m}:dt = frac{1}{a^m} int_0^x frac{t^k}{left(left(a^{-frac{1}{n}}tright)^n + 1right)^m}:dt
end{equation}
Here let $u = a^{-frac{1}{n}}t$ Thus,
begin{equation}
I(x,a,k,n,m) = frac{1}{a^m} int_0^{a^{-frac{1}{n}}x} frac{left(a^{frac{1}{n}}uright)^k}{left(u^n + 1right)^m}a^{frac{1}{n}}:du = a^{frac{k + 1}{n} - m}int_0^{a^{-frac{1}{n}}x} frac{u^k}{left(u^n + 1right)^m}:du = a^{frac{k + 1}{n} - m}I(a^{-frac{1}{n}}x,1,k,n,m)
end{equation}
From here I will use $I$ in place of $I(x,a,k,n,m)$ for ease of typing. The next step is to make the substitution $w = u^n$ to yield:
begin{equation}
I = a^{frac{k + 1}{n} - m}int_0^{ax^n} frac{w^frac{k}{n}}{left(w + 1right)^m}frac{:dw}{nw^{frac{n - 1}{n}}} = frac{1}{n}a^{frac{k + 1}{n} - m}int_0^{ax^n} frac{w^{frac{k + 1}{n} - 1}}{left(w + 1right)^m}:dw
end{equation}
Here make the substitution $z = frac{1}{1 + w}$ to yield:
begin{align}
I &= frac{1}{n}a^{frac{k + 1}{n} - m}int_1^{frac{1}{1 + ax^n}} z^m left(frac{1 - z}{z}right)^{frac{k + 1}{n} - 1}left(-frac{1}{z^2}right) :dz = frac{1}{n}a^{frac{k + 1}{n} - m}int_{frac{1}{1 + ax^n}}^1 z^{m - frac{k + 1}{n} - 1}left(1 - zright)^{frac{k + 1}{n} - 1}:dz \
&= frac{1}{n}a^{frac{k + 1}{n} - m} left[int_0^1 z^{m - frac{k + 1}{n} - 1}left(1 - zright)^{frac{k + 1}{n} - 1}:dz - int_0^{frac{1}{1 + ax^n}} z^{m - frac{k + 1}{n} - 1}left(1 - zright)^{frac{k + 1}{n} - 1}:dz right] \
&= frac{1}{n}a^{frac{k + 1}{n} - m} left[Bleft(m - frac{k + 1}{n}, frac{k + 1}{n}right) - Bleft( frac{1}{1 + ax^n}; m - frac{k + 1}{n}, frac{k + 1}{n} right)right]
end{align}
Where $B(a,b)$ is the Beta Function and $B(x; a,b)$ is the Incomplete Beta Function.
And so, we arrive at:
begin{equation}
int_0^x frac{t^k}{left(t^n + aright)^m}:dt = frac{1}{n}a^{frac{k + 1}{n} - m} left[Bleft(m - frac{k + 1}{n}, frac{k + 1}{n}right) - Bleft(frac{1}{1 + ax^n}; m - frac{k + 1}{n}, frac{k + 1}{n} right)right]
end{equation}
Here we observe that for convergence:
begin{equation}
m - frac{k + 1}{n} gt 0,quad frac{k + 1}{n} gt 0,quad n neq 0
end{equation}
Note: when $x rightarrow infty$ we have:
begin{equation}
int_0^infty frac{t^k}{left(t^n + aright)^m}:dt = frac{1}{n}a^{frac{k + 1}{n} - m} Bleft(m - frac{k + 1}{n}, frac{k + 1}{n}right)
end{equation}
Update: Today I realised that we can use this result for another integral:
begin{equation}
int_0^infty frac{ln(t)}{left(t^n + 1right)^m}:dt
end{equation}
This is achieved through a simple use of Feynman's Trick. Here we consider the case when $x rightarrow infty$ and $a = 1$. We see that
begin{align}
frac{d}{dk}left[ int_0^infty frac{t^k}{left(t^n + 1right)^m}:dt right]&= frac{d}{dk}left[frac{1}{n}Bleft(m - frac{k + 1}{n}, frac{k + 1}{n} right)right] \
int_0^infty frac{t^k ln(t)}{left(t^n + 1right)^m}:dt &= frac{1}{n^2}Bleft(m - frac{k + 1}{n}, frac{k + 1}{n} right)left[psi^{(0)}left(frac{k + 1}{n}right) - psi^{(0)}left(m - frac{k + 1}{n}right) right]
end{align}
Thus,
begin{equation}
lim_{k rightarrow 0} int_0^infty frac{t^k ln(t)}{left(t^n + 1right)^m}:dt = lim_{k rightarrow 0}frac{1}{n^2}Bleft(m - frac{k + 1}{n}, frac{k + 1}{n} right)left[psi^{(0)}left(frac{k + 1}{n}right) - psi^{(0)}left(m - frac{k + 1}{n}right) right]
end{equation}
And finally:
begin{equation}
int_0^infty frac{ ln(t)}{left(t^n + 1right)^m}:dt = frac{1}{n^2}Bleft(m - frac{1}{n}, frac{1}{n} right)left[psi^{(0)}left(frac{1}{n}right) - psi^{(0)}left(m - frac{1}{n}right) right]
end{equation}
Note: In the case where $m = 1$ we arrive:
begin{align}
int_0^infty frac{ ln(t)}{left(t^n + 1right)^1}:dt &= frac{1}{n^2}Bleft(1 - frac{1}{n}, frac{1}{n} right)left[psi^{(0)}left(frac{1}{n}right) - psi^{(0)}left(1 - frac{1}{n}right) right] \
&= frac{1}{n^2} Gammaleft(frac{1}{n} right)Gammaleft(1 - frac{1}{n} right) cdot -picotleft(frac{pi}{n}right) \
&= frac{1}{n^2} frac{pi}{sinleft(frac{pi}{n}right)}cdot -picotleft(frac{pi}{n}right)
end{align}
Thus:
begin{equation}
int_0^infty frac{ ln(t)}{t^n + 1}:dt = -frac{pi^2}{n^2} operatorname{cosec}left(frac{pi}{n} right)cotleft(frac{pi}{n}right)
end{equation}
real-analysis integration definite-integrals beta-function
real-analysis integration definite-integrals beta-function
edited Mar 13 at 3:46
NickD
1,1921512
asked Dec 30 '18 at 23:58
DavidGDavidG
1
edited Mar 13 at 3:46
NickD
1,1921512
asked Dec 30 '18 at 23:58
DavidGDavidG
1
edited Mar 13 at 3:46
NickD
1,1921512
edited Mar 13 at 3:46
NickD
1,1921512
edited Mar 13 at 3:46
NickD
1,1921512
1,1921512
asked Dec 30 '18 at 23:58
DavidGDavidG
1
asked Dec 30 '18 at 23:58
DavidGDavidG
1
asked Dec 30 '18 at 23:58
DavidGDavidG
1
1
1
$begingroup$
damn bro that's a crazy integral
$endgroup$
– clathratus
Dec 31 '18 at 6:07
$begingroup$
It's more that I see it so often I wanted to finally get it out in a closed form so I could be done with solving it 10 different ways haha. Please if you have time look in detail. I do believe it's correct, but I need some wise eyes to have a look over.
$endgroup$
– DavidG
Dec 31 '18 at 6:15
$begingroup$
Would this work for $n=2$, $a=frac{1+b}{1-b}$ for some $|b|<1$, and some integer $mgeq1$, and $k=2K$ for integer $0leq Kleq m-1$? I need to know for $$F(m;b)=int_0^pi frac{mathrm dx}{(1+bcos x)^m}$$
$endgroup$
– clathratus
Jan 17 at 18:03
$begingroup$
@clathratus - Given that the integrand is discontinuous at $x = frac{pi}{2}$ - do you expect there to be an elementary primitive?. Regardless, if you changed the upper bound to $frac{pi}{2}$ then the half-tangent substitution should yield an elementary form.
$endgroup$
– DavidG
Jan 18 at 9:36
1
$begingroup$
Check out this: math.stackexchange.com/q/3077345/583016 It's also on our document. I have made a desmos page and it seems as if it works.
$endgroup$
– clathratus
Jan 18 at 17:27
|
show 3 more comments
1
$begingroup$
damn bro that's a crazy integral
$endgroup$
– clathratus
Dec 31 '18 at 6:07
$begingroup$
It's more that I see it so often I wanted to finally get it out in a closed form so I could be done with solving it 10 different ways haha. Please if you have time look in detail. I do believe it's correct, but I need some wise eyes to have a look over.
$endgroup$
– DavidG
Dec 31 '18 at 6:15
$begingroup$
Would this work for $n=2$, $a=frac{1+b}{1-b}$ for some $|b|<1$, and some integer $mgeq1$, and $k=2K$ for integer $0leq Kleq m-1$? I need to know for $$F(m;b)=int_0^pi frac{mathrm dx}{(1+bcos x)^m}$$
$endgroup$
– clathratus
Jan 17 at 18:03
$begingroup$
@clathratus - Given that the integrand is discontinuous at $x = frac{pi}{2}$ - do you expect there to be an elementary primitive?. Regardless, if you changed the upper bound to $frac{pi}{2}$ then the half-tangent substitution should yield an elementary form.
$endgroup$
– DavidG
Jan 18 at 9:36
1
$begingroup$
Check out this: math.stackexchange.com/q/3077345/583016 It's also on our document. I have made a desmos page and it seems as if it works.
$endgroup$
– clathratus
Jan 18 at 17:27
1
1
$begingroup$
damn bro that's a crazy integral
$endgroup$
– clathratus
Dec 31 '18 at 6:07
$begingroup$
damn bro that's a crazy integral
$endgroup$
– clathratus
Dec 31 '18 at 6:07
$begingroup$
It's more that I see it so often I wanted to finally get it out in a closed form so I could be done with solving it 10 different ways haha. Please if you have time look in detail. I do believe it's correct, but I need some wise eyes to have a look over.
$endgroup$
– DavidG
Dec 31 '18 at 6:15
$begingroup$
It's more that I see it so often I wanted to finally get it out in a closed form so I could be done with solving it 10 different ways haha. Please if you have time look in detail. I do believe it's correct, but I need some wise eyes to have a look over.
$endgroup$
– DavidG
Dec 31 '18 at 6:15
$begingroup$
Would this work for $n=2$, $a=frac{1+b}{1-b}$ for some $|b|<1$, and some integer $mgeq1$, and $k=2K$ for integer $0leq Kleq m-1$? I need to know for $$F(m;b)=int_0^pi frac{mathrm dx}{(1+bcos x)^m}$$
$endgroup$
– clathratus
Jan 17 at 18:03
$begingroup$
Would this work for $n=2$, $a=frac{1+b}{1-b}$ for some $|b|<1$, and some integer $mgeq1$, and $k=2K$ for integer $0leq Kleq m-1$? I need to know for $$F(m;b)=int_0^pi frac{mathrm dx}{(1+bcos x)^m}$$
$endgroup$
– clathratus
Jan 17 at 18:03
$begingroup$
@clathratus - Given that the integrand is discontinuous at $x = frac{pi}{2}$ - do you expect there to be an elementary primitive?. Regardless, if you changed the upper bound to $frac{pi}{2}$ then the half-tangent substitution should yield an elementary form.
$endgroup$
– DavidG
Jan 18 at 9:36
$begingroup$
@clathratus - Given that the integrand is discontinuous at $x = frac{pi}{2}$ - do you expect there to be an elementary primitive?. Regardless, if you changed the upper bound to $frac{pi}{2}$ then the half-tangent substitution should yield an elementary form.
$endgroup$
– DavidG
Jan 18 at 9:36
1
1
$begingroup$
Check out this: math.stackexchange.com/q/3077345/583016 It's also on our document. I have made a desmos page and it seems as if it works.
$endgroup$
– clathratus
Jan 18 at 17:27
$begingroup$
Check out this: math.stackexchange.com/q/3077345/583016 It's also on our document. I have made a desmos page and it seems as if it works.
$endgroup$
– clathratus
Jan 18 at 17:27
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
NOT A SOLUTION:
I've found some special cases on this site that I will list (this will evolve as I find more special (but generalised) cases:
- Closed form for $ int_0^infty {frac{{{x^n}}}{{1 + {x^m}}}dx }$
Evaluate the integral $ int _0^{+infty} frac{x^m}{(a+bx^n)^p}$
answered Jan 2 at 0:29
DavidGDavidG
1
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
NOT A SOLUTION:
I've found some special cases on this site that I will list (this will evolve as I find more special (but generalised) cases:
- Closed form for $ int_0^infty {frac{{{x^n}}}{{1 + {x^m}}}dx }$
Evaluate the integral $ int _0^{+infty} frac{x^m}{(a+bx^n)^p}$
answered Jan 2 at 0:29
DavidGDavidG
1
$endgroup$
add a comment |
$begingroup$
NOT A SOLUTION:
I've found some special cases on this site that I will list (this will evolve as I find more special (but generalised) cases:
- Closed form for $ int_0^infty {frac{{{x^n}}}{{1 + {x^m}}}dx }$
Evaluate the integral $ int _0^{+infty} frac{x^m}{(a+bx^n)^p}$
answered Jan 2 at 0:29
DavidGDavidG
1
$endgroup$
add a comment |
$begingroup$
NOT A SOLUTION:
I've found some special cases on this site that I will list (this will evolve as I find more special (but generalised) cases:
- Closed form for $ int_0^infty {frac{{{x^n}}}{{1 + {x^m}}}dx }$
Evaluate the integral $ int _0^{+infty} frac{x^m}{(a+bx^n)^p}$
answered Jan 2 at 0:29
DavidGDavidG
1
$endgroup$
NOT A SOLUTION:
I've found some special cases on this site that I will list (this will evolve as I find more special (but generalised) cases:
- Closed form for $ int_0^infty {frac{{{x^n}}}{{1 + {x^m}}}dx }$
Evaluate the integral $ int _0^{+infty} frac{x^m}{(a+bx^n)^p}$
answered Jan 2 at 0:29
DavidGDavidG
1
edited Jan 9 at 0:19
answered Jan 2 at 0:29
DavidGDavidG
1
answered Jan 2 at 0:29
DavidGDavidG
1
answered Jan 2 at 0:29
DavidGDavidG
1
1
add a comment |
add a comment |
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1
$begingroup$
damn bro that's a crazy integral
$endgroup$
– clathratus
Dec 31 '18 at 6:07
$begingroup$
It's more that I see it so often I wanted to finally get it out in a closed form so I could be done with solving it 10 different ways haha. Please if you have time look in detail. I do believe it's correct, but I need some wise eyes to have a look over.
$endgroup$
– DavidG
Dec 31 '18 at 6:15
$begingroup$
Would this work for $n=2$, $a=frac{1+b}{1-b}$ for some $|b|<1$, and some integer $mgeq1$, and $k=2K$ for integer $0leq Kleq m-1$? I need to know for $$F(m;b)=int_0^pi frac{mathrm dx}{(1+bcos x)^m}$$
$endgroup$
– clathratus
Jan 17 at 18:03
$begingroup$
@clathratus - Given that the integrand is discontinuous at $x = frac{pi}{2}$ - do you expect there to be an elementary primitive?. Regardless, if you changed the upper bound to $frac{pi}{2}$ then the half-tangent substitution should yield an elementary form.
$endgroup$
– DavidG
Jan 18 at 9:36
1
$begingroup$
Check out this: math.stackexchange.com/q/3077345/583016 It's also on our document. I have made a desmos page and it seems as if it works.
$endgroup$
– clathratus
Jan 18 at 17:27