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Evaluate $int_0^{infty}frac{x^2}{(e^x-1)^2}dx$
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I want to do
$$int_0^infty!!! frac{x^2}{(e^x-1)^2}text{d}x$$
I know how to do $int_0^infty frac{x}{(e^x-1)}text{d}x$ by using $-frac{ln(1-x)}{x} = sum_{n=1}^infty frac{x^{n-1}}{n}$. But when I try to do the same trick I get a double sum which I have no idea of how to simply.
The result should be able to be presented as an expression involving $zeta(3)$.
calculus
asked Oct 17 '18 at 21:01
AwooAwoo
448
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add a comment |
$begingroup$
I want to do
$$int_0^infty!!! frac{x^2}{(e^x-1)^2}text{d}x$$
I know how to do $int_0^infty frac{x}{(e^x-1)}text{d}x$ by using $-frac{ln(1-x)}{x} = sum_{n=1}^infty frac{x^{n-1}}{n}$. But when I try to do the same trick I get a double sum which I have no idea of how to simply.
The result should be able to be presented as an expression involving $zeta(3)$.
calculus
asked Oct 17 '18 at 21:01
AwooAwoo
448
$endgroup$
add a comment |
$begingroup$
I want to do
$$int_0^infty!!! frac{x^2}{(e^x-1)^2}text{d}x$$
I know how to do $int_0^infty frac{x}{(e^x-1)}text{d}x$ by using $-frac{ln(1-x)}{x} = sum_{n=1}^infty frac{x^{n-1}}{n}$. But when I try to do the same trick I get a double sum which I have no idea of how to simply.
The result should be able to be presented as an expression involving $zeta(3)$.
calculus
asked Oct 17 '18 at 21:01
AwooAwoo
448
$endgroup$
I want to do
$$int_0^infty!!! frac{x^2}{(e^x-1)^2}text{d}x$$
I know how to do $int_0^infty frac{x}{(e^x-1)}text{d}x$ by using $-frac{ln(1-x)}{x} = sum_{n=1}^infty frac{x^{n-1}}{n}$. But when I try to do the same trick I get a double sum which I have no idea of how to simply.
The result should be able to be presented as an expression involving $zeta(3)$.
calculus
calculus
asked Oct 17 '18 at 21:01
AwooAwoo
448
asked Oct 17 '18 at 21:01
AwooAwoo
448
edited Oct 17 '18 at 22:12
Awoo
asked Oct 17 '18 at 21:01
AwooAwoo
448
asked Oct 17 '18 at 21:01
AwooAwoo
448
asked Oct 17 '18 at 21:01
AwooAwoo
448
448
add a comment |
add a comment |
3 Answers
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Lets start with the integral representation of the Polylogarithm $operatorname{Li}_s(z)$ which is given by
$$Gamma(s)operatorname{Li}_s(z)=int_0^{infty}frac{t^{s-1}}{e^t/z-1}mathrm dttag1$$
furthermore note the relations
$$frac{mathrm d}{mathrm dz}operatorname{Li}_s(z)=frac{operatorname{Li}_{s-1}(z)}{z}text{ and }operatorname{Li}_s(1)=zeta(s)tag2$$
Differentiate $(1)$ w.r.t. $z$ by using the first property of $(2)$ yields to
$$begin{align}
frac{mathrm d}{mathrm dz}Gamma(s)operatorname{Li}_s(z)&=frac {mathrm d}{mathrm dz}int_0^{infty}t^{s-1}frac{z}{e^t-z}mathrm dt\
Gamma(s)frac{operatorname{Li}_{s-1}(z)}{z}&=int_0^{infty}t^{s-1}frac{(e^t-z)+z}{(e^t-z)^2}mathrm dt\
Gamma(s)frac{operatorname{Li}_{s-1}(z)}{z}&=int_0^{infty}frac{t^{s-1}}{e^t-z}mathrm dt+zint_0^{infty}frac{t^{s-1}}{(e^t-z)^2}mathrm dt\
Gamma(s)frac{operatorname{Li}_{s-1}(z)}{z}&=Gamma(s)frac{operatorname{Li}_{s}(z)}{z}+zint_0^{infty}frac{t^{s-1}}{(e^t-z)^2}mathrm dt\
therefore~int_0^{infty}frac{t^{s-1}}{(e^t-z)^2}mathrm dt&=Gamma(s)left[frac{operatorname{Li}_{s-1}(z)}{z^2}+frac{operatorname{Li}_{s}(z)}{z^2}right]
end{align}$$
Setting $s=3$ and $z=1$ and by including the second relation of $(2)$ gives us
$$begin{align}
int_0^{infty}frac{t^{3-1}}{(e^t-1)^2}mathrm dt&=Gamma(3)left[frac{operatorname{Li}_{3-1}(1)}{1^2}+frac{operatorname{Li}_{3}(1)}{1^2}right]\
int_0^{infty}frac{t^{2}}{(e^t-1)^2}mathrm dt&=2[zeta(2)-zeta(3)]
end{align}$$
$$therefore~int_0^{infty}frac{t^{2}}{(e^t-1)^2}dt=frac{pi^2}3-2zeta(3)=0.885~754...$$
Where the integral on the left equals your integral by setting $t=x$. Numerically WolframAlpha gives the same solution.
Your approach to $int_0^{infty}frac x{e^x-1}dx$ is also given through the special case $s=2$. Note that the derivative of the Dilogarithm $operatorname{Li}_2(z)$ is given by the term $-ln(1-x)/x$. Therefore your attempt was not wrong at all.
answered Oct 17 '18 at 21:32
mrtaurhomrtaurho
6,04551641
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add a comment |
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One more approach: the integral is $$int_0^inftyfrac{x^2 e^{-2x}}{(1-e^{-x})^2}dx=sum_{n=2}^infty(n-1)int_0^infty x^2 e^{-nx}dx=2sum_{n=2}^infty (n^{-2}-n^{-3})=2(zeta(2)-zeta(3)).$$
answered Oct 18 '18 at 18:25
J.G.J.G.
31.2k23149
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Withouth invoking the polylogarithm, for any $ninmathbb{N}$ and any $a>0$ we have
$$begin{eqnarray*} I(n,a)=int_{0}^{+infty}frac{x^n}{e^{ax}-1},dx&=&sum_{mgeq 1}int_{0}^{+infty}x^n e^{-ma x},dx\&=&sum_{mgeq 1}frac{n!}{(ma)^{n+1}}=frac{n!}{a^{n+1}}zeta(n+1)end{eqnarray*}$$
and by differentiation under the integral sign
$$ J(n,a)=frac{partial}{partial a} I(n,a)=int_{0}^{+infty}frac{-x^{n+1} e^{ax}}{(e^{ax}-1)^2},dx = -frac{(n+1)!}{a^{n+2}}zeta(n+1)$$
such that
$$begin{eqnarray*} int_{0}^{+infty}frac{x^{n+1}}{(e^{ax}-1)^2},dx &=& -left[I(n+1,a)+J(n,a)right]\&=&frac{(n+1)!}{a^{n+1}}zeta(n+1)-frac{(n+1)!}{a^{n+2}}zeta(n+2)end{eqnarray*}$$
and by setting $a=n=1$
$$ int_{0}^{+infty}frac{x^{2}}{(e^{x}-1)^2},dx =color{red}{2left(zeta(2)-zeta(3)right)}=sum_{ngeq 1}frac{2n}{(n+1)^3}=sum_{ngeq 1}frac{6n+5(-1)^{n}}{n^3binom{2n}{n}}. $$
answered Oct 18 '18 at 18:11
Jack D'AurizioJack D'Aurizio
291k33284669
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3 Answers
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3 Answers
3
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$begingroup$
Lets start with the integral representation of the Polylogarithm $operatorname{Li}_s(z)$ which is given by
$$Gamma(s)operatorname{Li}_s(z)=int_0^{infty}frac{t^{s-1}}{e^t/z-1}mathrm dttag1$$
furthermore note the relations
$$frac{mathrm d}{mathrm dz}operatorname{Li}_s(z)=frac{operatorname{Li}_{s-1}(z)}{z}text{ and }operatorname{Li}_s(1)=zeta(s)tag2$$
Differentiate $(1)$ w.r.t. $z$ by using the first property of $(2)$ yields to
$$begin{align}
frac{mathrm d}{mathrm dz}Gamma(s)operatorname{Li}_s(z)&=frac {mathrm d}{mathrm dz}int_0^{infty}t^{s-1}frac{z}{e^t-z}mathrm dt\
Gamma(s)frac{operatorname{Li}_{s-1}(z)}{z}&=int_0^{infty}t^{s-1}frac{(e^t-z)+z}{(e^t-z)^2}mathrm dt\
Gamma(s)frac{operatorname{Li}_{s-1}(z)}{z}&=int_0^{infty}frac{t^{s-1}}{e^t-z}mathrm dt+zint_0^{infty}frac{t^{s-1}}{(e^t-z)^2}mathrm dt\
Gamma(s)frac{operatorname{Li}_{s-1}(z)}{z}&=Gamma(s)frac{operatorname{Li}_{s}(z)}{z}+zint_0^{infty}frac{t^{s-1}}{(e^t-z)^2}mathrm dt\
therefore~int_0^{infty}frac{t^{s-1}}{(e^t-z)^2}mathrm dt&=Gamma(s)left[frac{operatorname{Li}_{s-1}(z)}{z^2}+frac{operatorname{Li}_{s}(z)}{z^2}right]
end{align}$$
Setting $s=3$ and $z=1$ and by including the second relation of $(2)$ gives us
$$begin{align}
int_0^{infty}frac{t^{3-1}}{(e^t-1)^2}mathrm dt&=Gamma(3)left[frac{operatorname{Li}_{3-1}(1)}{1^2}+frac{operatorname{Li}_{3}(1)}{1^2}right]\
int_0^{infty}frac{t^{2}}{(e^t-1)^2}mathrm dt&=2[zeta(2)-zeta(3)]
end{align}$$
$$therefore~int_0^{infty}frac{t^{2}}{(e^t-1)^2}dt=frac{pi^2}3-2zeta(3)=0.885~754...$$
Where the integral on the left equals your integral by setting $t=x$. Numerically WolframAlpha gives the same solution.
Your approach to $int_0^{infty}frac x{e^x-1}dx$ is also given through the special case $s=2$. Note that the derivative of the Dilogarithm $operatorname{Li}_2(z)$ is given by the term $-ln(1-x)/x$. Therefore your attempt was not wrong at all.
answered Oct 17 '18 at 21:32
mrtaurhomrtaurho
6,04551641
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add a comment |
$begingroup$
Lets start with the integral representation of the Polylogarithm $operatorname{Li}_s(z)$ which is given by
$$Gamma(s)operatorname{Li}_s(z)=int_0^{infty}frac{t^{s-1}}{e^t/z-1}mathrm dttag1$$
furthermore note the relations
$$frac{mathrm d}{mathrm dz}operatorname{Li}_s(z)=frac{operatorname{Li}_{s-1}(z)}{z}text{ and }operatorname{Li}_s(1)=zeta(s)tag2$$
Differentiate $(1)$ w.r.t. $z$ by using the first property of $(2)$ yields to
$$begin{align}
frac{mathrm d}{mathrm dz}Gamma(s)operatorname{Li}_s(z)&=frac {mathrm d}{mathrm dz}int_0^{infty}t^{s-1}frac{z}{e^t-z}mathrm dt\
Gamma(s)frac{operatorname{Li}_{s-1}(z)}{z}&=int_0^{infty}t^{s-1}frac{(e^t-z)+z}{(e^t-z)^2}mathrm dt\
Gamma(s)frac{operatorname{Li}_{s-1}(z)}{z}&=int_0^{infty}frac{t^{s-1}}{e^t-z}mathrm dt+zint_0^{infty}frac{t^{s-1}}{(e^t-z)^2}mathrm dt\
Gamma(s)frac{operatorname{Li}_{s-1}(z)}{z}&=Gamma(s)frac{operatorname{Li}_{s}(z)}{z}+zint_0^{infty}frac{t^{s-1}}{(e^t-z)^2}mathrm dt\
therefore~int_0^{infty}frac{t^{s-1}}{(e^t-z)^2}mathrm dt&=Gamma(s)left[frac{operatorname{Li}_{s-1}(z)}{z^2}+frac{operatorname{Li}_{s}(z)}{z^2}right]
end{align}$$
Setting $s=3$ and $z=1$ and by including the second relation of $(2)$ gives us
$$begin{align}
int_0^{infty}frac{t^{3-1}}{(e^t-1)^2}mathrm dt&=Gamma(3)left[frac{operatorname{Li}_{3-1}(1)}{1^2}+frac{operatorname{Li}_{3}(1)}{1^2}right]\
int_0^{infty}frac{t^{2}}{(e^t-1)^2}mathrm dt&=2[zeta(2)-zeta(3)]
end{align}$$
$$therefore~int_0^{infty}frac{t^{2}}{(e^t-1)^2}dt=frac{pi^2}3-2zeta(3)=0.885~754...$$
Where the integral on the left equals your integral by setting $t=x$. Numerically WolframAlpha gives the same solution.
Your approach to $int_0^{infty}frac x{e^x-1}dx$ is also given through the special case $s=2$. Note that the derivative of the Dilogarithm $operatorname{Li}_2(z)$ is given by the term $-ln(1-x)/x$. Therefore your attempt was not wrong at all.
answered Oct 17 '18 at 21:32
mrtaurhomrtaurho
6,04551641
$endgroup$
add a comment |
$begingroup$
Lets start with the integral representation of the Polylogarithm $operatorname{Li}_s(z)$ which is given by
$$Gamma(s)operatorname{Li}_s(z)=int_0^{infty}frac{t^{s-1}}{e^t/z-1}mathrm dttag1$$
furthermore note the relations
$$frac{mathrm d}{mathrm dz}operatorname{Li}_s(z)=frac{operatorname{Li}_{s-1}(z)}{z}text{ and }operatorname{Li}_s(1)=zeta(s)tag2$$
Differentiate $(1)$ w.r.t. $z$ by using the first property of $(2)$ yields to
$$begin{align}
frac{mathrm d}{mathrm dz}Gamma(s)operatorname{Li}_s(z)&=frac {mathrm d}{mathrm dz}int_0^{infty}t^{s-1}frac{z}{e^t-z}mathrm dt\
Gamma(s)frac{operatorname{Li}_{s-1}(z)}{z}&=int_0^{infty}t^{s-1}frac{(e^t-z)+z}{(e^t-z)^2}mathrm dt\
Gamma(s)frac{operatorname{Li}_{s-1}(z)}{z}&=int_0^{infty}frac{t^{s-1}}{e^t-z}mathrm dt+zint_0^{infty}frac{t^{s-1}}{(e^t-z)^2}mathrm dt\
Gamma(s)frac{operatorname{Li}_{s-1}(z)}{z}&=Gamma(s)frac{operatorname{Li}_{s}(z)}{z}+zint_0^{infty}frac{t^{s-1}}{(e^t-z)^2}mathrm dt\
therefore~int_0^{infty}frac{t^{s-1}}{(e^t-z)^2}mathrm dt&=Gamma(s)left[frac{operatorname{Li}_{s-1}(z)}{z^2}+frac{operatorname{Li}_{s}(z)}{z^2}right]
end{align}$$
Setting $s=3$ and $z=1$ and by including the second relation of $(2)$ gives us
$$begin{align}
int_0^{infty}frac{t^{3-1}}{(e^t-1)^2}mathrm dt&=Gamma(3)left[frac{operatorname{Li}_{3-1}(1)}{1^2}+frac{operatorname{Li}_{3}(1)}{1^2}right]\
int_0^{infty}frac{t^{2}}{(e^t-1)^2}mathrm dt&=2[zeta(2)-zeta(3)]
end{align}$$
$$therefore~int_0^{infty}frac{t^{2}}{(e^t-1)^2}dt=frac{pi^2}3-2zeta(3)=0.885~754...$$
Where the integral on the left equals your integral by setting $t=x$. Numerically WolframAlpha gives the same solution.
Your approach to $int_0^{infty}frac x{e^x-1}dx$ is also given through the special case $s=2$. Note that the derivative of the Dilogarithm $operatorname{Li}_2(z)$ is given by the term $-ln(1-x)/x$. Therefore your attempt was not wrong at all.
answered Oct 17 '18 at 21:32
mrtaurhomrtaurho
6,04551641
$endgroup$
Lets start with the integral representation of the Polylogarithm $operatorname{Li}_s(z)$ which is given by
$$Gamma(s)operatorname{Li}_s(z)=int_0^{infty}frac{t^{s-1}}{e^t/z-1}mathrm dttag1$$
furthermore note the relations
$$frac{mathrm d}{mathrm dz}operatorname{Li}_s(z)=frac{operatorname{Li}_{s-1}(z)}{z}text{ and }operatorname{Li}_s(1)=zeta(s)tag2$$
Differentiate $(1)$ w.r.t. $z$ by using the first property of $(2)$ yields to
$$begin{align}
frac{mathrm d}{mathrm dz}Gamma(s)operatorname{Li}_s(z)&=frac {mathrm d}{mathrm dz}int_0^{infty}t^{s-1}frac{z}{e^t-z}mathrm dt\
Gamma(s)frac{operatorname{Li}_{s-1}(z)}{z}&=int_0^{infty}t^{s-1}frac{(e^t-z)+z}{(e^t-z)^2}mathrm dt\
Gamma(s)frac{operatorname{Li}_{s-1}(z)}{z}&=int_0^{infty}frac{t^{s-1}}{e^t-z}mathrm dt+zint_0^{infty}frac{t^{s-1}}{(e^t-z)^2}mathrm dt\
Gamma(s)frac{operatorname{Li}_{s-1}(z)}{z}&=Gamma(s)frac{operatorname{Li}_{s}(z)}{z}+zint_0^{infty}frac{t^{s-1}}{(e^t-z)^2}mathrm dt\
therefore~int_0^{infty}frac{t^{s-1}}{(e^t-z)^2}mathrm dt&=Gamma(s)left[frac{operatorname{Li}_{s-1}(z)}{z^2}+frac{operatorname{Li}_{s}(z)}{z^2}right]
end{align}$$
Setting $s=3$ and $z=1$ and by including the second relation of $(2)$ gives us
$$begin{align}
int_0^{infty}frac{t^{3-1}}{(e^t-1)^2}mathrm dt&=Gamma(3)left[frac{operatorname{Li}_{3-1}(1)}{1^2}+frac{operatorname{Li}_{3}(1)}{1^2}right]\
int_0^{infty}frac{t^{2}}{(e^t-1)^2}mathrm dt&=2[zeta(2)-zeta(3)]
end{align}$$
$$therefore~int_0^{infty}frac{t^{2}}{(e^t-1)^2}dt=frac{pi^2}3-2zeta(3)=0.885~754...$$
Where the integral on the left equals your integral by setting $t=x$. Numerically WolframAlpha gives the same solution.
Your approach to $int_0^{infty}frac x{e^x-1}dx$ is also given through the special case $s=2$. Note that the derivative of the Dilogarithm $operatorname{Li}_2(z)$ is given by the term $-ln(1-x)/x$. Therefore your attempt was not wrong at all.
answered Oct 17 '18 at 21:32
mrtaurhomrtaurho
6,04551641
edited Jan 26 at 12:01
answered Oct 17 '18 at 21:32
mrtaurhomrtaurho
6,04551641
answered Oct 17 '18 at 21:32
mrtaurhomrtaurho
6,04551641
answered Oct 17 '18 at 21:32
mrtaurhomrtaurho
6,04551641
6,04551641
add a comment |
add a comment |
$begingroup$
One more approach: the integral is $$int_0^inftyfrac{x^2 e^{-2x}}{(1-e^{-x})^2}dx=sum_{n=2}^infty(n-1)int_0^infty x^2 e^{-nx}dx=2sum_{n=2}^infty (n^{-2}-n^{-3})=2(zeta(2)-zeta(3)).$$
answered Oct 18 '18 at 18:25
J.G.J.G.
31.2k23149
$endgroup$
add a comment |
$begingroup$
One more approach: the integral is $$int_0^inftyfrac{x^2 e^{-2x}}{(1-e^{-x})^2}dx=sum_{n=2}^infty(n-1)int_0^infty x^2 e^{-nx}dx=2sum_{n=2}^infty (n^{-2}-n^{-3})=2(zeta(2)-zeta(3)).$$
answered Oct 18 '18 at 18:25
J.G.J.G.
31.2k23149
$endgroup$
add a comment |
$begingroup$
One more approach: the integral is $$int_0^inftyfrac{x^2 e^{-2x}}{(1-e^{-x})^2}dx=sum_{n=2}^infty(n-1)int_0^infty x^2 e^{-nx}dx=2sum_{n=2}^infty (n^{-2}-n^{-3})=2(zeta(2)-zeta(3)).$$
answered Oct 18 '18 at 18:25
J.G.J.G.
31.2k23149
$endgroup$
One more approach: the integral is $$int_0^inftyfrac{x^2 e^{-2x}}{(1-e^{-x})^2}dx=sum_{n=2}^infty(n-1)int_0^infty x^2 e^{-nx}dx=2sum_{n=2}^infty (n^{-2}-n^{-3})=2(zeta(2)-zeta(3)).$$
answered Oct 18 '18 at 18:25
J.G.J.G.
31.2k23149
answered Oct 18 '18 at 18:25
J.G.J.G.
31.2k23149
answered Oct 18 '18 at 18:25
J.G.J.G.
31.2k23149
answered Oct 18 '18 at 18:25
J.G.J.G.
31.2k23149
31.2k23149
add a comment |
add a comment |
$begingroup$
Withouth invoking the polylogarithm, for any $ninmathbb{N}$ and any $a>0$ we have
$$begin{eqnarray*} I(n,a)=int_{0}^{+infty}frac{x^n}{e^{ax}-1},dx&=&sum_{mgeq 1}int_{0}^{+infty}x^n e^{-ma x},dx\&=&sum_{mgeq 1}frac{n!}{(ma)^{n+1}}=frac{n!}{a^{n+1}}zeta(n+1)end{eqnarray*}$$
and by differentiation under the integral sign
$$ J(n,a)=frac{partial}{partial a} I(n,a)=int_{0}^{+infty}frac{-x^{n+1} e^{ax}}{(e^{ax}-1)^2},dx = -frac{(n+1)!}{a^{n+2}}zeta(n+1)$$
such that
$$begin{eqnarray*} int_{0}^{+infty}frac{x^{n+1}}{(e^{ax}-1)^2},dx &=& -left[I(n+1,a)+J(n,a)right]\&=&frac{(n+1)!}{a^{n+1}}zeta(n+1)-frac{(n+1)!}{a^{n+2}}zeta(n+2)end{eqnarray*}$$
and by setting $a=n=1$
$$ int_{0}^{+infty}frac{x^{2}}{(e^{x}-1)^2},dx =color{red}{2left(zeta(2)-zeta(3)right)}=sum_{ngeq 1}frac{2n}{(n+1)^3}=sum_{ngeq 1}frac{6n+5(-1)^{n}}{n^3binom{2n}{n}}. $$
answered Oct 18 '18 at 18:11
Jack D'AurizioJack D'Aurizio
291k33284669
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add a comment |
$begingroup$
Withouth invoking the polylogarithm, for any $ninmathbb{N}$ and any $a>0$ we have
$$begin{eqnarray*} I(n,a)=int_{0}^{+infty}frac{x^n}{e^{ax}-1},dx&=&sum_{mgeq 1}int_{0}^{+infty}x^n e^{-ma x},dx\&=&sum_{mgeq 1}frac{n!}{(ma)^{n+1}}=frac{n!}{a^{n+1}}zeta(n+1)end{eqnarray*}$$
and by differentiation under the integral sign
$$ J(n,a)=frac{partial}{partial a} I(n,a)=int_{0}^{+infty}frac{-x^{n+1} e^{ax}}{(e^{ax}-1)^2},dx = -frac{(n+1)!}{a^{n+2}}zeta(n+1)$$
such that
$$begin{eqnarray*} int_{0}^{+infty}frac{x^{n+1}}{(e^{ax}-1)^2},dx &=& -left[I(n+1,a)+J(n,a)right]\&=&frac{(n+1)!}{a^{n+1}}zeta(n+1)-frac{(n+1)!}{a^{n+2}}zeta(n+2)end{eqnarray*}$$
and by setting $a=n=1$
$$ int_{0}^{+infty}frac{x^{2}}{(e^{x}-1)^2},dx =color{red}{2left(zeta(2)-zeta(3)right)}=sum_{ngeq 1}frac{2n}{(n+1)^3}=sum_{ngeq 1}frac{6n+5(-1)^{n}}{n^3binom{2n}{n}}. $$
answered Oct 18 '18 at 18:11
Jack D'AurizioJack D'Aurizio
291k33284669
$endgroup$
add a comment |
$begingroup$
Withouth invoking the polylogarithm, for any $ninmathbb{N}$ and any $a>0$ we have
$$begin{eqnarray*} I(n,a)=int_{0}^{+infty}frac{x^n}{e^{ax}-1},dx&=&sum_{mgeq 1}int_{0}^{+infty}x^n e^{-ma x},dx\&=&sum_{mgeq 1}frac{n!}{(ma)^{n+1}}=frac{n!}{a^{n+1}}zeta(n+1)end{eqnarray*}$$
and by differentiation under the integral sign
$$ J(n,a)=frac{partial}{partial a} I(n,a)=int_{0}^{+infty}frac{-x^{n+1} e^{ax}}{(e^{ax}-1)^2},dx = -frac{(n+1)!}{a^{n+2}}zeta(n+1)$$
such that
$$begin{eqnarray*} int_{0}^{+infty}frac{x^{n+1}}{(e^{ax}-1)^2},dx &=& -left[I(n+1,a)+J(n,a)right]\&=&frac{(n+1)!}{a^{n+1}}zeta(n+1)-frac{(n+1)!}{a^{n+2}}zeta(n+2)end{eqnarray*}$$
and by setting $a=n=1$
$$ int_{0}^{+infty}frac{x^{2}}{(e^{x}-1)^2},dx =color{red}{2left(zeta(2)-zeta(3)right)}=sum_{ngeq 1}frac{2n}{(n+1)^3}=sum_{ngeq 1}frac{6n+5(-1)^{n}}{n^3binom{2n}{n}}. $$
answered Oct 18 '18 at 18:11
Jack D'AurizioJack D'Aurizio
291k33284669
$endgroup$
Withouth invoking the polylogarithm, for any $ninmathbb{N}$ and any $a>0$ we have
$$begin{eqnarray*} I(n,a)=int_{0}^{+infty}frac{x^n}{e^{ax}-1},dx&=&sum_{mgeq 1}int_{0}^{+infty}x^n e^{-ma x},dx\&=&sum_{mgeq 1}frac{n!}{(ma)^{n+1}}=frac{n!}{a^{n+1}}zeta(n+1)end{eqnarray*}$$
and by differentiation under the integral sign
$$ J(n,a)=frac{partial}{partial a} I(n,a)=int_{0}^{+infty}frac{-x^{n+1} e^{ax}}{(e^{ax}-1)^2},dx = -frac{(n+1)!}{a^{n+2}}zeta(n+1)$$
such that
$$begin{eqnarray*} int_{0}^{+infty}frac{x^{n+1}}{(e^{ax}-1)^2},dx &=& -left[I(n+1,a)+J(n,a)right]\&=&frac{(n+1)!}{a^{n+1}}zeta(n+1)-frac{(n+1)!}{a^{n+2}}zeta(n+2)end{eqnarray*}$$
and by setting $a=n=1$
$$ int_{0}^{+infty}frac{x^{2}}{(e^{x}-1)^2},dx =color{red}{2left(zeta(2)-zeta(3)right)}=sum_{ngeq 1}frac{2n}{(n+1)^3}=sum_{ngeq 1}frac{6n+5(-1)^{n}}{n^3binom{2n}{n}}. $$
answered Oct 18 '18 at 18:11
Jack D'AurizioJack D'Aurizio
291k33284669
edited Oct 18 '18 at 18:17
answered Oct 18 '18 at 18:11
Jack D'AurizioJack D'Aurizio
291k33284669
answered Oct 18 '18 at 18:11
Jack D'AurizioJack D'Aurizio
291k33284669
answered Oct 18 '18 at 18:11
Jack D'AurizioJack D'Aurizio
291k33284669
291k33284669
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