Integrating $ \int_0^\infty \frac{x^5}{e^x+1} \, dx $

Question

This improper integral has stumped me and not many integration problems give me problems. However, this one made me think over my limit. Finally, as I could not even get started on this problem, I thought to post it here, and have some talented mathematicians look at it. The problem is to integrate: $$ \int_0^\infty \frac{x^5}{e^x+1} \, dx $$

Answer 1

$$I=\int _{ 0 }^{ \infty }{ \cfrac { { x }^{ 5 } }{ { e }^{ x }+1 } dx= } \int _{ 0 }^{ \infty }{ \cfrac { { e }^{ -x } }{ { 1+e }^{ -x } } { x }^{ 5 }dx } =\int _{ 0 }^{ \infty }{ \left( \sum _{ r=1 }^{ \infty }{ { \left( -1 \right) }^{ r-1 }{ e }^{ -rx } } { x }^{ 5 } \right) dx } \\ =\sum _{ r=1 }^{ \infty }{ \left\{ \int _{ 0 }^{ \infty }{ { \left( -1 \right) }^{ r-1 }{ e }^{ -rx }{ x }^{ 5 }dx } \right\} }$$

Let $rx=t$: $$\int _{ 0 }^{ \infty }{ { \left( -1 \right) }^{ r-1 }{ e }^{ -rx }{ x }^{ 5 }dx } ={ \left( -1 \right) }^{ r-1 }\int _{ 0 }^{ \infty }{ { e }^{ -t }{ \left( \cfrac { t }{ r } \right) }^{ 6-1 }\cfrac { dt }{ r } } =\cfrac { { \left( -1 \right) }^{ r-1 } }{ { r }^{ 6 } } \Gamma (6)=120\cfrac { { \left( -1 \right) }^{ r-1 } }{ { r }^{ 6 } } \\ \\ \therefore I=\sum _{ r=1 }^{ \infty }{ 120\cfrac { { \left( -1 \right) }^{ r-1 } }{ { r }^{ 6 } } } =120\eta \left( 6 \right) =120\left\{ \left( 1-{ 2 }^{ 1-6 } \right) \right\} \zeta \left( 6 \right) \\ \\ =120\times \cfrac { 31 }{ 32 } \zeta \left( 6 \right) =\cfrac { 465 }{ 4 } \zeta \left( 6 \right)$$

$$\zeta \left( 6 \right) =\zeta \left( 2\times 3 \right) =\cfrac { { \left( -1 \right) }^{ 3+1 }{ B }_{ 6 }{ \left( 2\pi \right) }^{ 6 } }{ 2\left( 6 \right) ! } =\cfrac { { \pi }^{ 6 } }{ 945 }$$ Note that $B_6$ refers to the $6$th Bernoulli Number $\left(B_6=\frac{1}{42}\right)$:

$$\therefore I=\cfrac { 465 }{ 4 } \times \cfrac { { \pi }^{ 6 } }{ 945 } =\cfrac { { 31\pi }^{ 6 } }{ 252 }$$

Answer 2

The key is to express the integrand as the sum of a geometric series, and then switch the order of the integral and the summation.

$\displaystyle\int_{0}^{\infty}\dfrac{x^5}{e^x+1}\,dx$ $= \displaystyle\int_{0}^{\infty}\dfrac{x^5e^{-x}}{1+e^{-x}}\,dx$ $= \displaystyle\int_{0}^{\infty}\sum_{n = 1}^{\infty}x^5e^{-x}(-e^{-x})^{n-1}\,dx$ $= \displaystyle\int_{0}^{\infty}\sum_{n = 1}^{\infty}(-1)^{n-1}x^5e^{-nx}\,dx$ $= \displaystyle\sum_{n = 1}^{\infty}(-1)^{n-1}\int_{0}^{\infty}x^5e^{-nx}\,dx$ $= \displaystyle\sum_{n = 1}^{\infty}(-1)^{n-1}\int_{0}^{\infty}\left(\dfrac{u}{n}\right)^5e^{-u}\cdot \dfrac{1}{n}\,du$ $= \displaystyle\sum_{n = 1}^{\infty}\dfrac{(-1)^{n-1}}{n^6}\int_{0}^{\infty}u^5e^{-u}\,du$ $= \displaystyle\left(\sum_{n = 1}^{\infty}\dfrac{(-1)^{n-1}}{n^6}\right)\cdot \left(\int_{0}^{\infty}u^5e^{-u}\,du\right)$ $= \left(1-\dfrac{2}{2^6}\right)\zeta(6) \cdot \Gamma(6)$ $= \dfrac{31}{32} \cdot \dfrac{\pi^6}{945} \cdot 120 = \dfrac{31\pi^6}{252}$.

Here, we have used the following formulas involving the Gamma function, and the Riemann Zeta function: $\Gamma(n+1) = \displaystyle\int_{0}^{\infty}x^ne^{-x}\,dx$ and $(1-2^{1-s})\zeta(s) = \displaystyle\sum_{n = 1}^{\infty}\dfrac{(-1)^{n-1}}{n^s}$.

Answer 3

Here is yet another slightly different approach. Here we enforce the substitution $e^{-x}\to x$ and obtain

$$\begin{align} \int_0^{\infty} \frac{x^5}{1+e^{x}}dx&=-\int_0^1 \frac{\left(\log x\right)^5}{1+x}dx\\\\ &=-\frac{d^5}{dy^5}\left.\left(\int_0^1\frac{x^y}{1+x}dx\right)\right|_{y=0} \end{align}$$

Expanding $\frac{1}{1+x}$ and integrating term by term reveals

$$\begin{align} -\frac{d^5}{dy^5}\left.\left(\int_0^1\frac{x^y}{1+x}dx\right)\right|_{y=0}&=-\frac{d^5}{dy^5}\left.\left(\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{n+y}\right)\right|_{y=0}\\\\ &=(5!)\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^6}\\\\ &=(5!)\eta(6)\\\\ &=(120)\frac{31\pi^6}{30240}\\\\ &=\frac{31\pi^6}{252} \end{align}$$

which agrees with the results reported by others as expected!

Answer 4

Complex-Analytic Method

Note that, for a real parameter $k$, $$\int_0^{\infty}\,\frac{x^5}{\exp(x)+1}\,\text{d}x=\left[k^5\right]\left(5!\int_0^\infty\,\frac{\sin(kx)}{\exp(x)+1}\,\text{d}x\right)\,,$$ where $\left[k^r\right]\big(f(k)\big)$ denotes the coefficient of the $k^r$-term in the Laurent expansion of $f(k)$ around $k=0$. Furthermore, $$\int_0^\infty\,\frac{\sin(kx)}{\exp(x)+1}\,\text{d}x=\text{Im}\left(\int_0^\infty\,\frac{\exp(\text{i}kx)}{\exp(x)+1}\,\text{d}x\right)\,.$$ For $R>0$ and $\epsilon\in(0,\pi)$, let $Q_{R,\epsilon}$ be the positively oriented rectangle with vertices $0$, $R$, $R+2\pi\text{i}$, and $2\pi\text{i}$, where the line segment $[\pi\text{i}+\epsilon\text{i},\pi\text{i}-\epsilon\text{i}]$ is replaced by a semicircular arc with radius $\epsilon$ centered at $\pi\text{i}$ oriented clockwise going from $\pi\text{i}+\epsilon\text{i}$ to $\pi\text{i}-\epsilon\text{i}$. If $I(k)$ represents the integral $\displaystyle\int_0^\infty\,\frac{\exp(\text{i}kx)}{\exp(x)+1}\,\text{d}x$, then $$\lim_{R\to\infty}\oint\limits_{Q_{R,\epsilon}}\,\frac{\exp(\text{i}kz)}{\exp(z)+1}\,\text{d}z=\big(1-\exp(-2\pi k)\big)\,I(k)-\text{i}\int\limits_{[0,2\pi]\setminus(\pi-\epsilon,\pi+\epsilon)}\,\frac{\exp(-ky)}{\exp(\text{i}y)+1}\,\text{d}y+J_\epsilon\,,$$ where $J_\epsilon$ is the integral of $\frac{\exp(\text{i}kz)}{\exp(z)+1}$ along the clockwise semicircle of radius $\epsilon$ around $z=\pi\text{i}$. By the Residue Theorem, $$J_\epsilon=-{\pi\text{i}}\,\text{Res}_{z=\text{i}\pi}\left(\frac{\exp(\text{i}kz)}{\exp(z)+1}\right)+O\left(\epsilon\right)=\pi\text{i}\exp(-\pi k)+O\left(\epsilon\right)\,.$$ Because $Q_{R,\epsilon}$ encloses no poles of $\frac{\exp(\text{i}kz)}{\exp(z)+1}$, we have $$\oint\limits_{Q_{R,\epsilon}}\,\frac{\exp(\text{i}kz)}{1+\exp(z)}\,\text{d}z=0\,.$$ Also, $$\frac{\exp(-ky)}{1+\exp(\text{i}y)}=\frac{\exp(-ky)\big(\cos(y/2)-\text{i}\sin(y/2)\big)}{2\cos(y/2)}\,.$$ Hence, by taking $\epsilon\to 0$ and looking only at the imaginary part, we have $$\begin{align} 2\sinh(\pi k)\exp(-\pi k)\,\text{Im}\big(I(k)\big) &=-\pi\exp(-\pi k)+\frac12\,\int_{0}^{2\pi}\,\exp(-ky)\,\text{d}y \\ &=\exp(-\pi k)\left(-\pi+\frac{\sinh(\pi k)}{k}\right)\,. \end{align}$$ That is, $$\text{Im}\big(I(k)\big)=\frac{1}{2k}-\frac{\pi}{2}\,\text{csch}(\pi k)\,.$$ From $\text{csch}(x)=\frac{2}{\exp(x)-\exp(-x)}$, we can show that $$\text{csch}(x)=\frac{1}{x}-\frac{x}{6}+\frac{7x^3}{360}-\frac{31x^5}{15120}+O\left(x^7\right)\,.$$ Ergo, $$\left[k^5\right]\Big(\text{Im}\big(I(k)\big)\Big)=\frac{31\pi^6}{30240}\,.$$ Consequently, $$\int_0^{\infty}\,\frac{x^5}{\exp(x)+1}\,\text{d}x=5!\left(\frac{31\pi^6}{30240}\right)=\frac{31\pi^6}{252}\,.$$

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In general, $$\int_0^\infty\,\frac{x^{2r+1}}{\exp(x)+1}\,\text{d}x=\frac{(-1)^{r+1}\,(2r+1)!\,\pi^{2r+2}}{2}\,\Big(\left[x^{2r+1}\right]\big(\text{csch}(x)\big)\Big)\,,$$ for every $r\in\mathbb{N}_0$. If $\eta$ is the Dirichlet eta function, then $$\eta(2r+2)=\frac{(-1)^{r+1}\,\pi^{2r+2}}{2}\,\Big(\left[x^{2r+1}\right]\big(\text{csch}(x)\big)\Big)\,,$$ for every $r\in\mathbb{N}_0$. A similar technique yields $$\int_0^\infty\,\frac{x^{2r+1}}{\exp(x)-1}\,\text{d}x=\frac{(-1)^r\,(2r+1)!\,\pi^{2r+2}}{2}\,\Big(\left[x^{2r+1}\right]\big(\text{coth}(x)\big)\Big)\,,$$ which also means $$\zeta(2r+2)=\frac{(-1)^r\,\pi^{2r+2}}{2}\,\Big(\left[x^{2r+1}\right]\big(\text{coth}(x)\big)\Big)\,,$$ for every $r\in\mathbb{N}_0$, where $\zeta$ is the Riemann zeta function. This shows that $$\left[x^{2r+1}\right]\big(\text{csch}(x)\big)=-\left(1-\frac{1}{2^{2r+1}}\right)\,\Big(\left[x^{2r+1}\right]\big(\text{coth}(x)\big)\Big)$$ for all $r=0,1,2,\ldots$.

Answer 5

\begin{align} \int_0^\infty \frac{x^5}{e^x+1}\, dx&=\int_0^\infty x^5 \Big(\frac{1}{e^x-1}-\frac{2}{e^{2 x}-1}\Big)\, dx\\ &=\int_0^\infty \frac{x^5}{e^x-1} \, dx-2\int_0^\infty \frac{x^5}{e^{2 x}-1} \, dx\\ &=\int_0^\infty \frac{x^5}{e^x-1} \, dx-2^{-5}\int_0^\infty \frac{x^5}{e^{ x}-1} \, dx\\ &=(1-2^{-5})\int_0^\infty \frac{x^5}{e^{ x}-1} \, dx\\ &=(1-2^{-5})\Gamma(6)\zeta(6)\\ \end{align} note that $\int_0^\infty \frac{x^{s-1}}{e^{ x}-1}\,dx=\Gamma(s)\zeta(s)$. In general then it holds that $$\int_0^\infty \frac{x^s}{e^x+1} \, dx=(1-2^{-s})\Gamma(s+1)\zeta(s+1)$$