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I need to find a closed-form for the following integral. Please give me some ideas how to approach it: $$\int_0^{\infty}\frac{x^3}{(x^4+1)(e^x-1)}\mathrm dx$$

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up vote 19 down vote accepted

$$\int_0^{\infty}\frac{x^3}{(x^4+1)(e^x-1)}\mathrm dx=-\frac{\pi}{\sqrt8}-\log\sqrt{2\pi}-\frac{1}{2}\Re\ \psi\left(\frac{\sqrt[4]{-1}}{2\pi}\right),$$ where $\Re\ \psi(z)$ denotes the real part of the digamma function, $\psi(z)=\frac{\Gamma'(z)}{\Gamma(z)}$.

Solution: Use the approach from sos440's answer. To calculate the infinite sum and simplify the result we need the following: $$\frac{1}{n\left((2\,\pi\,n)^4+1\right)}=\frac{1}{n}-\frac{1}{4}\left(\frac{1}{n+\frac{(-1)^{1/4}}{2 \pi }}+\frac{1}{n+\frac{(-1)^{-1/4}}{2\pi}}+\frac{1}{n+\frac{(-1)^{3/4}}{2 \pi }}+\frac{1}{n+\frac{(-1)^{-3/4}}{2\pi}}\right),$$ $$\sum_{n=1}^\infty\left(\frac{1}{n}-\frac{1}{n+x}\right)=\gamma+\psi(1+x),$$ and the formulas (8), (9) from here.

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A clean answer! – Sangchul Lee May 17 '13 at 6:18
Anyway, could you show me how you derived this neat formula? – Sangchul Lee May 17 '13 at 14:33

My calculation shows that

\begin{align*} \int_{0}^{\infty} \frac{x^3}{(x^4 + 1)(e^x - 1)} \, dx &= \frac{\gamma}{2} - \log\sqrt{2\pi} + \frac{\pi}{4} \frac{\sin\frac{1}{\sqrt{2}}}{\cosh\frac{1}{\sqrt{2}} - \cos\frac{1}{\sqrt{2}}} \\ &\quad - \frac{1}{2}\sum_{n=1}^{\infty} \frac{1}{n\left(1 + (2\pi n)^4\right)} \\ &\approx 0.389075976914101580976629 \cdots. \end{align*}

My solution is divided into several steps:

Step 1. Let us introduce the function

$$ I(s) = \int_{0}^{\infty} \frac{x^{s}}{(x^4+1)(e^{x}-1)} \, dx $$

It is easy to see that

$$ I(s) + I(s+4) = \Gamma(s+1)\zeta(s+1). \tag{1} $$

This allows us to extend $I(s)$ as a meromorphic function on $\Bbb{C}$.

Step 2. Consider a contour $C$ starting from $\infty + \epsilon i$, making a counter-clockwise turn around $z = 0$ and going back to $\infty - \epsilon i$ as follows:

enter image description here

If we use a logarithm function with the branch cut $[0, \infty)$, we see that

$$ (e^{2\pi i s} - 1)I(s) = \int_{C} \frac{z^{s}}{(z^4+1)(e^z - 1)} \, dz. \tag{2} $$

Also, for $ 1 < s < 2$ we can confirm that $(2)$ is rewritten as

$$ (e^{2\pi i s} - 1)I(s) = \lim_{n\to\infty} \int_{C_{n}} \frac{z^{s}}{(z^4+1)(e^z - 1)} \, dz, $$

where $C_n$ is the contour given by

enter image description here

Here, the condition $1 < s < 2$ is introduced in order to create an appropriate decay speed for the integral along the contour $C - C_{n}$. By applying the Cauchy integration formula, we have

$$ (e^{2\pi i s} - 1)I(s) = -2\pi i \sum_{\omega^4 = -1} \operatorname{Res}_{z=\omega} \frac{z^{s}}{(z^4+1)(e^z - 1)} - 2\pi i \sum_{n\neq 0} \operatorname{Res}_{z=2\pi i n} \frac{z^{s}}{(z^4+1)(e^z - 1)}. $$


$$ I(s) = \frac{\pi}{\sin \pi s} \frac{e^{-i\pi s}}{4} \sum_{\omega^4 = -1} \frac{\omega^{s+1}}{e^{\omega} - 1} - \frac{2^{s}\pi^{s+1}}{\sin \frac{\pi s}{2}} \sum_{n=1}^{\infty} \frac{n^s}{1+(2\pi n)^4} \tag{3} $$

Since both sides define a meromorphic function for $\Re s < 3$, they coincide on this range.

Step 3. Combining $(1)$ and $(3)$, we have

\begin{align*} I(s+3) &= \Gamma(s)\zeta(s) - I(s-1) \\ &= \Gamma(s)\zeta(s) - \frac{\pi}{\sin \pi s} \frac{e^{-i\pi s}}{4} \sum_{\omega^4 = -1} \frac{\omega^{s}}{e^{\omega} - 1} - \frac{(2\pi)^{s}}{2\cos \frac{\pi s}{2}} \sum_{n=1}^{\infty} \frac{n^s}{n \left( 1+(2\pi n)^4 \right)} \end{align*}

Taking $s \to 0$, we obtain the desired result.

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Nice technique! (+1) Apparently Mathematica can express the remaining infinite sum as a sum of digamma functions over the roots of some 4th order polynomial. – Start wearing purple May 16 '13 at 8:58
@O.L.: what version Mathematica? V. 8.0.4 could't do anything with this. – Ron Gordon May 16 '13 at 21:21
Nice solution. How did you generate the contour diagram, if you don't mind? – Jonathan May 17 '13 at 16:55
@Jonathan, I drew them using Mathematica. – Sangchul Lee May 18 '13 at 0:22
@sos440 Could you please tell what Mathematica functions do you use to draw contour diagrams like this?… – Vladimir Reshetnikov May 22 '13 at 0:16

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\infty}{x^{3} \over \pars{x^{4} + 1}\pars{\expo{x} - 1}}\,\dd x: \ {\large ?}}$

\begin{align}&\color{#c00000}{% \int_{0}^{\infty}{x^{3} \over \pars{x^{4} + 1}\pars{\expo{x} - 1}}\,\dd x} =\half\int_{0}^{\infty}\pars{{1 \over x^{2} + \ic} + {1 \over x^{2} - \ic}}\, {x \over \expo{x} - 1}\,\dd x \\[3mm]&=\Re\int_{0}^{\infty}{x\,\dd x \over \pars{x^{2} + \ic}\pars{\expo{x} - 1}} =\Re\int_{0}^{\infty}{x\,\dd x \over \bracks{x^{2} + \ic\,/\pars{4\pi^{2}}}\pars{\expo{2\pi x} - 1}}\tag{1} \end{align}

The last integral in $\pars{1}$ is related to the Digamma function $\ds{\Psi\pars{z}}$ by means of the identity ${\bf\mbox{6.3.21}}$: $$ \Psi\pars{z} = \ln\pars{z} - {1 \over 2z} -2\int_{0}^{\infty} {t\,\dd t \over \pars{t^{2} + z^{2}}\pars{\expo{2\pi t} - 1}}\,,\qquad \verts{{\rm arg}\pars{z}} < {\pi \over 2} $$

Then, $\pars{1}$ is reduced to $\ds{\pars{~\mbox{with}\ \root{\ic} = \expo{\ic\pi/4} ={1 \over \root{2}}\pars{1 + \ic}~}}$: \begin{align} &\color{#44f}{\large% \!\!\!\!\!\int_{0}^{\infty}{x^{3} \over \pars{x^{4} + 1}\pars{\expo{x} - 1}}\,\dd x} =\Re\bracks{-\,\half\,\Psi\pars{\root{\ic} \over 2\pi} +\half\,\ln\pars{\root{\ic} \over 2\pi} - {1 \over 4}\,{2\pi \over \root{\ic}}} \\[3mm]&=\color{#44f}{\large% -\,\half\,\Re\Psi\pars{\root{\ic} \over 2\pi} -\half\,\ln\pars{2\pi} -{\root{2} \over 4}\,\pi} \approx 0.3891 \end{align}

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Is there a way to simplify the $\Re\Psi(\cdot)$? – Akiva Weinberger Aug 27 '14 at 14:12
@columbus8myhw I don't think so as far as I know. I was checking this page but it seems there isn't any thing simpler than that. Thanks. – Felix Marin Aug 27 '14 at 17:19

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