Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question

3 Answers 3

up vote 17 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.

share|improve this answer
13  
A clean answer! –  sos440 May 17 '13 at 6:18
    
Anyway, could you show me how you derived this neat formula? –  sos440 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

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

Simplifying,

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

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.

share|improve this answer
3  
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. –  O.L. May 16 '13 at 8:58
1  
@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. –  sos440 May 18 '13 at 0:22
1  
@sos440 Could you please tell what Mathematica functions do you use to draw contour diagrams like this? mathematica.stackexchange.com/questions/25626/… –  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}

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.