Find the value of $\int_0^{\infty}\frac{x^3}{(x^4+1)(e^x-1)}\mathrm dx$

Question

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

Answer 1

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

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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.

Answer 2

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:

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

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.