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

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

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

$$\int_0^{\infty}\frac{x^3}{(x^4+1)(e^x-1)}\mathrm dx=-\frac{\pi}{\sqrt8}-\log\sqrt{2\pi}-\frac{1}{2}\Re\ \psi\left(\frac{\sqrt[4]{-1}}{2\pi}\right),$$
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.

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.

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$\ds{\int_{0}^{\infty}{x^{3} \over \pars{x^{4} + 1}\pars{\expo{x} – 1}}\,\dd x:
\ {\large ?}}$

\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}