Encountered $\int_0^{\infty} \frac{x^3}{e^{bx}+1}dx $ in statistical mechanics, curious how to solve it According to Wolfram Alpha, this integral = $\frac{7\pi^4}{120 b^4}$. I tried solving it using the residue theorem, but there are infinite residues which I believe means you can't apply the theorem.
Edit: I forgot to mention that b>0.
 A: Using the power series
$$
\frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^n x^n \text { for }| x |<1,
$$
we can transform the integrand into a power series too.
For $b>0$, we have
$$
\begin{aligned}
I & =\int_0^{\infty} \frac{x^3}{e^{b x}+1} d x \\
& =\int_0^{\infty} \frac{e^{-b x} x^3}{1+e^{-b x}} d x \\
& =\sum_{n=0}^{\infty}(-1)^n \int_0^{\infty} e^{-b x} \cdot x^3 e^{-b n x} d x \\
& =\sum_{n=0}^{\infty}(-1)^n \int_0^{\infty} x^3 e^{-b(n+1) x} d x
\end{aligned}
$$
Letting $y=b(n+1)x$ yields
$$
\begin{aligned}
\int_0^{\infty} x^3 e^{-b(n+1) x} d x
& =\frac{1}{b^4(n+1)^4} \int_0^{\infty} y^3 e^{-y} d y \\
& =\frac{1}{b^4(n+1)^4} \Gamma(4) \\
& =\frac{6}{b^4(n+1)^4}
\end{aligned}
$$
We can now conclude that
$$
\begin{aligned}
I & =\frac{6}{b^4} \sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1)^4} \\
& =\frac{6}{b^4}\left[\sum_{n=1}^{\infty} \frac{1}{n^4}-2 \sum_{n=1}^{\infty} \frac{1}{(2 n)^4}\right]\\& =\frac{21}{4b^4} \zeta(4)\\&=\frac{7\pi^4}{120b^4} \end{aligned}\\
$$
Furthermore, replacing the power $3$ of $x$ by $m\in R^+$ gives a more general integral for $b>0$,
$$\boxed{\int_0^{\infty} \frac{x^m}{e^{b x}+1} d x= \frac{\Gamma(m+1)}{b^{m+1} }\sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1)^{m+1}}= \frac{\Gamma(m+1)}{b^{m+1}}\left(1-\frac{1}{2^{m}}\right)\zeta(m+1) }$$
A: Utilizing K.defaoite's hint it is also useful to multiply top and bottom by $e^{-bx}$. We have
\begin{align}
I(b) &= \int_0^\infty x^3 e^{-bx} \sum_{k=0}^\infty (-1)^ke^{-bkx} \\
&= \sum_{k=0}^\infty (-1)^k \int_0^\infty x^3 e^{-bx(k+1)} \ dx \\
& = \frac{6}{b^4} \sum_{k=0}^\infty \frac{(-1)^{k+1}}{(k+1)^4} = \frac{6}{b^4} \frac{7\pi^4}{720} = \frac{7\pi^4}{120 b^4}
\end{align}
