I need some help with the integral $\int_0^\infty \frac{x^3}{e^x+1} \, dx$ I've been stuck on an integral for days now and would love to get some help with it:
$$\int_0^\infty \frac{x^3}{e^x+1} \, dx$$
My teacher was also "kind" enough to give me the answer to another similiar integral: 
$$\int_0^\infty \frac{x^3}{e^x-1}dx=\frac{\pi^4}{15}$$
So that I should (based on that answer) calculate the integral:
$$\int_0^\infty \frac{x^3}{e^x+1} \, dx$$
I can't see how we can simplify/substitute anything so that it'll match the "help integral" so that we can use it's value to calculate the actual integral.
First post on this page, so please be kind! :)
 A: Note that $$\dfrac{x^3}{e^x+1} - \dfrac{x^3}{e^x-1} = x^3 \cdot \dfrac{-2}{e^{2x}-1} = -\dfrac{1}{4} \dfrac{(2x)^3}{e^{2x}-1},$$
so
$$\begin{gather}\int_0^\infty \dfrac{x^3}{e^x+1} \, dx - \int_0^\infty \dfrac{x^3}{e^x-1} \, dx = \\[6pt]
=-\dfrac{1}{4} \int_0^\infty \dfrac{(2x)^3}{e^{2x}-1}\,dx =  -\dfrac{1}{8} \int_0^\infty \dfrac{(2x)^3}{e^{2x}-1}\,d(2x) = -\dfrac{1}{8} \int_0^\infty \dfrac{u^3}{e^{u}-1}\,du.\end{gather}  $$
A: An approach using series.
Note that 
$$f(x)=\frac{x^3}{e^x+1}=\frac{x^3}{2e^{x/2}}\text{sech }\frac{x}2$$
And then recall that for $x>0$
$$\text{sech }x=-2\sum_{k\geq1}(-1)^ke^{(1-2k)x}$$
So $$f(x)=\sum_{k\geq1}(-1)^{k+1}x^3e^{-kx}$$
Hence 
$$J=\int_0^\infty\frac{x^3}{e^x+1}dx=\sum_{k\geq1}(-1)^{k+1}\int_0^\infty x^3e^{-kx}dx$$
The final integral:
$$\begin{align}
q_k&=\int_0^\infty x^{3}e^{-kx}dx\\
&\overset{kx\mapsto x}=\frac1{k^4}\int_0^\infty x^{4-1}e^{-x}dx
\end{align}$$
Then recall the definition of the Gamma function:
$$\Gamma(s)=(s-1)!=\int_0^{\infty}x^{s-1}e^{-x}dx\qquad \text{Re }s>0$$
Hence 
$$q_k=\frac6{k^4}$$
Which gives
$$J=6\sum_{k\geq1}\frac{(-1)^{k+1}}{k^4}=6\eta(4)=\frac{7\pi^4}{120}$$
Where $\eta(s)$ is the Dirichlet Eta Function.
In fact, we can show that 
$$\int_0^\infty \frac{x^s}{e^x+1}dx=(1-2^{2-s})\Gamma(s+1)\zeta(s-1)$$
With the same approach.
A: If you're interested in an approach to get both integrals without knowing either, $$\int_0^\infty\frac{x^3e^{-x}dx}{1\pm e^{-x}}=\sum_{n\ge 1}(\mp 1)^{n-1}\int_0^\infty x^3e^{-nx}dx=6\sum_{n\ge 1}\frac{(\mp 1)^{n-1}}{n^4},$$giving either $6\zeta(4)$ or $6\eta(4)$.
A: When $A=\int\frac{x^3}{e^{x}+1}dx,\ B= \int\frac{x^3}{e^{x}-1}dx$,
then $$ A-B = \int\frac{-2x^3\,dx}{e^{2x}-1} =\int\frac{-
t^3/4}{e^t-1} \frac{dt}{2} =-\frac{1}{8}B $$
A: Integration by parts with $u=x^3$ and $dv=\frac {dx}{e^x+1}$ may start you on the right track. 
