How To Prove:$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^4} = -\frac{7}{{720}}{\pi ^4}$ When I tried to solve this integral:
$$\int_0^\infty  {\frac{{{x^3}}}{{1 + {e^x}}}} \;{\rm{d}}x$$
I had trouble computing the sieries:
$$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^4}$$
Thanks.

From the definition of the polylogarithm, we have that
$$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^4}=\mathrm{Li}_4(-1).$$
Since $\mathrm{Li}_s(-1)=-\eta(s)$, we then seek to evaluate $-\eta(4)$. This is easy, as
$$\eta(s)=(1-2^{1-s})\zeta(s).$$
This would imply that 
$$-\eta(4)=-(1-2^{-3})\zeta(4),$$
and since $\zeta(4) = \pi^4/90$, we have
$$-\eta(4)=-\frac{7\pi^4}{720}.$$
Note that $\eta(s)$ denotes the Dirichlet eta function.
