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.

$$\sum_{n=1}^{+\infty}\frac{(-1)^n}{n^4} = -\sum_{n=1}^{+\infty}\frac{1}{n^4}+2\sum_{m=1}^{+\infty}\frac{1}{(2m)^4} = \left(-1+\frac{2}{16}\right)\sum_{n=1}^{+\infty}\frac{1}{n^4}=-\frac{7}{8}\zeta(4) = -\frac{7\pi^4}{720}.$$
