I have encountered the following identity on Wolfram alpha and I fail proving it (with $n \in \mathbb N^*$) $$\int_0^{\infty} \frac{t^n} {1+e^t}dt=(1-2^{-n})n! \zeta(n+1)$$
I tried to rewrite the term inside the integral as a power series, without success.
I know an identity that may be useful though $$(1-\frac{1}{2^n})\zeta(n+1)=\sum_{k=1}^{\infty} \frac {(-1)^{k-1}}{k^{n+1}}$$ (see one of my previous post).
using Geometric series
$$\frac{1}{1+e^t}=\frac{e^{-t}}{1+e^{-t}}=e^{-t}\sum_{m=0}^{\infty}(-1)^{m}e^{-mt}$$
so $$I=\int_{0}^{\infty}e^{-t}\sum_{m=0}^{\infty}(-1)^{m}e^{-mt}t^ndt$$
$$I=\sum_{m=0}^{\infty}(-1)^{m}\int_{0}^{\infty}e^{-t(m+1)}t^ndt$$
$$t(m+1)= u$$
$$I=\sum_{m=0}^{\infty}\frac{(-1)^{m}}{(m+1)^{n+1}}\int_{0}^{\infty}e^{-u}u^ndu$$
$$I=(1-2^{-n})\zeta{(n+1)}\Gamma{(n+1)}$$