Evaulate: $$\frac{1}{4}\sum_{k=0}^{\infty} \frac {1}{4^k}\int_{0}^{\infty}\frac{x^ne^{-x}}{e^{kx}}(1+e^x)^{2k}dx$$ Where $n$ is a natural number.
I tried using the binomial theorem on $(1+e^x)^{2k}$ which gave:
$$=\frac{1}{4}\sum_{k=0}^{\infty} \frac {1}{4^k}\int_{0}^{\infty}\frac{x^ne^{-x}}{e^{kx}}\sum_{r=0}^{2k} \binom {2k}{r} e^{rx}dx$$ After which I interchanged the integration and summation (I'm not sure if this is allowed in this case)
$$=\frac{1}{4}\sum_{k=0}^{\infty} \frac {1}{4^k}\sum_{r=0}^{2k} \binom {2k}{r}\int_{0}^{\infty}x^n e^{-(k+1-r)x}dx$$ Substituing $v = (k+1-r)x$,
$$=\frac{1}{4}\sum_{k=0}^{\infty} \frac {1}{4^k}\sum_{r=0}^{2k} \binom {2k}{r}\int_{0}^{\infty}\frac{v^n}{(k+1-r)^{n+1}}e^{-v}dv$$
Now as $n$ is a natural number, using the gamma function, this becomes:
$$=\frac{n!}{4}\sum_{k=0}^{\infty} \frac {1}{4^k}\sum_{r=0}^{2k} \binom {2k}{r}\frac{1}{(k+1-r)^{n+1}}$$
I have absolutely no idea on how to continue further and I'm not even sure about the steps I've done so far. Any help with the original expression or with this new expression would be appreciated!
