Is it possible to compute $\int_1^{+ \infty} \frac{dx}{x^2(1+e^x)}$?

Question

Essentially, all is in the title: Is it possible to compute the integral $$\int_1^{+ \infty} \frac{dx}{x^2(1+e^x)} \hspace{1cm} ?$$

I suspected some relation with a polygamma function, but I was not able to find something explicit.

Answer 1

Substituting the series $\frac{1}{1+e^{-x}}=\sum_{n=0}^{\infty}(-1)^n e^{-nx}$ and interchanging the order of summation and integration, \begin{align} \int_1^{+ \infty} \frac{dx}{x^2(1+e^x)} &= \int_{1}^{\infty}\frac{e^{-x}\,dx}{x^2(1+e^{-x})}\\ &=\int_{1}^{\infty}\frac{1}{x^2}e^{-x}\sum_{n=0}^{\infty}(-1)^n e^{-nx}\,dx\\ &=\sum_{n=0}^{\infty}(-1)^n\int_{1}^{\infty}\frac{1}{x^2}e^{-x} e^{-nx}\,dx\\ &=\sum_{n=0}^{\infty}(-1)^n\int_{1}^{\infty}\frac{1}{x^2} e^{-(n+1)x}\,dx\\ &=\sum_{n=0}^{\infty}(-1)^n \left((n+1)\operatorname{Ei}(-(n+1))+e^{-(n+1)}\right)\\ &=\sum_{n=0}^{\infty}(-1)^n e^{-(n+1)} + \sum_{n=0}^{\infty}(-1)^n (n+1)\operatorname{Ei}(-(n+1))\\ &=\frac{1}{1+e}+\sum_{n=0}^{\infty}(-1)^n (n+1)\operatorname{Ei}(-(n+1))\\ &=\frac{1}{1+e}-\sum_{n=1}^{\infty}(-1)^{n} n\operatorname{Ei}(-n). \end{align}

Note: using, $u=(n+1)x$,

$$\int_{1}^{\infty}\frac{1}{x^2} e^{-(n+1)x}\,dx = \int_{n+1}^{\infty}\frac{(n+1)}{u^2} e^{-u}\,du\\ =(n+1)\operatorname{Ei}(-(n+1))+e^{-(n+1)}$$