Solving the integral $\int_1^\infty \frac{1}{e^x-2^x}dx$ Is there a way to solve this integral explicitly. I have never seen a solution before
$$\int_1^\infty \frac{1}{e^x-2^x}dx$$
 A: Hint. One may expand the integrand and integrate termwise as follows
$$
\begin{align}
\int_1^\infty \frac{e^{-x}}{1-e^{- (1-\ln 2)x}} dx&=\int_1^\infty \sum_{n=0}^\infty {e^{-(n(1-\ln 2)+1)x}} dx
\\\\&=\sum_{n=0}^\infty \int_1^\infty {e^{-(n(1-\ln 2)+1)x}} dx
\\\\&=\sum_{n=1}^\infty \frac{e^{-(n-\ln 2)}}{n-\ln 2}
\\\\&=\frac2e \psi\left(e^{-1},1,1-\ln 2\right)
\end{align}
$$ where $\psi$ is a special function: the Lerch transcendent.
A: $\displaystyle\int_1^\infty \frac{e^{-x}}{1-e^{x (-1+\ln 2)}} dx$
Let $t=exp(-x)$. Then, the above is 
$\displaystyle\int_0^{1/e} \frac{1}{1-t^{1-\ln 2}} dt$
. 
Let $c = 1-\ln 2>0$. To find this integral, we have
$\displaystyle\int_0^{1/e} \frac{1}{1-t^c} dt = \displaystyle\int_0^{1/e} \sum_{i=0}^\infty t^{ic} dt = \sum_{i=0}^\infty \int_0^{1/e}  t^{ic} dt =  \sum_{i=0}^\infty \int_0^{1/e}  t^{ic} dt = \sum_{i=0}^\infty   \frac{e^{-ic-1}}{ic+1}$. 
(Note that the sum and integral could be interchanged, since the conditions of dominated convergence theorem apply. The upper bound on the limit is less than 1. )
