proof that, $ \int_0^\infty \frac{e^{-x}}{x}\left(a+x^se^{-ax}-\frac{1-e^{-ax}}{1-e^{-x}}\right)dx=\ln\Gamma(a+1)+\frac{\Gamma(s)}{(a+1)^s}$ proof that,
(1)
Where $\Gamma(n)=(n-1)! $ is a gamma function
$$
\int_0^\infty \frac{e^{-x}}{x}\left(a+x^se^{-ax}-\frac{1-e^{-ax}}{1-e^{-x}}\right)dx=\ln\Gamma(a+1)+\frac{\Gamma(s)}{(a+1)^s}
$$
 A: Split the integral into two parts:
$$ \int_0^{\infty} x^{s-1}e^{-(a+1)x}\hspace{2pt}\mathrm{d}x +\int_0^{\infty}e^{-x}\left(\frac{e^{-ax}-ae^{-x}-1+a}{x(1-e^{-x})}\right)\hspace{2pt}\mathrm{d}x $$
The first integral can be evaluated by differentiating the following integral $s-1$ times with respect to $a$:
$$ I(a) = \int_0^{\infty}e^{-(a+1)x}\hspace{2pt}\mathrm{d}x = \frac{1}{a+1} $$
$$ I^{(s-1)}(a) = \int_0^{\infty}x^{s-1}e^{-(a+1)x}\hspace{2pt}\mathrm{d}x = \frac{(s-1)!}{(a+1)^s} = \frac{\Gamma(s)}{(a+1)^s} $$
Which is the second term of the RHS in the question.
The second integral requires very minor manipulations to put it in a useful form:
$$ \int_0^{\infty}e^{-x}\left(\frac{e^{-ax}-ae^{-x}-1+a}{x\left(1-\frac{1}{e^x}\right)}\right)\hspace{2pt}\mathrm{d}x = \int_0^{\infty}\frac{e^{-ax}-ae^{-x}-1+a}{x(e^x -1)}\hspace{2pt}\mathrm{d}x$$
The integral in this form has been found to be the log-gamma function (performing the derivation for this is quite long):
$$ \ln\Gamma(1+z)=\int_0^{\infty}\frac{e^{-zt}-ze^{-t}-1+z}{t(e^t -1)}\hspace{2pt}\mathrm{d}t $$
Therefore, we have found the first term in the RHS of the question, solving the integral shown in the LHS.
