During my studies, I've encountered the following integral where $a \in \mathbb{R}_{\ge0}$
$$\int_{a}^{\infty}\frac{\sqrt{x}}{e^{x}-1} dx$$
I know if $a=0$, this evaluates to $\Gamma(\frac{3}{2})\zeta(\frac{3}{2})$, but I'm not sure how to handle the generalized lower bound.
Appreciated.
I shall assume $a>0$.
Write $$\frac{\sqrt{x}}{e^{x}-1}=\sum_{n=0}^\infty \sqrt{x}\, e^{-(n+1) x}$$ $$\int_a^\infty \sqrt{x}\, e^{-(n+1) x}\,dx=\sqrt{a}\,\,\frac{ e^{-a (n+1)}}{n+1}+\frac{\sqrt{\pi }}{2}\frac{\text{erfc}\left(\sqrt{a (n+1)}\right)}{(n+1)^{3/2}}$$
$$\color{red}{\int_a^\infty \frac{\sqrt{x}}{e^{x}-1}= \Gamma \left(\frac{3}{2}\right)\zeta \left(\frac{3}{2}\right)-\sqrt{a} \log \left(1-e^{-a}\right)-}$$ $$\color{red}{\frac{\sqrt{\pi }}{2}\sum_{n=0}^\infty \frac{\text{erf}\left(\sqrt{a (n+1)}\right)}{(n+1)^{3/2}}}$$