I was trying to compute: $$ I_{1/2}=\int_0 ^\infty \frac{\sqrt{x}}{e^x-1}dx. $$ I know it can be recast as follows $$ I_{\alpha}=\int_0^\infty \frac{x^\alpha}{e^x-1}\ dx= \int_0^\infty \frac{x^\alpha e^{-x}}{1-e^{-x}}\ dx =\\ = \sum_{n=0}^\infty\int_0^\infty x^\alpha\ e^{-(n+1)x} dx=\sum_{n=1}^\infty\frac{1}{n^{\alpha+1}} \int_0^\infty t^\alpha\ e^{-t} dt=\zeta(\alpha+1)\Gamma(\alpha+1). $$ so $$ I_{1/2}=\zeta(3/2)\Gamma(3/2). $$ But I was wondering if there is any direct contour integration solution. I tried using a rectangle of vertices $0, M, M+i2\pi, i2\pi$ with indented quarters of circle at $0$ and $i2\pi$, but I'm unsure on how to proceed.
In the limit $M\to \infty$ and of small radius of the circles I get $$ I_{1/2}=\int_0^\infty dx\frac{\sqrt{x+i2\pi}}{e^x-1}+i\int_0^{2\pi}dy\frac{\sqrt{iy}}{e^{iy}-1}+i\frac{\pi}{2}\sqrt{i2\pi} $$