The question come from a Mutusbara Sum like this
$${ \sum _{ { z=i\omega }_{ n } } { \frac { -\alpha E\pi }{ 4{ z }^{ 3 }\sqrt { -\alpha -z } } } }$$
it equal a contour integral around Imaginary axis with pole($\omega_n=\frac{(2n+1)\pi}{\beta}$,Fermion)
$${ { \frac { 1 }{ 2\pi i } \oint { \frac { -\beta }{ { e }^{ \beta z }+1 } \quad \frac { -\alpha E\pi }{ 4{ z }^{ 3 }\sqrt { -\alpha -z } } } } }-{ Res[\frac { -\beta }{ { e }^{ \beta z }+1 } \quad \frac { -\alpha E\pi }{ 4{ z }^{ 3 }\sqrt { -\alpha -z } } ] }_{ z=0 }$$
**the 1st integral have Branch point at $-\alpha$($\alpha \in Reals$)**,when I inflate original imaginary path to Infinite diameter, only the path around the pole $z=-\alpha$ and two parallel(but opposite direction) paths survive. Further more, **I reverse the direction of these paths, and the 1st integral convert to**
$$-{ { \frac { 1 }{ 2\pi i } \oint { \frac { -\beta }{ { e }^{ \beta z }+1 } \quad \frac { -\alpha E\pi }{ 4{ z }^{ 3 }\sqrt { -\alpha -z } } } } }$$
on a path around the branch point $z=-\alpha$.
**My question is How to calsulate this contour integral? If any one give some help or hint, I would appreciate!!**

This post imported from StackExchange Physics at 2016-06-17 12:17 (UTC), posted by SE-user alxandernashzhang