Complex analysis question: can't find similar question on internet

[runforest]

this question doesn't seem tough but i can't find anything like it.

\int\frac{e^{ax}}{1+e^{x}}dx along the real line (a is between 1 and 0).

I know this is a complex analysis question, so i took the complex integral (along a semicircle where the diameter is the real numbers). by residue theorem, this integral is (2(Pi)i)*[sum of residues].

But now I am stuck, how do i get just the line integral on the real line, given the integral in the closed complex circle?

 

[jackmell]

It is tough. Gotta' bunch of residues in there right? When is 1+e^z=0? Well, that's easy, it's when \log(z)=-1 all the way up there. Then have to sum all those residues up like
2\pi i\sum_{n=0}^{\infty}e^{(2n+1) \pi i a}

Part of the problem is also showing what the value of the integral over the half-circular arc is. Because |a|<1, then the denominator denominates so it's zero but really should prove that rigorously.

Oh yeah, to answer your question, if the integral over the closed half-washer contour is equal to that sum of residues and the integral over the half-circle part is (probalby) zero as it's radius goes to infinity, then what's left is the integral over the real line and then that's equal to that expression up there. Gotta' figure out what it is.

 

[runforest]

Nevermind actually it is really easy. Just take a rectangle contour with one of the sides along the real line and a height of 2*Pi*i. Stretch the rectangle from -infinity to infinity on the real line and take the contour integral.