Hi, this is the integral and how much I managed to solve
http://img689.imageshack.us/img689/4719/img8366.jpg
So at this point I have two problems:
Firstly finding a limit around zero pole and secondly finding a residue around pole which is inside contour. While I had limits in my high maths course, I still struggle with this one; as for residue, have no idea how to find it and at the moment I don't have any good maths books with myself and even more important - time to learn it, so I'm in a pretty bad situation, any help would more than grateful!
Firstly I thought It wasn't useful, but no I think that I can convert limit (D part) into semicircular integral (which equals "-i*pi", "-" because I'm going clockwise) and multiplicate it of function residue at 1, so the limit then equals to "-i*pi/b^2 "which seams to be reasonable.
I managed to solve E part, which equals to "2i*pi*e^(iaib)/ib2ib = -i*pi*e^(-ab)/b^2" .
After simple algebra, my integral equals "pi(1-e^(-ab))/2b^2"
Thanks!
I think that this is what your integral is (I'm sorry my comp. has bad resolution)
Problem: Compute
Solution: Think for a second about the complex integral where that is the semi-circular countour of radius centered at , which is oriented counterclockwise. Notice that the integrand of this integral as simple poles at both and . Also, notice that . By Jordan's lemma (Jordan's Lemma -- from Wolfram MathWorld) we can see that (where is the bigger semi-circle). Therefore, at the limit the contribution from the integral with respect to the bigger semi-circle approaches zero. Lastly, using the residue theorem we can see that which reduces to and using the fact that the integral is even we may deduce that
Remark: I have no idea how to draw stuff on here, but if the countour I used isn't clear I can try to figure it out. Also, you can do this another way considering the function which makes the pole at the origin removable.