[tex]\int[/tex]u^-B sin(u) du, 0<B<2 integrating from 0 to infinity. What is really throwing me off is the condition, I'm not sure why it's there or really what to do with it. Can I just solve this the same way I'd solve sin(x)/x?
Answers and Replies
Related Calculus and Beyond Homework Help News on Phys.org
Had I not something strange like u^B, I'd solve the problem as follows.
[tex]\frac{1}{u^B}[/tex](u-[tex]\frac{z^3}{6}[/tex]+...)
I'd then look for C_-1, the residue. Then it's just a simple matter of 2pi(i)*res. But again, having u^B does not allow me to do this. There is a systematic formula I can apply to find the residue, but it does not allow me to eliminate i.