Integrate Sin(u) over [0, ∞] with Residue Calculus

[kuahji]

$$\int$$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?

 

[kuahji]

Had I not something strange like u^B, I'd solve the problem as follows.
$$\frac{1}{u^B}$$(u-$$\frac{z^3}{6}$$+...)
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.