# 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?

$$\frac{1}{u^B}$$(u-$$\frac{z^3}{6}$$+...)