For which values of ''$a$'', does the integral $$\int_{0}^{\infty} \frac{\sin x}{x^a}\;\mathrm dx$$ converges?
I have shown that $$\left| \int_{0}^{\infty} \frac{\sin x}{x^a}\;\mathrm dx \right|\le \lim_{n\to\infty} \int_{0}^{n} \frac 1{x^a}\;\mathrm dx,$$ which converges if $a > 1$. Are there other values of $a$ for which the integral converges?
Both integrability near 0 and integrability near $\infty$ have to be taken into account. The integral exists for $1<a<2$ and the improper integral $\int_0 ^{\infty} \frac {\sin (x)} x \, dx$ exists and the value is $\frac {\pi} 2$. [ This is proved in almost all texts on Complex Analysis]. For other values of $a$ the integral does not exist.