Evaluate$$\int_0^{\infty} \frac{\sin x}{x^p} dx$$
I tried to use same trick as we do to evaluate ${sinx\over x}$ . But i couldn't find a suitable partial function to differentiate and then integrate .
Also tried writing it as $\Im[ \int_0^{\infty} \frac{e^{ix}}{x^p} dx]$ and use gamma function but powers of iota is confusing me .
I don't want to consider convergence of function here since there are already many similar questions . So p is such that integral very well converges .
Best Answer
$$\begin{split} \int_0^{+\infty}\frac{\sin x}{x^p}dx &= \int_0^{+\infty}\frac{\sin x }{\Gamma(p)}\int_0^{+\infty}t^{p-1}e^{-xt}dtdx\\ &= \int_0^{+\infty}\frac{t^{p-1}}{\Gamma(p)}\int_0^{+\infty}\sin(x)e^{-xt}dxdt\\ &= \frac 1 {\Gamma(p)}\int_0^{+\infty}\frac{t^{p-1}}{1+t^2}dt\\ &= \frac 1 {\Gamma(p)}\frac{\pi}{2\sin \left(\frac {p\pi} 2\right) } \end{split}$$ where the last equality comes from this answer.