TITLE: How to evaluate the integral $\int_{0}^{\infty}\frac{\sin{(ax)}}{x^b} dx$
QUESTION [6 upvotes]: Evaluate the integral
$$\int_0^\infty \dfrac{\sin(ax)}{x^b} \, dx,\quad a\in \mathbb{R},\quad 0<b<2.$$
I know $a=1$ ,and $ b\in \mathbb{N}$, I can find the value,
How to evaluate this integral for general $a,b$?  Thank you.

REPLY [9 votes]: Process 1:$$\int_0^\infty \dfrac{\sin(ax)}{x^b} \, dx.$$$$=-\Im\left(\int_0^\infty\large e^{-iax}x^{-b}\, dx\right)$$$$=-\Im\left(\frac{1}{(ia)^{1-b}}\int_0^\infty\large e^{-t}t^{-b}\, dt\right)$$$$=-\Im\left(\frac{1}{(ia)^{1-b}}\Gamma(1-b)\right)=a^{b-1}\Gamma(1-b)\cos \left({\pi b\over 2}\right)$$
Process 2: Use Mellin Transformation.
Let, $$I(a,b)=\int_0^\infty \dfrac{\sin(ax)}{x^b} \, dx=a^{b-1}\int_0^\infty \sin(x)x^{-b}\,dx$$
Now, using Mellin transformation of $\sin x$, we will have, $$I(a,b)=a^{b-1}\Gamma(1-b)\sin\left(\frac{\pi(1-b)}{2}\right)=a^{b-1}\Gamma(1-b)\cos \left({\pi b\over 2}\right)$$