Strategy for integrating product of powers and trigonometric functions

Question

I want to compute the simple-looking definite integral $$\int_0^\infty \frac{\cos\omega t}{t^a}\,dt,\quad 0<a<1\,,$$ but I have no idea where to begin. Mathematica and integration tables give the final result but I want to know the precise steps leading up to it, be it a clever choice of contour or substitution. Many thanks to everyone!

Answer 1

Using the Gamma function $$ \Gamma(x)=\int_0^\infty t^{x-1}e^{-t}dt $$ for $s=\sigma+i\omega$, one has $$ \int_0^\infty t^{-a}e^{-st}dt=s^{a-1}\Gamma(1-a)=(\sigma+i\omega)^{a-1}\Gamma(1-a). $$ Letting $\sigma\to0$ gives \begin{eqnarray} &&\int_0^\infty t^{-a}e^{-i\omega t}dt\\ &=&(i\omega)^{a-1}\Gamma(1-a)\\ &=&e^{\frac{(a-1)\pi i}{2}}\omega^{a-1}\Gamma(1-a)\\ &=&\omega^{a-1}\bigg(\cos\left(\frac{(a-1)\pi }{2}\right)+i\sin\left(\frac{(a-1)\pi }{2}\right)\bigg) \Gamma(1-a) \end{eqnarray} and hence $$ \int_0^\infty t^{-a}\cos(\omega t)dt=\omega^{a-1}\cos\left(\frac{(a-1)\pi }{2}\right)\Gamma(1-a). $$