I am trying to solve a exercise from my Complex Analysis book. The exercise says:
Show that $\displaystyle\int_0^\infty \frac {\cos(x)}{x^a} \, dx=\Gamma(1-a) \sin \left(\frac{\pi a} 2 \right)$ where $0<a<1.$
I am having trouble in choosing proper contour for this integral. Could someone please explain me how should i choose contour for computing the same integral?
Since $f(z) = e^{-z} z^{s-1}$ is holomorphic on $Re(z) > 0$ with exponential decay as $Re(z) \to +\infty$, with the Cauchy integral theorem you get that for every $Re(b) > 0$ : $$Re(s) > 0, \qquad\qquad\Gamma(s) = \int_0^\infty e^{-z}z^{s-1}dz=\int_0^{b \infty} e^{-z}z^{s-1}dz $$ $$ = \int_0^\infty e^{-bx}(bx)^{s-1}d(bx) = b^{s} \int_0^\infty e^{-bx}x^{s-1}dx$$
Now for $Re(s) \in (0,1)$, you can extend it by continuity to $b = \pm i$ and get $$\Gamma(s) =e^{i \pi s/2} \int_0^\infty e^{-ix}x^{s-1}dx=e^{-i \pi s/2} \int_0^\infty e^{ix}x^{s-1}dx$$ i.e. for $Re(a) \in (-1,0)$ : $$\boxed{\int_0^\infty \cos(x) x^{-a}dx = \frac{e^{ i\pi (1- a)/ 2}+e^{- i \pi(1-a) / 2}}{2}\Gamma(1-a) = \sin( \pi a/ 2)\Gamma(1-a)}$$