How to Express $\Gamma (x)$ in terms of $\cos$ I'm trying to figure out how to express the integral $$\int\limits_{0}^{\infty} \cos(x) \times x^{-a} \rm{dx}$$ as
$$\cos\frac{\pi-a\pi}{2}\times \int\limits_{0}^{\infty} e^{-x} x^{-a} \ \rm{dx} \qquad \text{where} \ 0 < a <1$$
I'm fairly sure it requires the calculation of a residue, and I've tried using a Fourier transform but it doesn't seem to get me anywhere.  Sorry about the poor formatting, it's my first time posting and I can't figure out how to express the integrals nicely.
Any hints on how to proceed? 
 A: It doesn't require the calculation of a residue, but it does involve contour integrals :-)
You can write the integral as
$$\int_0^\infty\cos (x)x^{-a}\mathrm dx=\Re\int_0^\infty\mathrm e^{\mathrm ix}x^{-a}\mathrm dx\;.$$
Then you can extend the contour on the positive real axis to a closed contour by integrating along the positive imaginary axis and along a quarter circle at infinity. Since $a>0$, the contribution from the circle vanishes in the limit. Since the integrand has no poles inside the contour, the two integrals along the positive real and imaginary axis must cancel each other. (To be rigorous, you'd have to also take a limit at the origin, since the function isn't holomorphic there; you can integrate on a small quarter circle around the origin to link the real axis and the imaginary axis; since $a<1$, the contribution from that circle also vanishes in the limit.)
So your integral is just the real part of the integral along the positive imaginary axis, which is
$$
\begin{eqnarray}
\Re\int_0^\infty\mathrm e^{\mathrm i\mathrm i x}(\mathrm ix)^{-a}\mathrm d(\mathrm ix)
&=&
\Re\int_0^\infty\mathrm e^{-x}x^{-a}\mathrm i^{1-a}\mathrm dx
\\
&=&
\Re\left(\mathrm i^{1-a}\right)\int_0^\infty\mathrm e^{-x}x^{-a}\mathrm dx
\\
&=&
\Re\left(\mathrm e^{(1-a)\frac{\pi}{2}\mathrm i}\right)\int_0^\infty\mathrm e^{-x}x^{-a}\mathrm dx
\\
&=&
\cos\left((1-a)\frac{\pi}{2}\right)\int_0^\infty\mathrm e^{-x}x^{-a}
\mathrm dx\;.
\end{eqnarray}
$$
