Evaluation of definite integral using complex analysis I want to evaluate the following indefinite integral
$$ \int_0^{\infty} x^{p - 1} \cos (ax) dx$$ where $0 < p < 1$ and $a > 0$. I was considering the function $f(z) = z^{p - 1} e^{iaz}$ and integrate it over the contour $\gamma = [-R, -\epsilon] - C_{\epsilon}^+ + [\epsilon, R] + C_R^+$. (Here $C_r^+$ denote the upper half circle centered at $0$ with radius $r$). However I only get 
$$ \int_0^{\infty} x^{p - 1} e^{iax} dx - e^{i \pi p} \int_0^{\infty} x^{p - 1} e^{-iax} dx = 0$$ instead, which only gives some relation between $\int_0^{\infty} x^{p - 1} \sin (ax) dx$ and $\int_0^{\infty} x^{p - 1} \cos (ax) dx$. I know the result is somehow related to gamma function that is $$ \int_0^{\infty} x^{p - 1} \cos(x) dx  =\frac{ \pi }{ 2 \Gamma(1 - p) \sin( (1 - p) \pi / 2) }$$ So would the regular method using complex analysis can still evaluate this integral?
 A: Use a contour in the first quadrant, that is, consider
$$\oint_C dz \, z^{p-1} \, e^{i a z}$$
where $C$ is the contour consisting of $[\epsilon,R]$, the quarter-circle of radius $R$ in the first quadrant centered at the origin, $[i R,i \epsilon]$, and a quarter circle of radius $\epsilon$ about the origin.
The integral then becomes
$$\int_{\epsilon}^R dx \, x^{p-1} \, e^{i a x} + i R^p \int_0^{\pi/2} d\theta \,  e^{i p \theta} \, e^{i a R e^{i \theta}} \\ + i^p \int_R^{\epsilon} dy \, y^{p-1} \, e^{-a y} + i \epsilon^p \int_{\pi/2}^0 d\phi \, e^{i p \phi} \, e^{i a \epsilon e^{i \phi}}$$
The fourth integral vanishes as $\epsilon \to 0$ as $0 \lt p \lt 1$.  Also, the second integral vanishes as $R \to \infty$, as its magnitude is bounded by
$$R^p \int_0^{\pi/2} d\theta \, e^{-a R \sin{\theta}} \le R^p \int_0^{\pi/2} d\theta \, e^{-2 a R \theta/\pi} \le \frac{\pi}{2 a R^{1-p}}$$
and, again, $0\lt p \lt1$.  
The contour integral is zero by Cauchy's theorem as there are no poles inside the contour.  Therefore, in the above limits, we have
$$\int_0^{\infty} dx \, x^{p-1} \, e^{i a x} = i^p \int_0^{\infty} dy \, y^{p-1} \, e^{-a y} = e^{i p \pi/2} \frac{\Gamma(p)}{a^p}$$
Use the relationship
$$\Gamma(p) \Gamma(1-p) = \frac{\pi}{\sin{\pi p}}$$
and we have, taking real parts of the above,
$$\int_0^{\infty} dx \, x^{p-1} \, \cos{(a x)} = \frac{\pi \cos{(\pi p/2)}}{a^p \sin{(\pi p)} \Gamma(1-p)} = \frac{\pi}{2 a^p \sin{(\pi p/2)} \Gamma(1-p)}$$
