Calculating $\int_{0}^{\infty} x^{a-1} \cos(x) \ \mathrm dx = \Gamma(a) \cos (\pi a/2)$ My goal is to calculate the integral
$\int_{0}^{\infty} x^{a-1} \cos(x) dx = \Gamma(a) \cos (\pi a/2)$,
where $0<a<1$,
and my textbook provides the hint: integrate $z^{a-1} e^{iz}$ around the boundary of a quarter disk. 
However, I couldn't figure out how to control the integral over the quarter arc. Any hints? 
 A: Consider the contour integral
$$\oint_C dz \, z^{a-1} \, e^{i z}$$
where $C$ is a quarter circle of radius $R$ in the 1st quadrant (real and imaginary $> 0$), with a small quarter circle of radius $\epsilon$ about the origin cut out (to avoid the branch point at the origin).
This integral is equal to
$$\int_{\epsilon}^R dx \, x^{a-1} \, e^{i x} + i R^a \int_0^{\pi/2} d\theta \, e^{i a \theta} e^{i R \cos{\theta}} \, e^{-R \sin{\theta}}\\+ i \int_R^{\epsilon} dy \, e^{i \pi (a-1)/2} y^{a-1} e^{-y} + i \epsilon^a \int_{\pi/2}^0 d\phi \, e^{i a \phi} \, e^{i \epsilon e^{i \phi}}$$
We note that the second integral vanishes as $R\to\infty$ because $\sin{\theta} \gt 2 \theta/\pi$, so that the magnitude of that integral is bounded by
$$R^a \int_0^{\pi/2} d\theta \, e^{-R \sin{\theta}} \le R^a \int_0^{\pi/2} d\theta \, e^{-2 R \theta/\pi} \le \frac{2}{\pi R^{1-a}}$$
We also note that the fourth integral vanishes as $\epsilon^a$ as $\epsilon \to 0$.  In the third integral, we write $i=e^{i \pi/2}$ to make a simplification.
The contour integral is zero by Cauchy's Theorem (no poles in the interior of $C$).  Ths we have (+)
$$\int_{0}^{\infty} dx \, x^{a-1} \, e^{i x} - e^{i \pi a/2} \int_0^{\infty} dy \, y^{a-1} \, e^{-y}=0$$
We use the definition of the gamma function:
$$\Gamma(a) = \int_0^{\infty} dy \, y^{a-1} \, e^{-y}$$
and take real parts of (+) to obtain the sought-after result.
A: $\newcommand{\+}{^{\dagger}}%
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$\ds{\bbox[5px,#ffd]{\int_{0}^{\infty}
x^{a - 1}\cos\pars{x}\,\dd x =
\Gamma\pars{a}\cos\pars{\pi a \over 2}}:\ {\large ?}}$.

Note that $\ds{\int_{0}^{\infty}x^{a - 1}
\cos\pars{x}\,\dd x
=
\Re\int_{0}^{\infty}x^{a - 1}\expo{\ic x}\,\dd x}$.

With the change of variable
$x \equiv \expo{\ic\pi/2}\,t$, we'll have:
\begin{align}
&\int_{0}^{\infty}x^{a - 1}\cos\pars{x}\,\dd x
=
\Re\int_{0}^{-\ic\infty}\pars{\expo{\ic\pi/2}t}^{a - 1}
\expo{-t}\,\ic\,\dd x
\\[3mm] = &\
\Re\bracks{\expo{\ic\pi a/2}
\int_{0}^{-\ic\infty}t^{a - 1}
\expo{-t}\,\dd x}
\\[3mm] = &\
\Re\braces{\expo{\ic\pi a/2}\bracks{-\int_{\infty}^{0}t^{a - 1}\expo{-t}\,\dd x}}
\\[3mm] = &\
\Re\bracks{\expo{\ic\pi a/2}\
\overbrace{\int_{0}^{\infty}t^{a - 1}\expo{-t}\,\dd x\ }^{\ds{=\ \Gamma\pars{a}}}}
\\[5mm] = &\
\underbrace{\ \Re\bracks{\expo{\ic\pi a/2}}\ }
_{\ds{=\ \cos\pars{\pi a/2}}}
\Gamma\pars{a}
\end{align}

\begin{align}
&\mbox{}
\\
&\bbox[10px,border:1px groove navy]{%
\int_{0}^{\infty}x^{a - 1}\cos\pars{x}\,\dd x
=
\Gamma\pars{a}\cos\pars{\pi a \over 2}} \\ &
\end{align}
A: I recently read a short, yet interesting, article by Boas and Friedman where the authors calculated this same integral by using a different kind of contour:  
a triangle with vertices $-p_{1}, \;\ p_{2}, \;\ (p_{1}+p_{2})i$.
Consider $R(z)e^{iz}$, where R(z) is a rational function that vanishes at $\infty$
Then, $p_{1}, \;\ p_{2}$ are taken to be sufficiently large as to enclose the poles in the UHP.
Instead of using the arc of a circle in the first quadrant, they use a straight line segment.
By letting $p_{1}, \;\ p_{2} \to \infty$ the integral over the real axis is equal to $2\pi i$ times the sum of the residues in the UHP. 
This computation is claimed to be simpler because the slope of the triangular contour is bounded away from 0 in the UHP. 
Connect the points $z=p, \;\ z=pi$ with a line rather than the arc of a circle.
On this line, $|dz|=\sqrt{2}dy$ and $\frac{1}{|z|}\leq \frac{\sqrt{2}}{p}$.
It follows immediately that the integral over the line is bounded by a constant times 
$p^{s-1}\int_{0}^{p}e^{-y}dy\leq p^{s-1}$ which tends to 0 as $p\to \infty$ since $s<1$.
I have not included all of the intricacies, so if any one is interested in looking over this article it is in JSTOR.  Look for "a simplification in certain contour integrals". 
