Proof of: $\int_0^\infty x^{m-1}e^{-ax} \cos bx \ dx = \frac{\Gamma(m)}{(a^{2} + b^{2})^{m/2}}\cos\left(m\tan^{-1}\left(\frac{b}{a}\right)\right)$ Where can I find a proof or how do you prove the following: 
$$\int_0^\infty x^{m-1}e^{-ax} \cos bx  \ dx = \frac{\Gamma(m)}{(a^{2} + b^{2})^{m/2}}\cos\left(m\tan^{-1}\left(\frac{b}{a}\right)\right)$$

Edit: 
I think I see the identity now
For a cascade of integration by parts let
$$\int e^{-ax}\sin(bx)=\frac{1}{a^2+b^2}e^{-ax}(-a\sin(bx)-b\cos(bx))$$
$$\int e^{-ax}\cos(bx)dx=\frac{1}{a^2+b^2}e^{-ax}(b\sin(bx)-a\cos(bx))$$
 A: Here's how Euler did this (from E675, translation available here). We start with
$$ \int_0^{\infty} x^{m-1} e^{-x} \, dx = \Gamma(m). $$
Changing variables gives
$$ \int_0^{\infty} x^{m-1} e^{-kx} \, dx = \frac{\Gamma(m)}{k^m}. $$
Euler now assumes that this still works if $k=p \pm iq$ is complex, provided $p>0$. (It does, but this needs an application of Cauchy's theorem.) We then have
$$ \int_0^{\infty} x^{m-1} e^{-(p \pm iq)x} \, dx = \frac{\Gamma(m)}{(p \pm iq)^m}, $$
and if we write $p=f\cos{\theta}$, $q=f\sin{\theta}$, we can apply Euler's formula to obtain
$$ \int_0^{\infty} x^{m-1} e^{-(p \pm iq)x} \, dx = \frac{\Gamma(m)}{f^m}(\cos{m\theta} \mp i\sin{m\theta}). $$
Taking the real part gives us
$$ \int_0^{\infty} x^{m-1} e^{-px} \cos{qx} \, dx = \frac{\Gamma(m)}{f^m}\cos{m\theta}, $$
and the result follows by inverting the expressions of $p$ and $q$ in terms of $f$ and $\theta$.
A: Recalling
$$ \Gamma(z)=\int_0^\infty x^{z-1}e^{-x}dx　$$
one has
\begin{eqnarray}
&&\int_0^\infty x^{m-1}e^{-ax} \cos bx  \; dx\\
&=&\Re\int_0^\infty x^{m-1}e^{-ax} e^{ibx}  \; dx\\
&=&\Re\int_0^\infty x^{m-1}e^{-(a-ib)x}  \; dx\\
&=&\Re\frac{1}{(a-ib)^{m}}\int_0^\infty x^{m-1}e^{x}  \; dx\\
&=&\Re\frac{1}{(a-ib)^{m}}\Gamma(m)\\
&=&\frac{1}{(a^2+b^2)^{m/2}}\Gamma(m)\cos(m\tan^{-1}(\frac{b}{a}))\\
\end{eqnarray}
