Proof of: $\int_0^\infty x^{m-1}e^{-ax} \cos bx \ dx = \frac{\Gamma(m)}{(a^{2} + b^{2})^{m/2}}\cos\left(m\tan^{-1}\left(\frac{b}{a}\right)\right)$

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 $\theta$.

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}