I was trying to solve the following integral $$ G(\alpha,m,n)=\int_0^{\infty}\cos(2nx)e^{-\alpha x}x^{m-1}dx;n\in N,\alpha>0,m\ge1. $$ By doing a change of variable I brought it to the integral $$ F(a,m)=\int_0^{\infty}\cos(x)e^{-a x}x^{m-1}dx. $$ Then I did not know how to proceed further and give it to Wolfram Mathematics to find the antiderivative and it found the following function $$ U(a,m,x)=-\frac{1}{2}\left[\frac{\Gamma(m,(a-i)x)}{(a-i)^m}+\frac{\Gamma(m,(a+i)x)}{(a+i)^m}\right], $$ where $\Gamma(s,t)$ is the incomlete Gamma function. By reading about some of its properties I assumed that $\Gamma(s,\infty)=0$ and $\Gamma(s,0)=\Gamma(s)$,so I got $$ F(a,m)=\frac{\Gamma(m)}{2}\left[\frac{1}{(a-i)^m}+\frac{1}{(a+i)^m}\right]=\Gamma(m)\frac{\cos\left[m\arccos\left(\frac{a}{\sqrt{a^2+1}}\right)\right]}{(a^2+1)^{\frac{m}{2}}}. $$ Of course I had doubts about the correctness of the last equality and tried to compute my integral by both using the last equality and approximated methods in Matlab. And the funniest part of this story is that for several combinations of $\alpha$,$m$ and $n$ the two results matched.
Now my question is how the last equality can be proved more rigorously(of course if it is correct). Any hints and suggestions would be appreciated.
Thanks!
You can perform your integration of $F(a,m)$ via the following:
$$F(a,m) = Re\left( \int_{0}^\infty e^{-(a-i)x}x^{m-1} dx \right) =$$ $$= Re\left((a-i)^{-m} \int_0^\infty e^{-x}x^{m-1} dx\right)=Re\left((a-i)^{-m} \Gamma(m)\right).$$
To compute the real part of a complex number $z$ we can write $Re(z) = \frac{z+\bar z}{2}$. Since $\Gamma(m)$ is real, $\Gamma(m)=\overline{ \Gamma(m)}$.
Thus $$F(a,m)= \Gamma(m) \frac12 \left(\frac1{(a-i)^m} + \frac1{(a+i)^m}\right).$$ $$=\frac{\Gamma(m)}{2}\left(\frac{(a+i)^m+(a-i)^m}{(a^2+1)^m} \right).$$
Now if we write $(a+i)=\sqrt{a^2+1}e^{i\arccos(a/\sqrt{a^2+1})}$ we see that $$(a+i)^m + (a-i)^m = (a^2+1)^{m/2}(e^{im\arccos(a/\sqrt{a^2+1})}+e^{-im\arccos(a/\sqrt{a^2+1})})$$ $$=(a^2+1)^{m/2}\cos(m\arccos(a/\sqrt{a^2+1})).$$
Finally we conclude: $$F(a,m)=\frac{\Gamma(m)}{2}\left(\frac{\cos(m\arccos(a/\sqrt{a^2+1}))}{(a^2+1)^{m/2}} \right).$$