Evaluate integral in terms of Gamma function Need help evaluating the following integral in terms of the gamma function where gamma function is: $$\Gamma(x)=\int_0^\infty e^{-t}t^{x-1}dt.$$  The integral is the following: $$\large{\int_0^\infty t^{x-1}e^{-\lambda t\cos(\theta)}\cos(\lambda t \sin(\theta))dt}$$
 A: You just rewrite the cosine in terms of the complex exponential:
$$ \cos(\lambda t \sin\theta) = \frac 12\left[\exp(i\lambda t \sin\theta)+\exp(-i\lambda t \sin \theta)\right] $$
Your integral is then 
$$ \frac 12\int_0^\infty dt\,t^{x-1}
\left(
\exp(-t(\lambda\cos\theta-i\lambda \sin\theta))+
\exp(-t(\lambda\cos\theta+i\lambda \sin\theta))
\right)$$
That's a sum of two similar terms. In each of them, one only rescales $t$ by the factor of
$$\lambda\cos\theta\mp i\lambda \sin\theta = \lambda\exp(\mp i \theta) $$
so that the exponentials become $\exp(-T)$ while a factor of the coefficient above to the $(-x)$-th power is picked from $dt\,t^{x-1}$. So the result is
$$ \dots = \frac {\Gamma(x)}{2\lambda^x} (\exp(ix\theta)+\exp(-ix\theta  )) = \frac{\Gamma(x)}{\lambda^x}\cos(x\theta) $$
I am going to fix the minor mistakes now. Note that the limits of the definite integral aren't really changed even though $t$ was rescaled by a complex factor – as long as the phase is within some limits.
A: This is the same as
$$\frac12\left(\int_0^{\infty}t^{x-1}\exp\left\{-\lambda t e^{i\theta}\right\}dt+\int_0^{\infty}t^{x-1}\exp\left\{-\lambda t e^{-i\theta}\right\}dt\right).$$
Making the obvious change of variables $s=\lambda t e^{\pm i\theta}$ in the 1st and 2nd integrals, we find
$$\frac{\lambda^{-x}}{2}\left(e^{-ix\theta}+e^{ix\theta}\right)\int_0^{\infty}s^{x-1}e^{-s}ds=\lambda^{-x}\cos x\theta\, \Gamma(x).$$
