Evaluate integral in terms of Gamma function

Question

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: $$\large{\int_0^\infty t^{x-1}e^{-\lambda t\cos(\theta)}\cos(\lambda t \sin(\theta))dt}$$

Answer 1

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.

Answer 2

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 $$\frac{\lambda^{-x}}{2}\left(e^{-ix\theta}+e^{ix\theta}\right)\int_0^{\infty}s^{x-1}e^{-s}ds=\lambda^{-x}\cos x\theta\, \Gamma(x).$$