A definite integral identity

Question

While reading a physics paper, I encountered the following identity: $$ \int_{0}^\infty t^p \cos(Lat) \,e^{-Lbt} \mathrm{d}t = (-1)^{p+1} \frac{p!}{L\sqrt{a^2+b^2}} \cos{((p+1)\alpha)}, $$ where $\alpha = \arctan\left( \frac{a}{b} \right)$ and $a,b,p,L$ are real constants, $b,L>0$. However, I don't know how to prove this, I'm at loss even for a starting point. Could someone give a hint for how to prove this?

Answer 1

Assuming $p>-1$ and $b,L>0$ to ensure convergence, your integral is the real part of $$ \int_{0}^{+\infty}t^p e^{-L(b-ia)t}\,dt =\frac{\Gamma(p+1)}{L^{p+1}(b-ia)^{p+1}}$$ and to compute the real part of $\frac{1}{(b-ia)^{p+1}}$ is not a difficult task.