Prove that $\int_{0}^{+\infty} u^{s-1} \cos (a u) \:e^{-b u}\:du=\frac{\Gamma(s)\cos\left(s\arctan \left(\frac{a}{b}\right)\right)}{(a^2+b^2)^{s/2}}$ From the answer of this OP: Ramanujan log-trigonometric integrals, I found the following formula
$$\begin{align}
& \int_{0}^{+\infty} u^{s-1}  \cos (a u) \:e^{-b u}\:\mathrm{d}u = \Gamma (s)\frac{\cos \left(\! s \arctan \left(\frac{a}{b}\right)\right)}{(a^2+b^2)^{s/2}}, \, \left(\Re(s)>0, b>0, a>0 \right)
\end{align}$$
How one proves that formula? I am interested in knowing the approach without using contour integration method. Thanks in advance.
 A: Using $a+i b = \sqrt{a^{2} + b^{2}} \, e^{i \tan^{-1}(b/a)}$ then from the Gamma function, 
\begin{align}
\int_{0}^{\infty} e^{-st} \, t^{x-1} \, dt = \frac{\Gamma(x)}{s^{x}}
\end{align}
let $s = a + i b$ to obtain
\begin{align}
\int_{0}^{\infty} e^{-(a+ib)t} \, t^{x-1} \, dt &= \Gamma(x) \, (a+ i b)^{-x} = \Gamma(x) \, (a^{2} + b^{2} )^{-x/2} \, e^{-i x \tan^{-1}(b/a)}.
\end{align}
From this result it is seen that
\begin{align}
\int_{0}^{\infty} e^{-a t} \, \cos(b t) \, t^{x-1} \, dt &= \frac{\Gamma(x)}{(a^{2} + b^{2})^{x/2}} \, \cos\left( x \, \tan^{-1}\left(\frac{b}{a}\right) \right) \\
\int_{0}^{\infty} e^{-a t} \, \sin(b t) \, t^{x-1} \, dt &= \frac{\Gamma(x)}{(a^{2} + b^{2})^{x/2}} \, \sin\left( x \, \tan^{-1}\left(\frac{b}{a}\right) \right) \\
\end{align}
A: It is sufficient to write the cosine as the real part of a complex exponential. From:
$$\int_{0}^{+\infty} u^{s-1} e^{(ia-b) u}\,du = \frac{1}{(b-ia)^s}\Gamma(s) $$
it is straightforward to derive the wanted identity, by just considering the real part of both terms.
