Interesting integral involving $\Gamma (z)$. Find the value of $$\int_0^\infty t^{x-1}e^{-\lambda t \cos(\theta)} \cos(\lambda t \sin (\theta)) dt$$ where $\lambda >0$, $x>0$, and ${-1\over 2}\pi < \theta < {1\over 2}\pi$ in terms of the Gamma function.
I'm assuming this requires at least two different substitutions in order to evaluate it.
 A: Generally speaking, when $\Re{(a)}>0$,
$$\int_0^{\infty} dt \, t^{x-1} \, e^{-a t} = \frac{\Gamma(x)}{a^x}$$
Note that the above integral may be written as
$$\Re{\left [\int_0^{\infty} dt \, t^{x-1} \, e^{-\lambda t e^{i \theta}} \right ]} $$
Therefore, as $\cos{\theta}>0$, the integral is
$$\Re{\left [\frac{\Gamma(x)}{\left( \lambda e^{i \theta}\right )^x} \right ]} = \frac{\Gamma(x)}{\lambda^x} \cos{(x \theta)}$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
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\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}t^{x - 1}\,\,
\expo{-\lambda t\cos\pars{\theta}}\,\,
\cos\pars{\lambda t\sin\pars{\theta}}\,\dd t}
\\[5mm] = &\
\Re\int_{0}^{\infty}t^{x - 1}\,\,
\expo{-\lambda t\cos\pars{\theta}}\,\,
\expo{\ic\lambda t\sin\pars{\theta}}\,\,\,\dd t
\\[5mm] = &\
\left.\Re\int_{0}^{\infty}t^{\color{red}{x} - 1}\,\,
\expo{-\lambda zt}\,\dd t\,\right\vert_{\,z\ =\ \exp\pars{-\ic\theta}}
\end{align}
Note that
$\ds{\expo{-\lambda zt} =
\sum_{k = 0}^{\infty}{\pars{-\lambda zt}^{k} \over k!} =
\sum_{k = 0}^{\infty}\color{red}{\pars{\lambda z}^{k}}\,\,{\pars{-t}^{k} \over k!}}$.
Then,
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}t^{x - 1}\,\,
\expo{-\lambda t\cos\pars{\theta}}\,\,
\cos\pars{\lambda t\sin\pars{\theta}}\,\dd t}
\\[5mm] = &\
\Re\bracks{\Gamma\pars{\color{red}{x}}
\pars{\lambda z}^{-\color{red}{x}}}\quad
\pars{\substack{\ds{Ramanujan's}\\[1mm] \ds{Master}
\\[1mm] \ds{Theorem}}}
\\[5mm] = &\
\Gamma\pars{x}\lambda^{-x}\,\,
\Re\bracks{\pars{\expo{-\ic\theta}}^{-x}} =
\bbx{\Gamma\pars{x}\lambda^{-x}\cos\pars{\theta x}} \\ &
\end{align}
