Interesting integral involving $\Gamma (z)$.

Question

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.

Answer 1

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

$$\Re{\left [\frac{\Gamma(x)}{\left( \lambda e^{i \theta}\right )^x} \right ]} = \frac{\Gamma(x)}{\lambda^x} \cos{(x \theta)}$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}