Calculating a complex definite improper integral: $I= \int_{0}^\infty x^{it}\,\mathrm{e}^{-ax}\, dx$ Does anyone know how to find the value of this integral:
$$I= \int_{0}^\infty x^{it}\,\mathrm{e}^{-ax}\, dx,$$
where $i=\sqrt{-1}$ and $t$, $a$ are real. 
Please give me a hint.
Thank you. 
 A: Ok, Let's start from the beginning. Do case $t=0$ easy. Then induction continuously. Easy to show $I=$ $$-\frac{1}{2} e^{-\frac{3 \pi  t}{2}} \left(e^{2 \pi  t}-1\right)
   (\text{sgn}(a)+i) \left(-a^2\right)^{-\frac{1}{2}-\frac{i t}{2}} \Gamma (i
   t+1)$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,{\rm Li}_{#1}}
 \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
The integration is, indeed, $\ds{\sf\mbox{quite simple}}$ and it can be exppressed in terms of the $\ds{\sf Gamma}$ $\ds{\Gamma}$ function:

\begin{align}&\overbrace{\color{#66f}{\large%
\int_{0}^{\infty}x^{\ic t}\expo{-ax}\,\dd x}}
^{\ds{\dsc{ax}\ \mapsto\ \dsc{x}}}\ =\
a^{-1 - \ic t}\ \overbrace{\int_{0}^{\infty}x^{\ic t}\expo{-x}\,\dd x}
^{\dsc{\Gamma\pars{\ic t + 1}}}\ =\
\color{#66f}{\large a^{-1 - \ic t}\ \Gamma\pars{\ic t + 1}}
\end{align}

