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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.

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  • $\begingroup$ Is $a$ also a real number? $\endgroup$
    – Clayton
    Commented Oct 7, 2014 at 20:27
  • $\begingroup$ Why do you want to know this? $\endgroup$
    – Peter Halburt
    Commented Oct 7, 2014 at 20:27
  • $\begingroup$ For a general treatment of these types of integrals, you may be interested in learning about the Bilateral Laplace Transform: en.wikipedia.org/wiki/… $\endgroup$
    – pshmath0
    Commented Oct 7, 2014 at 20:30
  • $\begingroup$ If $a$ is real, then this integral does not converge, as $\lvert x^{it}\rvert=1$. $\endgroup$
    – Yiorgos S. Smyrlis
    Commented Oct 7, 2014 at 20:33
  • $\begingroup$ If $a>0$ (or Re$\,a>0$), then this integral would make sense if the limits were $\int_0^\infty$. $\endgroup$
    – Yiorgos S. Smyrlis
    Commented Oct 7, 2014 at 20:34

2 Answers 2

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Ok, Let's start from the beginning. Do case $t=0$ easy. Then induction continuously. Easy to show $I=$ $$-\frac{1}{2} e^{-\frac{3 \pi t}{2}} \left(e^{2 \pi t}-1\right) (\text{sgn}(a)+i) \left(-a^2\right)^{-\frac{1}{2}-\frac{i t}{2}} \Gamma (i t+1)$$

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    $\begingroup$ Is it possible to elaborate little bit of how you come up with the result ? Did you use integration by parts ? $\endgroup$
    – upol94
    Commented Oct 7, 2014 at 21:45
  • $\begingroup$ Maybe. It is a good exercise. $\endgroup$
    – Peter Halburt
    Commented Oct 11, 2014 at 20:45
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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ 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}

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