# 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.

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

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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