# Calculating improper integral

Does anyone know how to solve the following integral:

$$I =\int_{0}^\infty \cos(t \mathrm{log}( x))\,\mathrm{e}^{-ax}\, \mathrm{d}x,$$

where $t$ and $a$ are real.

Please show some intermediate steps if you can.

Thank you.

• why do you mention $i$? how does it relate to your question? Commented Oct 8, 2014 at 23:38
• @upol94 check out the relations on this Wolfram site functions.wolfram.com/GammaBetaErf/Gamma/19 Commented Oct 9, 2014 at 3:05

## 1 Answer

Consider the integral \begin{align} J = \int_{0}^{\infty} e^{-a x} \, x^{\mu} \, dt = \frac{\Gamma(\mu+1)}{a^{\mu+1}}. \end{align} Let $\mu = i t$, where $i = \sqrt{-1}$, for which \begin{align} \int_{0}^{\infty} e^{-a x} \, x^{i t} \, dt &= \int_{0}^{\infty} e^{-a x} \, e^{i t \ln(x)} \, dt = \frac{\Gamma(i t+1)}{a^{it+1}} \end{align} From this it is seen that \begin{align} \int_{0}^{\infty} \cos(t \, \ln(x) ) \, e^{-a x} \, dx &= \mathcal{R}\left( \frac{\Gamma(i t+1)}{a^{it+1}}\right) \\ &= \frac{1}{a} \mathcal{R} \left( e^{-i t \ln(a)} \, \Gamma(i t + 1) \right) \end{align}

• My understanding is the first integral result comes using integration by parts and when $\mu$ is real and positive. Integration by parts can not be applied when $\mu$ is complex. I dont think you can simply write second integral from the $J$ when $\mu$ is complex. Can you please justify ? Commented Oct 9, 2014 at 0:07
• In the integral for the Gamma function, namely, \begin{align} \int_{0}^{\infty} e^{-t} \, t^{z-1} \, dt \end{align} it is required that $\mathcal{R}(z) > 0$. In this integral $z$ can be $z = x + i y$. Commented Oct 9, 2014 at 1:04