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.
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$\begingroup$ why do you mention $i$? how does it relate to your question? $\endgroup$– ArianCommented Oct 8, 2014 at 23:38
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$\begingroup$ @upol94 check out the relations on this Wolfram site functions.wolfram.com/GammaBetaErf/Gamma/19 $\endgroup$– LeucippusCommented Oct 9, 2014 at 3:05
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1 Answer
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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}
answered Oct 8, 2014 at 23:58
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$\begingroup$ 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 ? $\endgroup$– upol94Commented Oct 9, 2014 at 0:07
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$\begingroup$ 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 $z = x + i y$. $\endgroup$ Commented Oct 9, 2014 at 1:04