My Question:

How do I complete the inverse Fourier Transform of:

$\displaystyle \int_{-\infty}^\infty F(\omega)e^{-k\omega^2t}e^{-\gamma t}e^{-i\omega x}\,d\omega$

I cant figure out quite how to use the convolution theorem/table of transforms here.

The Problem:


$\displaystyle \frac{\partial u}{\partial t} = k\frac{\partial^2 u}{\partial x^2}-\gamma u, -\infty \lt x \lt \infty$

$\displaystyle u(x,0)=f(x)$

What I have done so far:

$\displaystyle \mathcal{F}\left[\frac{\partial u}{\partial t}\right] = k\mathcal{F}\left[\frac{\partial^2 u}{\partial x^2}\right]-c\mathcal{F}\left[\gamma u\right]$


$\displaystyle \frac{dU}{dt}=-k\omega^2U-\gamma U$

$\displaystyle \implies U(\omega,t)=C(\omega)e^{-k\omega^2t}e^{-\gamma t}$

$\displaystyle u(x,0)=f(x)\implies U(\omega,t)=F(\omega)e^{-k\omega^2t}e^{-\gamma t}$


$\displaystyle u(x,t)=\mathcal{F}^{-1}[U(\omega,t)]$

$\displaystyle = \int_{-\infty}^\infty F(\omega)e^{-k\omega^2t}e^{-\gamma t}e^{-i\omega x}\,d\omega$ (Stuck here, when trying to do the inverse....)

I cant figure out if any of these apply? Also is Gamma just a constant here (the Euler-Mascheroni constant?)


  • 1
    $\begingroup$ You can factor our the $e^{\gamma t}$ term from the integral as it is independent of $\omega$ and then complete the square on $-kt \omega^{2} - i \omega x$. Also, I'm pretty sure the person who posed the question would just be implying $\gamma \in \mathbb{R}^{+}$ (can you see why $\gamma$ must be $\ge 0$?). $\endgroup$ – Mattos Mar 3 at 2:38
  • 1
    $\begingroup$ Shouldn't your integral in your first line be with respect to $\omega$ $\endgroup$ – Memeboy Inc. Mar 3 at 3:32
  • $\begingroup$ @MemeboyInc. ty fixed $\endgroup$ – LovesPeanutButter Mar 4 at 0:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.