# Inverse Fourier Transform involving Gamma

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:

Solve:

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

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

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?)

https://en.wikipedia.org/wiki/Fourier_transform#Tables_of_important_Fourier_transforms

• 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 $\ge 0$?). – Mattos Mar 3 at 2:38
• Shouldn't your integral in your first line be with respect to $\omega$ – Memeboy Inc. Mar 3 at 3:32
• @MemeboyInc. ty fixed – LovesPeanutButter Mar 4 at 0:55