6
$\begingroup$

I'm working on a PDE semigroup problem and I'm calculating some things using Fourier transform, now I ended up having to find the inverse Fourier transform of $e^{-(1+\omega^2)^2 t}$. Watch this:

$$ \mathcal{F}^{-1}\{e^{-(1+\omega^2)^2 t}\}(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{-(1+\omega^2)^2 t} e^{i\omega x} d\omega $$

I tried messing about with the exponentials a bit, but nothing came from it. The problem is that we have this huge expression as an exponent, that has both real and imaginary part, so we can not easily (in my eyes) reduce it to for example a Gaussian integral via substitution. I however found the following:

$$ \int_{-\infty}^\infty e^{-(1+\omega^2)^2 t} e^{i\omega x} d\omega = \int_{-\infty}^\infty e^{-(1+\omega^2)^2 t} \cos(\omega x) d\omega + i \int_{-\infty}^\infty e^{-(1+\omega^2)^2 t} \sin(\omega x ) d\omega$$

Now both integrals converge nicely, since the $e$-power kills everything off fast enough, and because $\sin(\omega x)$ is odd for all $x\in\mathbb{R}$, and $e^{-(1+\omega^2)^2 t}$ is even, we find:

$$ i \int_{-\infty}^\infty e^{-(1+\omega^2)^2 t} \sin(\omega x ) d\omega = 0 $$

So that:

$$\mathcal{F}^{-1}\{e^{-(1+\omega^2)^2 t}\}(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-(1+\omega^2)^2 t} \cos(\omega x) d\omega$$

But did we really make it easier on ourselves? I tried using partial integration on the above expression but got nowhere. I'm at a loss here. Can someone help me out?

$\endgroup$
  • $\begingroup$ You can try to compute this integral through complex analysis ? $\endgroup$ – Phoenix Jan 2 '18 at 18:32
  • 1
    $\begingroup$ This is tricky... you have to find a functional equation involving your function where poles appears, then integrate over the boundary of a rectangle and use the residue theorem... I'll work on it later :) $\endgroup$ – Phoenix Jan 2 '18 at 18:38
  • $\begingroup$ Thanks man... and to tell you the truth, my contour integration skills have gotten a bit rusty. Most integrals I work with are just a simple substitution away from one of the familiar integrals (Gaussian, Beta function, etc.) $\endgroup$ – The Brainlet Exterminator Jan 2 '18 at 18:47
  • $\begingroup$ $x$ and $t$ are real numbers ? $\endgroup$ – Phoenix Jan 2 '18 at 18:54
  • $\begingroup$ Yes, $x\in\mathbb{R}$ and $t > 0$. $\endgroup$ – The Brainlet Exterminator Jan 2 '18 at 18:56
1
$\begingroup$

It's not much help, but you can use the convolution theorem to separate out the "hard" part from the easier parts:

$$\begin{align*}\mathcal{F}^{-1}\left\{e^{-(1+\omega^2)^2 t}\right\}(x) &= e^{-t}\mathcal{F}^{-1}\left\{e^{-t\omega^2}\cdot e^{-t\omega^4}\right\}\\ \\ &= e^{-t} \left[\mathcal{F}^{-1}\left\{e^{-t\omega^2}\right\} * \mathcal{F}^{-1}\left\{e^{-t\omega^4}\right\}\right] \\ \\ &= e^{-t} \left[\dfrac{1}{\sqrt{2t}}e^{-\frac{1}{4t} x^2} * \mathcal{F}^{-1}\left\{e^{-t\omega^4}\right\}\right] \\ \\ \end{align*}$$

where '$*$' denotes convolution.

The remaining hard part is

$$\mathcal{F}^{-1}\left\{e^{-t\omega^4}\right\}$$

which I don't have a solution for right now.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.