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?
Best Answer
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.