How can I find the inverse Fourier transform of $e^{x^2}$ and similar functions?

Question

I'm honestly clueless; I tried writing it as ${\lmoustache}_{-\infty}^{+\infty} {e^{x^2}} {e^{ix}} {dx}$ , but I don't know where to go from there.

Answer 1

I think this integral does not exist as your function does not stay bounded for $x\to \infty$.

Maybe your are missing a minus sign? If it was a minus you could do the follwing. You are also missing a parameter if this is a fourrier inverse transform (something like time, I will also leave it out as it does not change the method).

But $-x^2+2\frac{i}{2}x=-x^2+2\frac{i}{2}x-\frac{i^2}{4}+\frac{i^2}{4}=-(x-\frac{i}{2})^2-\frac{1}{4}$

Introduce substitution $u=x-\frac{i}{2}$ and $du = dx$. Note also the changed integration bounds. Try to evaluate the integral using contour integration.

$$\int_{-\infty-i/2}^{\infty-i/2}e^{-u^2}e^{-1/4}du$$ This integral should be similar due, to Cauchy's Integral Theorem, to (use a rectangular contour along the real axis that goes down -i/2 and traces along an parallel to the real axis and gets back to the real axis): $$\int_{-\infty}^{\infty}e^{-u^2}e^{-1/4}du$$

Note that I am not sure if it is exactly the previous integral but it should be something similar.

This can be solved by multipling it by itself and using polar coordinates.