# What is the Fourier transform of $e^{(-a+bi)x^2}$?

Let $a>0$. Let $f:\mathbb{R}\rightarrow\mathbb{C}$ be $f(x)= e^{(-a+bi)x^2}$. What is the Fourier transform of $f$?

Here is what I have tried:

The exponential decay of $e^{(-a+bi)x^2}$ means that $f$ is in the Schwartz space. So we can talk about its Fourier transform. And since $f$ is in the Schwartz space, we know that $$(f')^{\wedge}(\xi)=i\xi\hat{f}(\xi)$$ and that $$(-ixf(x))^{\wedge}(\xi)=(\hat{f})'(\xi).$$ Since $$f'(x)=2(-a+bi)xf(x),$$ using the above identities, after some simplification, we have that $$(\hat{f})'(\xi)+\frac{\xi}{2(a-bi)}\hat{f}(\xi).$$ Solving this differential equation using the method of integrating factors, I obtain that $$\hat{f}(\xi)=\hat{f}(0)e^{-\frac{\xi^2}{4(a-bi)}}$$

But how can I find $\hat{f}(0)$ i.e. what is $\int_\mathbb{R}e^{(-a+bi)x^2}dx$?

• You don't have to choose $\hat f(0)$, you can choose any $\xi$ you like.. See here in the comments under my question. Alternatively, just go straight to the Gaussian integral here and equate coefficients. – Mattos Feb 13 '15 at 4:06

The function $$F(z) = \int_{-\infty}^{\infty}e^{-zx^{2}}dx$$ is holomorphic in the right half plane $\Re z > 0$. For $z=a$ where $a$ is real and positive, one has $$F(a) = \sqrt{\frac{\pi}{a}}.$$ Choose the branch of $\sqrt{\pi/z}$ which is holomorphic in the right half plane, and this chosen branch must equal $F(z)$ everywhere in the right half plane by the identity theorem.