1
$\begingroup$

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 $\int_\mathbb{R}e^{(-a+bi)x^2}dx$?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Mattos Feb 13 '15 at 4:06
1
$\begingroup$

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.

$\endgroup$

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.