Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is related to a separate post: Question about computing a Fourier transform of an integral transform related to fractional Brownian motion, but because the question is so basic, I thought I would give it some different tags.

How do you show the Fourier transform with repsect to $x$ of $ 2 \int_{0}^{\infty} \frac{e^{\frac{-x^2}{2y}}}{\sqrt{2 \pi y}} e^{-2y}\mathrm dy$ is $\frac{4}{k^2+4}$ ?

share|cite|improve this question
up vote 6 down vote accepted

Since the Fourier Transform of $e^{-\pi x^2}$ is $e^{-\pi x^2}$, we get by substituting $x\to x\sqrt{2\pi y}$ $$ \begin{align} \int_{-\infty}^\infty e^{\frac{-x^2}{2y}}e^{-2\pi ixt}dx&=\int_{-\infty}^\infty e^{-\pi x^2}e^{-2\pi ixt\sqrt{2\pi y}}\sqrt{2\pi y}\;dx\\ &=\sqrt{2\pi y}\;e^{-2\pi^2 y t^2} \end{align} $$ By linearity, the Fourier Transform of $2\int_{0}^{\infty} \frac{e^{\frac{-x^2}{2y}}}{\sqrt{2 \pi y}} e^{-2y} dy$ is $$ \begin{align} 2\int_{0}^{\infty} \frac{\sqrt{2\pi y}\;e^{-2\pi^2 y t^2}}{\sqrt{2 \pi y}} e^{-2y} dy&=2\int_{0}^{\infty} e^{-2\pi^2 y t^2} e^{-2y} dy\\ &=\int_{0}^{\infty} e^{-(1+\pi^2t^2)y} dy\\ &=\frac{1}{1+\pi^2t^2} \end{align} $$ Since this is not the form you are looking for, you are probably using the Fourier Transform using $e^{-ixt}$ instead of $e^{-2\pi ixt}$. Substituting $t\to\frac{t}{2\pi}$, we get $$ \int_{-\infty}^\infty e^{\frac{-x^2}{2y}}e^{-ixt}dx=\sqrt{2\pi y}\;e^{-y/2\;t^2} $$ So that by linearity, the Fourier Transform of $2\int_{0}^{\infty} \frac{e^{\frac{-x^2}{2y}}}{\sqrt{2 \pi y}} e^{-2y} dy$ is $$ \begin{align} 2\int_{0}^{\infty} \frac{\sqrt{2\pi y}\;e^{-y/2\;t^2}}{\sqrt{2 \pi y}} e^{-2y} dy&=2\int_{0}^{\infty} e^{-y/2\;t^2} e^{-2y} dy\\ &=\int_{0}^{\infty} e^{-(1+t^2/4)y} dy\\ &=\frac{4}{4+t^2} \end{align} $$

share|cite|improve this answer


Let $$ f(x) = 2 \int_0^{\infty} \frac{e^{-\frac{x^2}{2y}}}{\sqrt{2 \pi y}} e^{-2y} dy. $$

Assuming you define $$ \widehat{f}(\xi) = \int_{\mathbb{R}^d}^{} f(x) e^{- 2 \pi i x \cdot \xi} dx, $$ then: $$\begin{align} \widehat{f}(\xi) &= \int_{-\infty}^{\infty} f(x) e^{- 2 \pi i x \cdot \xi} dx \\ &= 2 \int_{-\infty}^{\infty} \int_0^{\infty} \frac{1}{\sqrt{2 \pi y}} e^{-\frac{1}{2y}(4y \pi i x\xi + x^2 +1)} dy dx \\ &=2 \int_0^{\infty} \frac{1}{\sqrt{2 \pi y}}e^{- \frac{1}{2y}(1 + 4 \pi^2 \xi^2 y^2)} \int_{- \infty}^{\infty} e^{- \frac{1}{2y}(x + 2 \pi i\xi y)^2} dx dy. \end{align}$$

The inner integral you can evaluate by a (complex) change of variables. The rest should come out (though admittedly, I haven't check this...)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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