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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 $\frac{4}{k^2+4}$ ?

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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 $$ \begin{align} 2\int_{0}^{\infty} \frac{\sqrt{2\pi y}\;e^{-y/2\;t^2}}{\sqrt{2 \pi y}} e^{-2y} dy&=2\int_{0}^{\infty} e^{-y/2\;t^2} e^{-2y} dy\\ &=\int_{0}^{\infty} e^{-(1+t^2/4)y} dy\\ &=\frac{4}{4+t^2} \end{align} $$

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HINT

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...)

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