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I have to calculate the Fourier transform of $G_s(x) = \dfrac{1}{\sqrt{s}} e^{-\dfrac{\pi x^2}{s}}$; $s > 0$.

I have proven that $G_s$ is a mollifier when $s \rightarrow 0^+$.

Have you an idea to calculate this transform ?

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3 Answers 3

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{\infty}{\expo{-\pi x^{2}/s} \over \root{s}}\,\expo{\ic ks}\,\dd s & = \int_{0}^{\infty}\exp\pars{-\,{\pi x^{2} \over s} + \ic ks} {\dd s \over \root{s}} \end{align}

Set $\ds{s = A\expo{\theta}}$. I'll choose $\ds{A}$ in a 'convenient way' later on. Namely,

\begin{align} \int_{0}^{\infty}{\expo{-\pi x^{2}/s} \over \root{s}}\,\expo{\ic ks}\,\dd s & = \int_{0}^{\infty}\exp\pars{-\pi x^{2}A^{-1}\expo{-\theta} - \bracks{-\ic k} A\expo{\theta}} {A\expo{\theta}\dd\theta \over \root{A\expo{\theta}}} \end{align}

I'll choose $\ds{A}$ with $\ds{\pi x^{2}A^{-1} = -\ic kA \implies A = \root{\pi x^{2} \over -\ic k} = \root{\pi \over k}\verts{x}\expo{\ic\pi/4}}$. Then,

\begin{align} &\int_{0}^{\infty}{\expo{-\pi x^{2}/s} \over \root{s}}\,\expo{\ic ks}\,\dd s \\[5mm] = &\ \int_{-\infty A}^{\infty A} \exp\pars{-\ic k \bracks{\root{\pi \over k}\verts{x}\expo{\ic\pi/4}}\bracks{2\cosh\pars{\theta}}} \pars{\pi \over k}^{1/4}\root{\verts{x}}\expo{\ic\pi/8}\expo{\theta/2} \,\dd\theta \\[5mm] = &\ \pars{\pi \over k}^{1/4}\root{\verts{x}}\expo{\ic\pi/8} \int_{-\infty A}^{\infty A}\exp\pars{-2\ic \root{\pi k}\verts{x}\expo{\ic\pi/4}\cosh\pars{\theta}}\expo{\theta/2} \,\dd\theta \\[5mm] = &\ \pars{\pi \over k}^{1/4}\root{\verts{x}}\expo{\ic\pi/8} \int_{0}^{\infty A}\exp\pars{-2\ic\root{\pi k}\verts{x}\expo{\ic\pi/4} \bracks{2\sinh^{2}\pars{\theta \over 2} + 1}} \cosh\pars{\theta \over 2}\,\dd\theta \end{align}

With $\ds{t \equiv \sinh\pars{\theta/2}}$:

\begin{align} &\int_{0}^{\infty}{\expo{-\pi x^{2}/s} \over \root{s}}\,\expo{\ic ks}\,\dd s \\[5mm] = &\ \pars{\pi \over k}^{1/4}\root{\verts{x}}\expo{\ic\pi/8} \bracks{\exp\pars{-2\ic\root{\pi k}\verts{x}\expo{\ic\pi/4}}}\ \times \\[2mm] &\ \lim_{R \to \infty}\int_{0}^{\sinh\pars{RA/2}} \exp\pars{-4\ic\root{\pi k}\verts{x}\expo{\ic\pi/4}t^{2}}\,2\,\dd t \end{align}

I guess you can continue from the last expression.

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HINT:

Enforce the substitution $s\mapsto s^2$ to find

$$I=\int_0^\infty \frac{e^{-\pi x^2/s}}{\sqrt s}e^{iks}\,ds=2\int_0^\infty e^{-(\pi x^2/s^2 -iks^2)}\,ds$$

Find constants $a$ and $b$ such that

$$I =\int_{-\infty}^\infty e^{-a\left(\left(\frac{b}{s}\right)^2+\left( \frac{s}{b}\right)^2 \right)}\,ds$$

Substitute $s/b\mapsto s$ and deform the transformed integration path back onto the real line.

Note that $\left(\frac 1s\right)^2+s^2=\left(s-\frac1s\right)^2+2$.

Finally, exploit the Cauchy-Schlomilch transform, which is a special case of Glasser's Master Theorem.

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\begin{align} \widehat{G_s}(y)&= \frac{1}{\sqrt{2\pi s}} \int_{-\infty}^{\infty}e^{-\pi x^2/s}e^{-iyx}dx \\ \widehat{G_s}'(y)&=\frac{1}{\sqrt{2\pi s}}\int_{-\infty}^{\infty}-ixe^{-\pi x^2/s}e^{-iyx}dx \\ & = \frac{1}{\sqrt{2\pi s}}\int_{-\infty}^{\infty}\frac{is}{2\pi}\frac{d}{dx}(e^{-\pi x^2/s})e^{-iyx}dx \\ & = -\frac{1}{\sqrt{2\pi s}}\frac{is}{2\pi}\int_{-\infty}^{\infty}e^{-\pi x^2/2}\frac{d}{dx}e^{-iyx}dx \\ & = -\frac{1}{\sqrt{2\pi s}}\frac{sy}{2\pi}\int_{-\infty}^{\infty}e^{-\pi x^2/2}e^{-iyx}dx \\ & = -\frac{sy}{2\pi}\widehat{G_s}(y). \end{align} This is a first order ODE in $s$ with solution $$ \widehat{G_s}(y)=A(s)e^{-sy^2/4\pi} $$ $A(s)$ is determined by setting $y=0$: $$ \widehat{G_s}(0)=A(s) \\ A(s) = \frac{1}{\sqrt{2\pi s}}\int_{-\infty}^{\infty}e^{-\pi x^2/s}dx $$ You can compute $A(s)$; I believe it is constant. This is a standard normal distribution (Guassian) integral.

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