Question

fourier transform question
first question please
as soon as possible.

Answer 1

@dan815

Answer 2

Find the fourier Transform of \[e^{-a^{2}x^{2}} ,a>0\]Given \[\int\limits_{-\infty}^{\infty} e^{-t} dt = \sqrt{\pi}\]

Answer 3

@SolomonZelman

Answer 4

I think you meant \(e^{-\color{red}{t^2}}\) in that last integral?

Answer 5

\[\large\begin{align*}\mathcal{F}\{e^{-a^2x^2}\}&=\frac{1}{2\pi}\int_{-\infty}^\infty e^{-a^2x^2}e^{-i\xi x}\,dx\\[2ex]
&=\frac{1}{2\pi}\int_{-\infty}^\infty e^{-a^2\left(x^2+\frac{i\xi}{a^2}x\right)}\,dx
\end{align*}\]
Complete the square:
\[\begin{align*}-a^2\left(x^2+\frac{i\xi}{a^2}x\right)&=-a^2\left(x^2+\frac{i\xi}{a^2}x+\left(\frac{i\xi}{2a^2}\right)^2-\left(\frac{i\xi}{2a^2}\right)^2\right)\\[2ex]
&=-a^2\left(\left(x+\frac{i\xi}{2a^2}\right)^2+\frac{\xi^2}{4a^4}\right)\\[2ex]
&=-\left(ax+\frac{i\xi}{2a}\right)^2-\frac{\xi^2}{4a^2}\end{align*}\]
A substitution will do the rest: \(y=ax+\dfrac{i\xi}{2a}\) with \(dy=a\,dx\).\[\large\begin{align*}\mathcal{F}\{e^{-a^2x^2}\}&=\frac{1}{2a\pi}\int_{-\infty}^\infty e^{-y^2-\frac{\xi^2}{4a^2}}\,dy
\end{align*}\]

Answer 6

yes it was \[e^{t^{-2}}\]
isn't it \[\frac{ 1 }{ \sqrt{2\pi} }\] instead of \[\frac{ 1 }{ {2\pi} }\]
otherwise thank you very much... it had been bugging me for days...

Answer 7

I've seen a few variations of the transform's definition, things like \(\dfrac{1}{2\pi}\int e^{-i\xi x}\,dx\) and \(\int e^{-2\pi i\xi x}\,dx\) (which are identical) as well as \(\dfrac{1}{\sqrt{2\pi}}\) in place of \(\dfrac{1}{2\pi}\). (see here: http://mathworld.wolfram.com/FourierTransform.html )

I'm not sure myself about the details of the advantage to using one definition over another, but I think it has something to do with symmetry. As long as you're consistent, either way is fine.

Answer 8

cool.....
thanks again