## Anas.P one year ago fourier transform question first question please as soon as possible.

1. anas.p

@dan815

2. anas.p

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

3. anas.p

@SolomonZelman

4. anonymous

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

5. anonymous

\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*}

6. anas.p

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...

7. anonymous

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.

8. anas.p

cool..... thanks again