1. ## [SOLVED] Fourier transform

So I'm just sitting in the chair being bored and I decided to review Fourier transforms. I pick what I thought would be an easy function to transform: $e^{-x^2}$, so I have
$g(k) = \int_{-\infty}^{\infty}e^{-x^2} e^{-ikx}~dx$

$= e^{-k^2/4}\int_{-\infty}^{\infty}e^{-(x - ik/2)^2}~dx$

I have (somewhere) in my references how to do these, but I can't find it. (I really need to organize my references!) My initial impluse is to do a complex integration. Let $z = x + ik/2$. Then
$g(k) = \int_{-\infty + ik/2}^{\infty + ik/2}e^{-z^2}~dz$
which is a line integral on the whole line y = ik/2. I can't quite figure out a good contour for this.

Thanks!
-Dan

2. I think you made a typo there, $e^{-k^2/4}\int_{-\infty}^{\infty}e^{-(x - ik/2)^2}~dx$ makes $\int_{-\infty}^{\infty}e^{-x^2} e^{\color{red}+\color{black}ikx}~dx$

I used,

$e^{-k^2/4}\int_{-\infty}^{\infty}e^{-(x - ik/2)^2}~dx$

$e^{-k^2/4}\int_{-\infty}^{\infty}e^{-(x + ik/2)^2}~dx$

$e^{-k^2/4}\int_{-\infty+ ik/2}^{\infty+ ik/2}e^{-u^2}~du$

I don't think those ik/2's will mean anything even when summed by infinity. So I get

$e^{-k^2/4}\int_{-\infty}^{\infty}e^{-u^2}~du$

$e^{-k^2/4}\sqrt{\pi}$

I'm not sure about this result. I'll check it using contour integration as soon as I have time, (if someone doesn't find my error, lol)

3. Originally Posted by topsquark
So I'm just sitting in the chair being bored and I decided to review Fourier transforms. I pick what I thought would be an easy function to transform: $e^{-x^2}$, so I have
$g(k) = \int_{-\infty}^{\infty}e^{-x^2} e^{-ikx}~dx$
There's a good trick for doing this. Let $g(w) = \int_{-\infty}^{\infty}e^{-x^2} e^{-iwx}\,dx$. Differentiate under the integral sign to get $g'(w) = \int_{-\infty}^{\infty}-ixe^{-x^2} e^{-iwx}\,dx = \int_{-\infty}^{\infty}{\textstyle \frac{d}{dx}(\frac12e^{-x^2})} e^{-iwx}\,dx$. Now integrate by parts and you find that $g'(w) = -\textstyle\frac12wg(w)$, from which $g(w) = g(0)e^{-w^2/4}.$ But $g(0) = \int_{-\infty}^{\infty}e^{-x^2}\,dx = \sqrt\pi.$

4. Originally Posted by Opalg
There's a good trick for doing this. Let $g(w) = \int_{-\infty}^{\infty}e^{-x^2} e^{-iwx}\,dx$. Differentiate under the integral sign to get $g'(w) = \int_{-\infty}^{\infty}-ixe^{-x^2} e^{-iwx}\,dx = \int_{-\infty}^{\infty}{\textstyle \frac{d}{dx}(\frac12e^{-x^2})} e^{-iwx}\,dx$. Now integrate by parts and you find that $g'(w) = -\textstyle\frac12wg(w)$, from which $g(w) = g(0)e^{-w^2/4}.$ But $g(0) = \int_{-\infty}^{\infty}e^{-x^2}\,dx = \sqrt\pi.$
Oooh! I like that one.

-Dan

5. Originally Posted by Opalg
There's a good trick for doing this. Let $g(w) = \int_{-\infty}^{\infty}e^{-x^2} e^{-iwx}\,dx$. Differentiate under the integral sign to get $g'(w) = \int_{-\infty}^{\infty}-ixe^{-x^2} e^{-iwx}\,dx = {\color{red}i} \, \int_{-\infty}^{\infty}{\textstyle \frac{d}{dx}(\frac12e^{-x^2})} e^{-iwx}\,dx$. Now integrate by parts and you find that $g'(w) = -\textstyle\frac12wg(w)$, from which $g(w) = g(0)e^{-w^2/4}.$ But $g(0) = \int_{-\infty}^{\infty}e^{-x^2}\,dx = \sqrt\pi.$
Nice solution. i (ha ha) think you'll find a small typo (in red). The final result is unaffected.