I am looking for the fourier transform of $$F(x)=\exp\left(\frac{-x^2}{2a^2}\right)$$ where over $$-\infty<x<+\infty$$

I tried by definition $$f(u)={\int_{-\infty}^{+\infty} {\exp(-iux)\exp(\frac{-x^2}{2\sigma^2})}}dx$$ $$={\int_{-\infty}^{+\infty} {\exp(\frac{-x^2}{2\sigma^2})}[{\cos(ux)-i \sin(ux)}]}dx$$ $$={\int_{-\infty}^{+\infty} {\exp(\frac{-x^2}{2\sigma^2})}\cos(ux)}dx - i{\int_{-\infty}^{+\infty} {\exp(\frac{-x^2}{2\sigma^2})}\sin(ux)}dx$$ But we know

$$ \exp(\frac{-x^2}{2\sigma^2})\sin(ux)$$ is odd function and its integral over R is zero $${\int_{-\infty}^{+\infty} {\exp(\frac{-x^2}{2\sigma^2})}\sin(ux)}dx = 0$$ so that we get $$f(u)={\int_{-\infty}^{+\infty} {\exp(\frac{-x^2}{2\sigma^2})}\cos(ux)}dx = 2 {\int_{0}^{+\infty} {\exp(\frac{-x^2}{2\sigma^2})}\cos(ux)}dx = \sqrt{2\pi}\sigma \exp{(-\frac{1}{2}\sigma^2 u^2)}$$

BUT the problem is ..

when I calculate this transform by using wolframalpha... the result is only $$\sigma \exp{(-\frac{1}{2}\sigma^2 u^2)}$$

the result does not contain the part $$\sqrt{2\pi}$$

That's... where is the mistake or difference...?


1 Answer 1


The difference lies in the definition of the Fourier transform. Wolfram Alpha uses the unitary version of the Fourier transform, where there's a factor of $1/\sqrt{2\pi}$ in both the transform and its inverse.

  • $\begingroup$ does that mean, my solution is also acceptable if I use the def. of Fourier transform without constant preceding..?...and by the way when should we add constant such as $$\frac{1}{\sqrt{2\pi}}$$ $\endgroup$
    – leave2014
    Aug 26, 2014 at 9:56
  • $\begingroup$ @leave2014: Yes, your solution is acceptable if you just define which version of the Fourier transform you use. The differences lie only in the choice of constants and this choice is pure convention. Look at this table to see the relationships between the different definitions. $\endgroup$
    – Matt L.
    Aug 26, 2014 at 10:08
  • $\begingroup$ thank you, that's more clear. I think I'v confused "the Fourier Transform" with Fourier transform... $\endgroup$
    – leave2014
    Aug 26, 2014 at 10:15

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