Fourier transform of $F(x)=\exp(-x^2/(2 \sigma^2))$

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...?

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.
• 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 $$\frac{1}{\sqrt{2\pi}}$$ Aug 26 '14 at 9:56