Quick question on Fourier Transform of $\exp(-\frac{x^2}{2})$ I am currently looking at an example of how to calculate the Fourier Transform for the function
\begin{equation}
f(x) = \exp\left({-\frac{x^2}{2}}\right)
\end{equation}
Now $f$ solves the differential equation
\begin{equation}
f'(x) = -xf(x)
\end{equation}
and so, applying the FT to both sides gives
\begin{equation}
i \xi\,\hat{f}(\xi) = -i(\hat{f})'(\xi)
\end{equation}
Using these two equations we can derive that
\begin{equation}
\left(\frac{\hat{f}(x)}{f(x)}\right)' = 0
\end{equation}
hence 
\begin{equation}
\hat{f}(x) = c_0f(x) = c_0\exp\left({-\frac{x^2}{2}}\right)
\end{equation}
And here is where I have trouble, because the author of the notes that I am using says that $c_0 \geq 0$, which I can't see why .. I am sure it must be something obvious and I am just too blind to see it. Any hint would be highly appreciated, many thanks !
Edit: It is not the claim that I don't understand (I know it is right), it is just that I can't see why this is true looking solely at the derivations I have so far. 
 A: $$c_0=\hat{f}(0)=\int_{-\infty}^{+\infty}f(x)dx=\sqrt{2\pi}.$$
A: You might want to have another proof. Well, it does not hurt.
We can define, for any complex $\xi$,
$$\hat{f} (\xi) = \int_{-\infty}^{+\infty} e^{-i x \xi} e^{-\frac{x^2}{2}} dx.$$
This is well-defined, and you can check that $\hat{f}$ is analytic. There are some general results that you can apply to prove it; I don't remember them precisely, but with a gaussian, everything works well. Now, if we compute $\hat{f}$ for imaginary numbers, we have:
$$\hat{f} (iy) = \int_{-\infty}^{+\infty} e^{xy} e^{-\frac{x^2}{2}} dx = \int_{-\infty}^{+\infty} e^{\frac{y^2}{2}} e^{-\frac{(x-y)^2}{2}} dx = e^{\frac{y^2}{2}} \int_{-\infty}^{+\infty} e^{-\frac{x^2}{2}} dx = \sqrt{2\pi} e^{-\frac{(iy)^2}{2}}.$$
Hence, $\hat{f} (\xi) = \sqrt{2\pi} e^{-\frac{\xi^2}{2}}$ for imaginary $\xi$. By analytic continuation, this identity holds for any $\xi$.
A: Note that inverse Fourier transform is defined as $$\hat{f}(\zeta) =\int_{-\infty}^{\infty}f(x)e^{-j\zeta x}dx$$
From your solution we can see that $$c_0 = \hat{f}(0) = \int_{-\infty}^{\infty}f(x)dx$$ which is nothing but the area of the Gaussian which is always positive...
Thanks...
