Fourier transform of $f$ is $f$

Question

I want to prove that for $f(x)=e^{-\pi x^{2}}$ the Fourier transform of $f$ equals $f$ itselfs. I want to do it in three given steps (I got the three steps as a hint):

1. $h=\hat f$. Prove that $h$ is derivative and $h'(t)=-2\pi th(t)$
2. prove that the function $t\to h(t) e^{\pi t^{2}}$ is constant.
3. prove $h(0)=1$ and conclude that $\hat f=f$

Now for (1) I first tried to find $\hat f$. \begin{align} \hat f(t)&=\int_{\mathbb{R}}f(x) e^{-2\pi itx} \\ &= \int_{\mathbb{R}}e^{-\pi x^{2}}e^{-2\pi itx} \\ &=....=e^{-\pi t^{2}} \end{align} I couldn't solve the integral myself but I find on the internet that the equation should have this answer. Maybe someone can help me to find it?

Now I now that $h(t)=e^{-\pi t^{2}}$ I know that $h'(t)=-2\pi t e^{-\pi t^{2}}=-2\pi t h(t)$

For (2) I proved that the derivative of that function is zero. \begin{align} \frac{d(h(t)e^{\pi t^{2}})}{dt}&=\frac{d(h(t))}{dt}e^{\pi t^{2}}+\frac{d(e^{\pi t^{2}})}{dt}h(t)& \\ &= -2\pi t e^{-\pi t^{2}}e^{\pi t^{2}}+2\pi t e^{\pi t^{2}}e^{-\pi t^{2}} \\ &=0 \end{align}

For (3) $h(0)=e^{-2\pi 0^{2}}=1$. But I don't know how to conlcude that $\hat f=f$.

So in conclusion, I need some help to find the integral in (1) and to make the conclusion in (3).

EDIT: (2) you can also prove without knowing exactly what h(t) is, so if someone has a suggestion to prove (1) in an other way that calculating $\hat f$, then that would be helpfull too.

EDIT2: my question doesn't make sense if I would already find what equals to $h(t)$ so I would like to find another methode for option (1)

EDIT3: can I use for (1) that $h(t)=\hat f (t)= \int_{\mathbb{R}}f(x)e^{-2\pi itx}dx$. So $\frac{d(h(t))}{dt}=\frac{d(\int_{\mathbb{R}}f(x)e^{-2\pi itx}dx)}{dt}=\int_{\mathbb{R}}f(x)\frac{d(e^{-2\pi itx})}{dt}dx)=-2\pi t\int_{\mathbb{R}}f(x)e^{-2\pi itx}dx=-2\pi t h(t) $

Answer 1

As $t\to h(t) e^{\pi t^{2}}$ is constant, you have $h(t) = h(0) e^{-\pi t^{2}}$ (the constant value is $h(0)$).

And as $h(0) = 1$, $\hat f(t) =h(t) = e^{-\pi t^{2}} =f(t)$.

Answer 2

Since $(e^{\pi t^{2}} h(t))'=0$ it follows that $e^{\pi t^{2}} h(t)$ is a consatnt . Since it has the value $1$ when $t=0$ we get $e^{\pi t^{2}} h(t)=1$ for all $t$. So $\hat {f} (t)=h (t)=e^{-\pi t^{2}} $.

For (1) search the net for characteristic function of Gaussian distribution.