Let $\varphi (x) = e^{-\pi x^2}$ and let $F:=\tilde {\varphi}(\xi)$, the Fourier transform of $\varphi $. I must show that $F$ is a solution of these equations: $$F'(\xi)=-2\pi \xi F(\xi)$$ $$F (0)=1.$$ Now, $F(\xi)=\int e^{-\pi x^2-2\pi i x \xi}dx$, so we have the following:$F'(\xi)=\tilde {(-2\pi i x \varphi(x))}=\int e^{-\pi x^2-2\pi i x \xi}(-2\pi i x)dx$.
I am stuck here; probably I must do a proper substitution, however I tried a few and they didn't work. Can you give me a hint? Thanks
On
$$\varphi (x) = e^{-\pi x^2}$$ Differentiate: $$\varphi' (x) = e^{-\pi x^2}(-2\pi x)$$ $$\varphi' (x) = \varphi (x)(-2\pi x)$$ Fourier transform $$2\pi ik\hat {\varphi} (k) = -i\hat{\varphi}' (k)$$ $$ \hat{\varphi}' (k)=- {2\pi} k \hat {\varphi} (k) $$ This DE can easily be solved as it's separable: $$\frac { \hat{\varphi}(k)'}{\hat {\varphi} (k)}=-{2\pi} k $$
It's separable: $$(\ln \hat{\varphi} (k))'=-{2\pi} k$$ Integrate: $$ \hat{\varphi} (k)=Ce^{- {k^2}{\pi}} $$ $$C=\hat{\varphi} (0)=\int_{-\infty}^\infty e^{-\pi x^2} dx=I$$ Substitute $u=\sqrt {\pi}x \implies du=\sqrt {\pi}dx$: $$\implies I=\frac 1 {\sqrt {\pi}} \int_{-\infty}^\infty e^{-u^2} du=1$$ $$ \hat{\varphi} (k)=e^{- {k^2}{\pi}} \text{ , and :} \hat{\varphi} (0)=1 $$
Note that, in general, your function $\varphi$ solves $$\left(\frac{d}{dx}+2\pi x\right)\varphi=0.$$ If you take the Fourier transform of this equation and use how the Fourier transform intertwines differentiation and multiplication (although it's extra easy for this particular function), you get $$i\left(2\pi \xi+2\pi \frac{1}{2\pi}\frac{d}{d\xi}\right)\hat{\varphi}=0.$$ So, $$\hat{\varphi}(\xi)=\hat{\varphi}(0) e^{-\pi \xi^2},$$ and $$\hat{\varphi}(0)=\int\limits_{-\infty}^\infty e^{-\pi x^2}e^{-2\pi i x\cdot 0}\, dx=1,$$ which is a standard computation. So, $\hat{\varphi}(\xi)=e^{-\pi \xi^2}.$