Let $\varphi: \mathbb{R}^n \rightarrow \mathbb{R}$ be defined by $$ \varphi(x)=e ^ { - \frac {|x|^2 } { 2 } }, $$ Find $\varphi * \varphi$ by using Fourier transforms.
My attempt: By Convolution theorem, we have $$ \begin{array}{l} F(\varphi * \psi)(\xi)=F(\varphi)(\xi) \cdot F(\psi)(\xi) \\ \Rightarrow \varphi * \psi(x)=F^{-1}(F(\varphi)(\xi) \cdot F(\psi)(\xi)) \\ \Rightarrow \varphi * \varphi(x)=\int_{\Bbb R^n} e^{2 \pi i x \cdot \xi} \cdot e^{-|\xi^2} d \xi \\ =\int_{\Bbb R^2} e^{\left(2\pi i ( x _ { 1 } \xi_ { 1 } + \cdots + x _ { n } \xi _ { n }) \ - (\xi _ { 1 } ^ { 2 } + \cdots +\xi_ { n } ^ { 2 } )\right)}d \xi \end{array} $$ Now how to proceed.
HINT
$\phi(x)=e^{-\frac{|x|^2}{2}}=e^{-\pi|\frac{x}{\sqrt{2\pi}}|^2}$
$\widehat{e^{-\pi|\frac{x}{\sqrt{2\pi}}|^2}}(\xi)=\sqrt{2\pi}e^{-\pi|\sqrt{2\pi}\xi|^2}=\sqrt{2\pi}e^{-2\pi^2|\xi|^2}$
So $\widehat{(\phi \ast \phi)}(\xi)=\hat{\phi}^2(\xi)=2\pi e^{-4\pi^2|\xi|^2}$
Now use the fact that $\widehat{e^{-\pi|x|^2}}(\xi)=e^{-\pi|\xi|^2}$ and the Fourier inversion and some appropriate change of variables to find the convolution.