The Question: Prove that if $f(x)=e^{-\pi x^2}$ then $\hat{f}(x)=f(x)$
My Proof: To prove that $\hat{f}(x)=f(x)$, we need to show that the Fourier transform of $f(x)$ is equal to $f(x)$ itself, that is, $\hat{f}(x)=\int_{-\infty}^{\infty} f(t) e^{-2 \pi i x t} d t=f(x)$. Substituting $f(x)=e^{-\pi x^2}$ into the above formula, we obtain $\hat{f}(x)=\int_{-\infty}^{\infty} e^{-\pi t^2} e^{-2 \pi i x t} d t$. Using the definition of the Fourier transform, we have $\hat{f}(x)=\int_{-\infty}^{\infty} e^{-\pi t^2} e^{-2 \pi i x t} d t=\int_{-\infty}^{\infty} e^{-\pi(t+i x)^2} e^{-\pi x^2} d t$. Completing the square in the exponent of the integrand, we have $\hat{f}(x)=e^{-\pi x^2} \int_{-\infty}^{\infty} e^{-\pi(t+i x)^2-\pi x^2+\pi x^2} d t$. The exponent in the integral can be rewritten as \begin{align*} -\pi(t+ix)^2 - \pi x^2 + \pi x^2 &= -\pi (t^2+2ixt-x^2) \\ &= -\pi (t+ix)^2 + \pi x^2 \end{align*}
Substituting this expression back into the integral, we obtain $\hat{f}(x)=e^{-\pi x^2} \int_{-\infty}^{\infty} e^{-\pi(t+i x)^2} e^{\pi x^2} d t$. Since the integrand is the Gaussian function $e^{-\pi (t+ix)^2}$, the integral is equal to $\sqrt{\pi}$, and we have $\hat{f}(x)=e^{-\pi x^2} \int_{-\infty}^{\infty} e^{-\pi(t+i x)^2} e^{\pi x^2} d t=e^{-\pi x^2} \sqrt{\pi}=f(x)$. Therefore, we have shown that $\hat{f}(x) = f(x)$, as required.
Is it correct or any suggestions?
The solution text needs some cleanup, but $\int_{-\infty}^\infty e^{-\pi(t+ix)^2}\,dt=\color{red}{1}$ is the major step here. It requires some justification (unless this "Gaussian integral" is considered "known").
The fact that $\int_{-\infty}^\infty e^{-\pi(t+ix)^2}\,dt=\int_{-\infty}^\infty e^{-\pi z^2}\,dz$ is an application of Cauchy's integral theorem (to a rectangular contour enclosing $[-R,R]+i[0,x]$, followed by taking $R\to\infty$).