I'm working on a few problems in Reed & Simon, and I ran across this problem.
Compute the Fourier transform of $f(x) = e^{-\alpha x^2/2}$ via the following steps. (a) Prove that $-\lambda \hat{f}(\lambda)= \alpha \frac{d}{d\lambda}\hat{f}(\lambda)$ and conclude that $\hat{f}(\lambda) = ce^{-\lambda^2/(2\alpha)}$.
This is what I have so far (with help from first comment):
\begin{align*} -\frac{\alpha}{\lambda} \frac{d}{d\lambda} \hat{f}(\lambda) &= -\frac{\alpha}{\lambda \sqrt{2\pi}} \int \frac{d}{d\lambda} e^{-i\lambda x}e^{-\alpha x^2/2} dx \\ &= \frac{\alpha i}{\lambda\sqrt{2\pi}} \int x e^{-\alpha x^2/2} e^{\frac{2i\lambda}{\alpha}} dx \\ &= \frac{\alpha i}{\lambda\sqrt{2\pi}} \Big[-\frac{i\lambda}{\alpha} e^{-\frac{\lambda^2}{2\alpha}}\sqrt{\frac{2}{\alpha\pi}}\Big] \\ &=\frac{1}{\sqrt{\alpha}\pi}e^{-\frac{\lambda^2}{2\alpha}} \end{align*}
I'm not entirely sure where to go from here. I am going to keep thinking about it, but any hints would be appreciated.
Edits: I fixed the computations, all I have left to figure out on my own is the second portion of the problem, which probably isn't too bad using $u$-substitution and the fact that $e^{-x^2/2}$ is its own Fourier transform. The accepted answer provides a different and interesting way of looking at this problem as well.
Theorem: The characteristic function of a standard normal random variable $X$ is $ \varphi_{X}(t)=e^{-t^{2}/2} $.
Proof. Note that \begin{align*} \varphi_{X}(t)=\mathbb{E}\left[e^{itX}\right] & =\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}e^{-\frac{x^{2}}{2}}e^{itx}dx\\ & =\frac{1}{\sqrt{2\pi}}\left(\int_{-\infty}^{0}e^{-\frac{x^{2}}{2}}e^{itx}dx+\int_{0}^{\infty}e^{-\frac{x^{2}}{2}}e^{itx}dx\right)\\ & =\frac{1}{\sqrt{2\pi}}\left(\int_{0}^{\infty}e^{-\frac{x^{2}}{2}}e^{-itx}dx+\int_{0}^{\infty}e^{-\frac{x^{2}}{2}}e^{itx}dx\right)\\ & =\frac{2}{\sqrt{2\pi}}\int_{0}^{\infty}e^{-\frac{x^{2}}{2}}\cos(tx)dx. \end{align*} Now, take the derivative with respect to $t$ to get $$ \varphi_{X}^{\prime}(t)=-\frac{2}{\sqrt{2\pi}}\int_{0}^{\infty}e^{-\frac{x^{2}}{2}}x\sin(tx)dx. $$ Integrate by parts to get \begin{align*} \varphi_{X}^{\prime}(t) & =\frac{2}{\sqrt{2\pi}}\left(e^{-\frac{x^{2}}{2}}\sin(tx)\mid_{0}^{\infty}-t\int_{0}^{\infty}e^{-\frac{x^{2}}{2}}\cos(tx)dx\right)\\ & =-\frac{2t}{\sqrt{2\pi}}\int_{0}^{\infty}e^{-\frac{x^{2}}{2}}\cos(tx)dx\\ & =-t\varphi_{X}(t). \end{align*} Note that $$ \varphi_{X}^{\prime}(t)=-t\varphi_{X}(t) $$ is an ODE with solution $$ \varphi_{X}(t)=ce^{-t^{2}/2}. $$ We can figure out the value of the constant $c$ by recalling that the characteristic function has to satisfy $\varphi_{X}(0)=1$ so that $c=1$, as desired.