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The standard normal distribution $$f(x) = \frac{1}{\sqrt{2\pi}} e^{\frac{-x^2}{2}},$$ has the characteristic function $$\int_{-\infty}^\infty f(x) e^{itx} dx = e^{-\frac{t^2}{2}}$$ and this can be proved by obtaining the moments.

However, is there a more direct method of proving that the standard normal has the stated characteristic function? I got stuck on trying to show that

$$\int_{-\infty}^\infty e^{-\frac{1}{2}(x-it)^2} dx= \sqrt{2\pi}.$$

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    $\begingroup$ the standard way of proving your last equation is by Cauchy theorem (i.e. it is used to show that the integral doesn't depend on $t$; imagime integrating $e^{-x^2/2}$ over a long rectangle) $\endgroup$ – user8268 Mar 27 '11 at 16:03
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A simple change of variables allows you to compute $\mathbb{E}[e^{tX}]$ for real $t$ and standard normal $X$, $$ \begin{align} \mathbb{E}[e^{tX}]&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12x^2}e^{tx}\,dx\\ &= \frac{e^{\frac12t^2}}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(x-t)^2}\,dx\\ &=\frac{e^{\frac12t^2}}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12y^2}\,dy\\ &=e^{\frac12t^2}. \end{align} $$ Here, the substitution $y=x-t$ has been used. In fact, this identity holds for all complex $t$ as well by analytic continuation. The right hand side, $e^{\frac12t^2}$ is clearly analytic. The left hand side is analytic, as it has the derivative $\mathbb{E}[Xe^{tX}]$. The fact that you can commute differentiation and expectation follows from the dominated convergence theorem. Two analytic functions which agree on the real line must agree everywhere (by analytic continuation). So the identity holds for all complex $t$, and replacing $t$ by $it$ gives the expression you ask for.

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In Fristedt and Gray's A Modern Approach to Probability, the authors sketch a proof. Use integration by parts and dominated convergence to show that $\beta(t)=\int f(x) e^{itx} dx$ satisfies the differential equation $\beta^\prime(t)=-t \beta(t)$ with $\beta(0)=1$. You have to justify manipulating integrals of $\mathbb{C}$-valued functions.

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