Proof of the formula for the characteristic function of the normal distribution

Proof

We compute the moments of the standard normal law and expand the cosine in its Taylor polynomials.

1. Let $\mu \in \R$, $\sigma^2 > 0$, and $\sigma = \sqrt{\sigma^2} > 0$. Let $X \sim \mathcal{N}(\mu, \sigma^2)$ be normal, and fix $t \in \R$.
2. Let $f = \varphi_{\mu, \sigma^2}$ and $\varphi = \varphi_{0,1}$ be the normal densities. By definition, $f$ is a probability density function of $X$, and $\int_\R \varphi = 1$.
3. By definition, the characteristic function is $\varphi_X(t) = E[\cos(tX)] + i \, E[\sin(tX)]$. Both expectations exist, since the variables are bounded.
4. By the law of the unconscious statistician, $E[g(X)] = \int_\R g(x) f(x) \, dx$ for each bounded continuous $g$. These integrals converge absolutely as improper integrals.
5. For all real $z$, $\sigma f(\mu + \sigma z) = \varphi(z)$. With substitution $x = \mu + \sigma z$ on bounded intervals, step 4 gives $E[g(X)] = \int_\R g(\mu + \sigma z) \varphi(z) \, dz$.

First, the integrals $\int_\R \abs{z}^k \varphi(z) \, dz$ converge.

6. Let $k \ge 0$ be an integer. By the exponential series, $e^{z^2/4} \ge (z^2/4)^k / k!$. So $\abs{z}^k e^{-z^2/4} \le k! \, 4^k$ for $\abs{z} \ge 1$, and it is at most $1$ for $\abs{z} < 1$.
7. Since $(\abs{z}/2 - 1)^2 \ge 0$, we have $z^2/4 \ge \abs{z} - 1$. So $e^{-z^2/4} \le e \cdot e^{-\abs{z}}$.
8. By steps 6 and 7, $\abs{z}^k \varphi(z) \le C_k e^{-\abs{z}}$ with $C_k = k! \, 4^k e$. By the comparison test and the integral of $e^{-x}$, $\int_\R \abs{z}^k \varphi(z) \, dz$ converges.
9. Step 8 with $k + 1$ in place of $k$ gives $\abs{z}^k \varphi(z) \le C_{k+1} e^{-\abs{z}}/\abs{z}$ for $z \ne 0$. So $\abs{z}^k \varphi(z) \to 0$ as $\abs{z} \to \infty$.

Next, the even moments.

10. For $k \ge 0$, let $m_{2k} = \int_\R z^{2k} \varphi(z) \, dz$. Then $m_0 = 1$ by step 2.
11. By the chain rule and the derivative of the exponential, $\varphi'(z) = -z \varphi(z)$.
12. Integration by parts with $u(z) = z^{2k+1}$ and $v(z) = -\varphi(z)$ gives, for $R > 0$, $$\int_{-R}^{R} z^{2k+2} \varphi(z) \, dz = -2R^{2k+1} \varphi(R) + (2k + 1) \int_{-R}^{R} z^{2k} \varphi(z) \, dz.$$
13. Let $R \to \infty$. By step 9, the boundary term tends to $0$. So $m_{2k+2} = (2k + 1) \, m_{2k}$.
14. By induction on $k$, $m_{2k} = (2k)!/(2^k k!)$. Indeed, $(2k+1) (2k)!/(2^k k!) = (2k+2)!/(2^{k+1} (k+1)!)$, since $(2k+2)/(2(k+1)) = 1$.

Now the standard normal case.

15. For real $s$, let $S(s) = \int_\R \sin(sz) \varphi(z) \, dz$ and $C(s) = \int_\R \cos(sz) \varphi(z) \, dz$. Both converge absolutely by step 8 with $k = 0$.
16. The function $\sin(sz) \varphi(z)$ is odd in $z$. Substituting $z = -u$ on $(-\infty, 0]$ shows that $S(s) = 0$.
17. Fix $N \ge 0$. By Taylor's theorem with the Lagrange remainder for $\cos$ at $0$, of degree $2N + 1$, we get for real $y$ $$\abs{\cos y - \sum_{k=0}^{N} \frac{(-1)^k y^{2k}}{(2k)!}} \le \frac{y^{2N+2}}{(2N+2)!}.$$ Here we used that all derivatives of $\cos$ are $\pm\sin$ or $\pm\cos$, so they are bounded by $1$.
18. Put $y = sz$, multiply by $\varphi(z) > 0$, and integrate. By monotonicity of the integral on bounded intervals and a limit, step 14 gives $$\abs{C(s) - \sum_{k=0}^{N} \frac{(-1)^k s^{2k} m_{2k}}{(2k)!}} \le \frac{s^{2N+2} m_{2N+2}}{(2N+2)!} = \frac{(s^2/2)^{N+1}}{(N+1)!}.$$
19. By step 14, the sum in step 18 equals $\sum_{k=0}^{N} (-s^2/2)^k / k!$. By the exponential series, it tends to $e^{-s^2/2}$ as $N \to \infty$.
20. The right side of step 18 is a term of the convergent series for $e^{s^2/2}$. So it tends to $0$ as $N \to \infty$.
21. By steps 18 to 20, $C(s) = e^{-s^2/2}$ for all real $s$.

Finally, the general case.

22. By step 5 and the angle addition formulas, $E[\cos(tX)] = \cos(\mu t) \, C(\sigma t) - \sin(\mu t) \, S(\sigma t)$.
23. In the same way, $E[\sin(tX)] = \sin(\mu t) \, C(\sigma t) + \cos(\mu t) \, S(\sigma t)$.
24. By steps 16 and 21, $E[\cos(tX)] = \cos(\mu t) \, e^{-\sigma^2 t^2/2}$ and $E[\sin(tX)] = \sin(\mu t) \, e^{-\sigma^2 t^2/2}$.
25. By step 3 and Euler's formula, $\varphi_X(t) = (\cos(\mu t) + i \sin(\mu t)) \, e^{-\sigma^2 t^2/2} = e^{i\mu t} e^{-\sigma^2 t^2/2}$.
26. By the product of exponentials, $e^{w} e^{w'} = e^{w + w'}$ for complex $w$ and $w'$. So $\varphi_X(t) = \exp\left(i\mu t - \tfrac{1}{2}\sigma^2 t^2\right)$.
27. For $\mu = 0$ and $\sigma^2 = 1$, this gives $\varphi(t) = e^{-t^2/2}$ for the standard normal distribution.