I'm trying to derive the normal squared characteristic function, there's already a question on this but the answer has a part which is "proved as an excercise" which I try to do here. Is my proof correct? Also I tried to type the integral into mathematica but it only requires the condition $\text{Re} \frac{1}{r^2} > 0$ instead of $\text{Re} \frac{1}{r^2} > 0, \text{Re} \frac{1}{r} > 0$, is there any way to prove the former? Thanks!
Suppose $Z \sim \mathcal{N}(0, 1)$ and $X = Z^2$.
$\varphi (t) =\mathbf{E} [e^{i t X}] = \int_{- \infty}^{\infty}\frac{1}{\sqrt{2 \pi}} e^{- \frac{x^2}{2} + i t x^2} \text{d} x = r \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi} r} e^{- \frac{x^2}{2 r^2}} \text{d} x = r I$, where $r = 1 / \sqrt{1 - 2 i t}$. We now prove that if $\text{Re}\frac{1}{r^2} > 0, \text{Re} \frac{1}{r} > 0$ then the integral $I = 1$. Define
$$ \begin{array}{ll} I_L & = \int_{- L}^L \frac{1}{\sqrt{2 \pi} r} e^{- \frac{x^2}{2 r^2}} \text{d} x = \int_{- L / r}^{L / r} \frac{1}{\sqrt{2 \pi}} e^{- \frac{1}{2} z^2} \text{d} z\\ & = \left( \int_{- L / r}^{L / r} \frac{1}{\sqrt{2 \pi}} e^{- \frac{1}{2} z^2} \text{d} z - \int_{- L}^L \frac{1}{\sqrt{2 \pi}} e^{- \frac{1}{2} z^2} \text{d} z \right) + \int_{- L}^L \frac{1}{\sqrt{2 \pi}} e^{- \frac{1}{2} z^2} \text{d} z\\ & = \left( \int_{- L / r}^{L / r} \frac{1}{\sqrt{2 \pi}} e^{- \frac{1}{2} z^2} \text{d} z + \int_L^{- L} \frac{1}{\sqrt{2 \pi}} e^{- \frac{1}{2} z^2} \text{d} z \right) + \mathcal{I}_L \end{array} $$
Consider the contour $\gamma = (- L / r, L / r, L, - L)$. Since the integrand is holomorphic the integral is zero. Now
$$ \begin{array}{ll} & \int_{L / r}^L \frac{1}{\sqrt{2 \pi}} e^{- \frac{1}{2} z^2} \text{d} z + \int_{- L}^{- L / r} \frac{1}{\sqrt{2 \pi}} e^{- \frac{1}{2} z^2} \text{d} z\\ = & \int_0^1 \frac{L - L / r}{\sqrt{2 \pi}} e^{- \frac{1}{2} (L / r + (L - L / r) x)^2} \text{d} x + \int_0^1 \frac{L - L / r}{\sqrt{2 \pi}} e^{- \frac{1}{2} (- L + (L - L / r) x)^2} \text{d} x \end{array} $$
$\left| e^{- \frac{1}{2} (L / r + (L - L / r) x)^2} \right| = e^{\text{Re} \left( - \frac{1}{2} (L / r + (L - L / r) x)^2 \right)}$. Take $s = 1 / r$, then $$y := \text{Re} ((s + (1 - s) x)^2) = \text{Re} ((1 - s)^2 x^2 + 2 s (1 - s) x + s^2) = \text{Re} (s^2) (x^2 - 2 x + 1) + 2 \text{Re} (s) (- x^2 + x) + x^2$$ For $x = 0, 1$, $y > 0$. For $0 < x < 1$, all summands $> 0$ so $y > 0$. Since $y$ is continuous on a closed bounded interval, $\inf_x y$ is attained, so $y > \epsilon$ for some $\epsilon > 0$, and the integral is zero as $L \rightarrow \infty$.
$\left| e^{- \frac{1}{2} (- L + (- L / r + L) x)^2} \right| = e^{\text{Re} \left( - \frac{1}{2} (- L + (L - L / r) x)^2 \right)}$. Take $s = 1 / r$, then $$y := \text{Re} ((- 1 + (1 - s) x)^2) = \text{Re} ((1 - s)^2 x^2 - 2 (1 - s) x + 1) = \text{Re} (s^2) x^2 + 2 \text{Re} (s) (- x^2 + x) + (x^2 - 2 x + 1)$$ Similarly, $y > \epsilon$ for some $\epsilon > 0$, and the integral is zero as $L \rightarrow \infty$.
Hence $\lim_{L \rightarrow \infty} I_L = \mathcal{I}_L = 1$.
