If $X \sim N(0,1)$, then $\varphi_{X}(t) = e^{-t^{2}/2}$.
From the proof of the previous proposition, I have the following question: why is it enough to prove that $\max_{z\in I_2}|e^{-z^{2}/2}| \to 0 $ when $ n \to \infty$ and why is $0\leq s \leq t$ chosen? Also, at the end of the proof, when they take the maximum, I don't understand why $s$ is replaced by $t$. I would appreciate some justification on this problem.
Proof.
\begin{align} \varphi_{X}(t) &= \int{e^{itx}}dF_{X}(x)\\ &=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{e^{itx-x^{2}/2}}dx \\ &=e^{-t^{2}/2}\cdot \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-(x-it)^{2}/2}dx \hspace{0.4cm} \text{(completing squares)}\\ &=e^{-t^{2}/2} \end{align} where the last equation is verified as follows:
\begin{align} \int_{-\infty}^{\infty}{e^{-(x-it)^{2}/2}}dx &= \lim_{n \to \infty}{\int_{-n}^{n}{e^{-(x-it)^{2}/2}}}dx\\ &=\lim_{n \to \infty}\int_{-n-it}^{n-it}{e^{-z^{2}/2}}dz \hspace{0.4cm} (\text{where} \hspace{0.3cm}z = x-it) \end{align} where the integration interval in this integral (for $n$ fixed and $t>0$) is $I_1$:
Since the function $f(z)= e^{-z^{2}/2}$ is analytic in the complex plane $\mathbb{C}$, Cauchy's theorem says that the integral of $f$ over any closed curve is zero. Then for $t>0$,
$$\int_{I_1}{e^{-z^{2}/2}}dz+\int_{I_2}{e^{-z^{2}/2}}dz+\int_{I_3}{e^{-z^{2}/2}}dz+\int_{I_4}{e^{-z^{2}/2}}dz=0$$ i.e., $$\int_{-n-it}^{n-it}{e^{-z^{2}/2}}dz = -\int_{I_2}{e^{-z^{2}/2}}dz + \int_{-n}^{n}{e^{-t^{2}/2}}dt - \int_{I_4}{e^{-z^{2}/2}}dz$$
Since $$\int_{-n}^{n}{e^{-t^{2}/2}}dt \to \int_{-\infty}^{\infty}{e^{-t^{2}/2}}dt = \sqrt{2\pi} \hspace{0.4cm} ( \text{when} \hspace{0.2cm} n \to \infty)$$ it is enough to prove that $$\int_{I_2}{e^{-z^{2}/2}}dz \hspace{0.4cm} \text{and} \hspace{0.4cm} \int_{I_4}{e^{-z^{2}/2}}dz$$ tend to zero when $n \to \infty$ (thus it will be proven that $\varphi_{X} (t) = e^{-t^{2}/2}$, $t>0$. For $t<0$, the method is analogous, the only difference is that the intervals $I_2$ and $I_4$ invert the sense. Furthermore, it is enough to note that $X$ symmetric $\implies \varphi_{X}(t) = \varphi_{X}(-t)$ . For $t=0$, the result is obvious).
Let us prove that $$\int_{I_2}e^{-z^{2}/2}dz \to 0 $$ the proof that $$\int_{I_4}e^{-z^{2}/2}dz \to 0 $$is analogous.
Since the length of the interval $I_2$ is $t$, it suffices to prove that $$\max_{z\in I_2}|e^{-z^{2}/2}| \to 0 \hspace{0.4cm} \text{when} \hspace{0.2cm} n \to \infty$$ For $z=n-is$, where $0\leq s \leq t$, we have $$z^2 = n^2 - s^2 -2nsi \hspace{0.5cm} \text{and} \hspace{0.5cm} e^{-z^{2}/2} = e^{(s^2 - n^2)/2}e^{nsi}$$ Since $|e^{it}| =1 \hspace{0.1cm} \forall t$, we have $|e^{-z^{2}/2}| = e^{(s^2 - n^2)/2}$. Therefore $$\max_{z \in I_2}{|e^{-z^{2}/2}|} = e^{(t^2 - n^2)/2} \to 0 \hspace{0.4cm} \text{when} \hspace{0.3cm} n \to \infty$$
$\square$
Definition
Let $(\Omega, \mathcal{F}, P)$ be a probability space and $X$ be a random variable. The characteristic function of $X$, which we denote by $\varphi_{X}$, is:
$$\begin{array}{rccc}\varphi_X \colon&\mathbb{R}&\longrightarrow&\mathbb C\\&t&\mapsto&\displaystyle\varphi_{X}(t)\colon=\mathbb{E}[e^{itX}] = \int_{\mathbb{R}}e^{itx}dF_{X}(x)\end{array}$$
$|\int_{\gamma} f(z)dz| \leq Ml$ where $L$ is the length of $\gamma$ and $M$ is the maximum value of $|f(z)|$ on $\gamma$. In this case $t$ is fixed and $L=t$ so to show that the integral over $I_2$ tends to $0$ it is enough to show that $M \to 0$.
Points on the line segment $I_21$ are of the type $n-is$ with $0 \leq s \leq t$. That is why they are taking $0 \leq s \leq t $ in estimating $M$.