I have a physic paper to do and one of the question is to demonstrate this equality using Cauchy's theorem :
$$\int_{-\infty}^{\infty} e^{i\zeta^2 sgn \varphi''(k_0) } d\zeta = \sqrt{\Pi} e^{\frac{i\Pi}{4}*sng \varphi''(k_0)}$$
sng being the sign function
What I have used to try to solve this is this calculation :
With $x=\zeta$ and $a=-i*sng\varphi''(k_0)$, we have :
\begin{align*} I &= \int_{-\infty}^{\infty} e^{-ax^2 } dx \\ I^2 &= \int_{-\infty}^{\infty} e^{-ax^2 } dx \int_{-\infty}^{\infty} e^{-ay^2 } dy \\ &=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-ax^2 } e^{-ay^2 } dxdy \\ &=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-a(x^2+y^2) } dxdy \\ &=\int_{0}^{2\pi} \int_{0}^{\infty} e^{-ar^2} rdrd\theta \\ &=2\pi \int_{0}^{\infty} e^{-u} \frac{du}{2a} \\ I^2 &=\frac{\pi}{a} \end{align*} So $$I=\sqrt{\frac{\pi}{a}}$$
Therefore $$\int_{-\infty}^{\infty} e^{i\zeta^2 sgn \varphi''(k_0) } d\zeta = \sqrt{\frac{\pi}{-i*sgn \varphi''(k_0)}} = \sqrt{\pi * i * sgn \varphi''(k_0)}$$
The result I have is far from what I'm looking for. I also know that the fact that a is complex might make my calculations false, but I have to say that I'm quite lost. I didn't understand how to use Cauchy's theorem for this problem.
If anyone could help me I would be very greatful ! :)
Thanks in advance and have a good day
You are right about the fact that $a \notin \mathbb{R}$ change a few things. I will set $b = \mathrm{sgn}(\varphi''(k_0)) \in \{\pm 1\}$ hence $a = ib$. Moreover, the square root of a complex number that is not in $\mathbb{R}_+$ is not well defined. You have to choose between two possiblities and you have no canonical way to choose one more than the other. Therefore, unless you precise what you mean, $\sqrt{i\pi\mathrm{sgn}(\varphi''(k_0))}$ is non sens.
The main problem is that $\int_{\mathbb{R}} e^{ibx^2} \, dx$ exists in the sens that when $b \in \mathbb{R}^*$, $$ \int_A^B \cos(bx^2) \, dx \underset{A,B \rightarrow -\infty,+\infty}{\longrightarrow} I_1 \in \mathbb{R}, \qquad \int_A^B \sin(bx^2) \, dx \underset{A,B \rightarrow -\infty,+\infty}{\longrightarrow} I_2 \in \mathbb{R}, $$ but $x \mapsto e^{ibx^2}$ is not integrable on $\mathbb{R}$ in the sens of Lebesgue in the sens that, $$ \int_{\mathbb{R}} \left|e^{ibx^2}\right| \, dx = \int_{\mathbb{R}} 1 \, dx = +\infty. $$ It can cause some problems. For example, you can't use Fubini theorem so easily like you did when you computed $I^2$. For $b = 1$, you have, $$ \int_{\mathbb{R}} \cos(x^2) \, dx = \sqrt{\frac{\pi}{2}}, \qquad \int_{\mathbb{R}} \sin(x^2) \, dx = \sqrt{\frac{\pi}{2}}. $$ Look at https://en.wikipedia.org/wiki/Fresnel_integral for more details. They can be computed, as you said, using Cauchy formula (look at the paragraph "Limits as $x$ approaches infinity") by integrating $x \mapsto e^{ix^2}$ on a well chosen domain. Now, for parity reasons, $$ I = \int_{\mathbb{R}} e^{ibx^2} \, dx = \sqrt{\frac{\pi}{2}} + ib\sqrt{\frac{\pi}{2}} = \sqrt{\pi}\frac{1 + ib}{\sqrt{2}} = \sqrt{\pi}e^{ib\frac{\pi}{4}}. $$