Hi: I already asked this question on the complex analysis tag but nobody answered it so then I found this complex-integration tag and was hoping that someone might be able to answer it here. It is below and thanks.
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I'm reading some notes I found on complex analysis on the internet. In the example, they prove that
$$\int_{0}^{\infty} \cos(x^2) = \int_{0}^{\infty} \sin(x^2) = \frac{\sqrt{2\pi}}{4}.$$
I understand a lot of the proof so I don't think that I need to write down all the steps. I will just provide what I think is necessary for readers to understand my lack of understanding.
The example starts off with
"We shall consider the analytic function $f(z) = e^{(iz^2)}$ over a contour C of the form of a circular wedge in the first quadrant. Then contour has three edges:
(1) a line segment of length R from the $0$ to the point $A$.=,
(2) from $A$ to $B$ over a circular arc of angle $\frac{\pi}{4}$ (so $B = R e^{i \pi/4})$ and
(3) from $B$ to $0$.
That make sense. Then the author breaks the integral into three pieces so that
$$\int_{C} f(z) dz = \int_{OA} f(z) dz + \int_{AB} f(z) dz + \int_{BO} f(z) dz = 0.$$
by cauchy's theorem. That makes sense too.
But then, when evaluating the third integral, the author has
$$\begin{align} I_{3}(R) &= \int_{BO} e^{(i z^2)} dz \\&= \int_{R}^{0} e^{-r^2} e^{i \pi/4} dr \\&= -e^{i \pi/4} \int_{0}^{R} e^{-r^2} dr. \end{align}$$
Then the author explains
" for $I_{3}(R) $ we used $ z = r e^{i \pi/4}, 0 \le r \le R; z^2 = r^2e^{i \pi/4}; dz = e^{i \pi/4} dr $ ".
So, given what was used for $I_{3}(R)$ , I don't see how that integral is obtained. I must be missing something but could someone spell out how the negative comes about ? In fact, I don't even see where the $e^{i \pi/4}$ part of $z^2$ went ? Basically, I don't follow the integral construction at all.
There seems to be a misprint in the part which says what $z^2$ is supposed to be. Since $z=r{\rm e}^{{\rm i}\pi/4}$ on $BO$, you get that $$z^2=r^2{\rm e}^{{\rm i}\pi/2}={\rm i}r^2$$ and hence $f(z)={\rm e}^{-r^2}$.
Putting things together, just substitute ${\rm d}z={\rm e}^{{\rm i}\pi/4}{\rm d}r$ and remember that along the segment $BO$, the parameter $r$ will start at $R$ and decrease to $0$. The minus sign in the last expression for $I_3(R)$ comes from switching the direction and letting $r$ run from $0$ to $R$ instead.