I need to solve the integral $I=\int_{0}^{\infty}sin(u^2)du$(Fresnel Integral). I made some progress in this regard, but I had some doubts as to how to proceed from then on.
Here is the approach i used:
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Let x=$u^2$ $\implies dx=2udu \implies du=\frac{x^{-\frac{1}{2}}dx}{2}$, where the limits remain same. Thus $$I=1/2*\int_{0}^{\infty}x^{-\frac{1}{2}}sin(x)dx$$
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From here I consider $\oint_C{z^{-\frac{1}{2}}}e^{iz}dz$; where C is
$C_1=[Re^{i\theta}|0\le\theta\le\frac{\pi}{2}]$; a quarter circle of radius R, $C_2=[iy|R\le y\le r ]$; along the Imaginary axis, $C_3=[re^{i\theta}|\frac{\pi}{2}\le\theta\le0]$; quarter circle of radius r ,$C_4=[x|r\le x\le R ]$
Now, I have considered C to be broken into $C_1,C_2,C_3,C_4$, where these paths are also labeled in the diagram above.
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As $z^{-\frac{1}{2}}e^{iz}$ is analytic in the region described by C, we have that $$\oint_C{z^{-\frac{1}{2}}}e^{iz}dz=0;[Cauchy's Theorem]$$
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From Here I get the following: $$\int_{C_1}z^{-\frac{1}{2}}e^{iz}dz+\int_{C_2}z^{-\frac{1}{2}}e^{iz}dz+\int_{C_3}z^{-\frac{1}{2}}e^{iz}dz+\int_{C_4}z^{-\frac{1}{2}}e^{iz}dz=0$$
where $C_1, C_3$ are along the quarter circles of radii R and r respectively, and $C_2,C_4$ are along the coordinate axes. Now we would let $r\rightarrow0, R\rightarrow\infty$ in hopes of recovering our integral.
This brings me to my question:
a) How do I show that $\int_{C_1}z^{-\frac{1}{2}}e^{iz}dz=\int_{C_3}z^{-\frac{1}{2}}e^{iz}dz=0$ in the above limit? I tried using the Estimation Lemma to derive: $$|\int_{C_1}z^{-\frac{1}{2}}e^{iz}dz|\le \int_{C_1}|z^{-\frac{1}{2}}e^{iz}|=\frac{\pi}{2}R*\frac{1}{\sqrt{R}}dz $$ [As $C_1$ is a quarter circle of radius R, z=R on C] However in the limit R$\rightarrow\infty$, my integral doesn't tend to 0.
However, using the same procedure on $C_3$, I get that: $$|\int_{C_1}z^{-\frac{1}{2}}e^{iz}dz|\le \int_{C_1}|z^{-\frac{1}{2}}e^{iz}|=\frac{\pi}{2}r*\frac{1}{\sqrt{r}}dz $$ then in the limit r$\rightarrow0$ this integral does tend to 0.
b) Why couldn't the estimation lemma be used for the quarter circle of radius R then? How would i show that $\int_{C_1}z^{-\frac{1}{2}}e^{iz}dz=0$?
c) Jordan's Lemma seems like the perfect candidate to prove that the above integral is 0, however, as i deal with contours of quarter-circle like shape I cannot use this Lemma. Is it possible to use a variation of this lemma to get 0?, if so how?.
I only need to prove that this integral is indeed 0, then I can derive the Fresnel Integral's value.
Best Answer
The integral along $C_1$ can be estimated as follows $$\left |\int\limits_{C_1}z^{-1/2}e^{iz}\,dz\right |\le \int\limits_{C_1}|z|^{-1/2}|e^{iz}|\,|dz|=R^{-1/2}\int\limits_0^{\pi/2}e^{-R\sin\theta}\, R\,d\theta=R^{1/2}\int\limits_0^{\pi/2}e^{-R\sin\theta}\, d\theta$$ We have $\sin\theta\ge {2\over \pi}\theta.$ Hence $$\left |\int\limits_{C_1}z^{-1/2}e^{iz}\,dz\right |\le R^{1/2}\int\limits_0^{\pi/2}e^{-(2R/\pi)\theta}\, d\theta=R^{1/2}{\pi\over 2R}(1-e^{-R})\le {\pi\over 2}R^{-1/2}$$ The integral along $C_3$ is easier as $|e^{iz}|\le 1,$ hence $$\left |\int\limits_{C_3}z^{-1/2}e^{iz}\,dz\right |\le {\pi\over 2} r\cdot r^{-1/2}={\pi\over 2}r^{1/2}$$