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Prove that the above integral is equal to $\frac{\sqrt{2\pi}}{2}$

I have already tried expanding using $\cos$ identity and also taking Laplace for it. I am getting nowhere with this.


marked as duplicate by Tomás, Jack D'Aurizio, apnorton, mrf, user147263 Feb 11 '15 at 0:32

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    $\begingroup$ This is a Fresnel integral. It has been asked several times, see for example this answer. $\endgroup$ – mickep Feb 10 '15 at 19:34
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    $\begingroup$ Yeah i mean that@kobe. $\endgroup$ – vineetq Feb 10 '15 at 19:36
  • $\begingroup$ It should be $\frac{\sqrt{2\pi}}{4}$, no? Have you seen, or know the/some computation of $\int_{0}^{\infty}e^{-x^2}dx$? From this one you can compute yours, using that $\cos(a)=\frac{e^{ai}+e^{-ai}}{2}$. $\endgroup$ – Carol Feb 10 '15 at 20:21
  • $\begingroup$ The answer i wrote is the correct one and you can further look up Fresnel integral for the method to solve.The solution provided in one of the forward links is really elegant.Anyways Thanks for the help,Got what i needed. $\endgroup$ – vineetq Feb 10 '15 at 20:26

Let $f(z) = \exp( iz^2)$ be an entire function of a complex variable with $\Gamma$ the contour consisting of the circle wedge of angle $\pi/4$. In other words, it consists of three pieces. First the real segment $[0,R]$, followed by the circular arc $Re^{i\theta}$ where $0\leq \theta \leq \frac{\pi}{4}$ and then returns back, in a straight line, to the origin. So it looks like a pizza.

Thus, $$ \oint_{\Gamma} f(z) ~ dz = 0 $$ Now break up this integral into three pieces. The circular arc goes to zero as $R\to \infty$ by a standard estimation argument. The two segments of the integral can be computed.


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