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Compute $$\int_0^\infty \sin x^2dx.$$

The source of this question is here. I'm trying to solve this question using robjohn's method. The integral we need to consider is $$\int_\gamma e^{-z^2}dz$$

Consider the contour $\gamma$ consisting of the line from $0$ to $R$, then counterclockwise along the circular arc from $R$ to $Re^{i\pi/4}$, then back along the line from $Re^{i\pi /4}$ to $0$.

My problem is proving that the integral over the arc is $0$, here is what I've tried so far:

Take $\gamma(t)=Re^{it}$, where $0\leq t\leq \frac{\pi}{4}$. \begin{align*} \int_0^{\frac{\pi}{4}}e^{-z^2}dz&=\int_0^{\frac{\pi}{4}}e^{-R^2e^{2it}}\cdot Rie^{it}dt\\&\leq \int_0^{\frac{\pi}{4}}\vert{e^{-R^2e^{2it}}\vert\cdot \vert Rie^{it}}\vert dt\\ &= \int_0^{\frac{\pi}{4}}\vert{e^{-R^2(\cos 2t+i\sin 2t)}}\vert \cdot Rdt\\ &=\int_0^{\frac{\pi}{4}}R{e^{-R^2\cos 2t}}dt. \end{align*} Then how to prove that $$\lim_{R\to\infty}R\int_0^{\frac\pi 4}e^{-R^2\cos 2t}dt=0?$$

I know someone has provided a solution in the original question, but I wonder if there is a better way to solve it. Also, we used different parametrizations.

Bowei Tang
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  • Do you intend for the limiting quantity to be an integral? Is the $dt$ a typo? – Sammy Black Nov 24 '24 at 03:59
  • @SammyBlack Thanks. Fixed now. – Bowei Tang Nov 24 '24 at 04:01
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    Check this: https://math.stackexchange.com/q/442950/42969 – Martin R Nov 24 '24 at 04:06
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    Your first line of bounding the integral is incorrect. It should be $\left|\int_{\gamma}e^{-z^{2}}dz\right| = \left|\int_{0}^{\frac{\pi}{4}}e^{-R^{2}e^{2it}}\cdot Rie^{it}dt\right| \leq \int_{0}^{\frac{\pi}{4}}\left|e^{-R^{2}e^{2it}}\right|\cdot\left|Rie^{it}\right|dt$. – Accelerator Nov 24 '24 at 05:05

2 Answers2

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You can use a similar trick as in your linked post and in the answer linked in the comment.

Because of the concavity of $\cos x$ on $[0,\pi/2]$ you have $$\cos x \geq 1-\frac 2\pi x \text{ on } [0,\pi/2]$$ Hence, $$\cos 2t \geq 1-\frac 4\pi t \text{ on } [0,\pi/4]$$ Using this, you can estimate your integral from above:

$$R\int_0^{\frac\pi 4}e^{-R^2\cos 2t}dt\leq R\int_0^{\frac\pi 4}e^{-R^2(1-\frac 4\pi t)}\, dt = \frac{\pi}{4R}(1 - e^{-R^2})$$

trancelocation
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$$I=R\int_0^{\frac{\pi}{4}}e^{-R^2\cos (2t)}\,dt=\frac R 2 \int_0^{\frac{\pi}{2}}e^{-R^2\cos (x)}\,dx$$ $$I=\frac \pi 4 R \left(I_0\left(R^2\right)-\pmb{L}_0\left(R^2\right)\right) $$ where appear Bessel and Struve function.

Using their asymptotics $$\color{blue}{\large I=\frac 1{2\pi}\sum_{n=0}^\infty \frac{2^{2 n}\,\, \Gamma\left(n+\frac{1}{2}\right)^2}{R^{4 n+1} }}$$ which is $$I=\frac{1}{2 R}+\frac{1}{2R^5}+O\left(\frac{1}{R^9}\right)$$

Claude Leibovici
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