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Use Cauchy's theorem to prove $$\displaystyle\int_0^\infty\sin(x^2)\,dx=\int_0^\infty\cos(x^2)\,dx=\frac{\sqrt{2\pi}}{4}~.$$

This is an exercise in Stein's Complex Analysis. He hints that integrate the funtion $e^{-z^2}$ over the path in the figure as follows:

enter image description here

So I get \begin{equation} \int_0^R e^{-x^2}\,dx+\int_{0}^{\frac{\pi}{4}}e^{-R^2\cos(2\theta)-iR^2\sin(2\theta)+i\theta}Ri\,d\theta-\int_{0}^{R}e^{-x^2}e^{i\frac{\pi}{4}}\,dx=0. \end{equation}

If the middle term of the above formula converges to $0$ as $R\to\infty$, then we are done.

I get $$ \bigg|\int_{0}^{\frac{\pi}{4}}e^{-R^2\cos(2\theta)-iR^2\sin(2\theta)+i\theta}Ri\,d\theta\bigg|\leq R\int_0^{\frac{\pi}{4}}e^{-R^2\cos(2\theta)}\,d\theta. $$ And since $\cos\theta\geq1-\theta^2$, we have $$ R\int_0^{\frac{\pi}{4}}e^{-R^2\cos(2\theta)}\,d\theta\leq Re^{-R^2}\int_0^{\frac{\pi}{4}}e^{2R^2\theta^2}\,d\theta. $$ But I don't know how to prove this integral goes to $0$ as $R\to\infty$.

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    $\begingroup$ That last expression doesn''t go to zero. Your cosine estimate is too loose. $\endgroup$ – Angina Seng Aug 4 '19 at 4:50
  • $\begingroup$ Then how to prove the original expression goes to zero? $\endgroup$ – Xie Aug 4 '19 at 5:40
  • $\begingroup$ Just an idle though, but could you use L'Hôpital's rule on $\lim_{R \to \infty} \frac{\int_0^{\frac{\pi}{4}}e^{-R^2\cos(2\theta)}\,d\theta}{1/R}$? $\endgroup$ – Richard D. James Aug 5 '19 at 5:26
  • $\begingroup$ The convergence is not uniform, so perhaps the integration sign and the limit sign cannot be interchanged. $\endgroup$ – Xie Aug 5 '19 at 5:37
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This solution doesn't follow your approach, but is still valid. We actually seek the evaluation of the following integral:

$$I=\int_0^\infty x^\mu e^{-ax}\sin(x)dx=\Im\int_0^\infty x^\mu e^{(i-a)x}dx$$

for $\mu>-1$ and $a>0$. The integral will be evaluated with a complex contour, the path of which will make sense in hindsight. We take as a contour a sector of an annulus with outer radius $R$ and inner radius $\epsilon$, and an angle of $\varphi=\arctan(1/a)$. This angle gives us the useful relationship of

$$(i-a)=\sqrt{a^2+1}e^{i(\pi-\varphi)}=-\sqrt{a^2+1}e^{-i\varphi}$$

which will come in handy later. The four pieces of the arc enclose no singularities, so by the Cauchy Integral Theorem, its integral is $0$. We were also careful to avoid the essential singularity at $0$, and must be wise to consider branch cuts. We then have

$$0=\oint z^\mu e^{(i-a)z}dz=\int_\epsilon^R x^\mu e^{(i-a)x}dx+\int_0^\varphi\left(Re^{i\theta}\right)^\mu e^{(i-a)Re^{i\varphi}}iRe^{i\theta}d\theta$$

$$+\int_R^\epsilon\left(xe^{i\varphi}\right)^\mu e^{(i-a)xe^{i\varphi}}e^{i\varphi}dx+\int_\varphi^0\left(\epsilon e^{i\theta}\right)^\mu e^{(i-a)\epsilon e^{i\theta}}i\epsilon e^{i\theta}d\theta$$

The second integral goes to $0$ as $R\to\infty$, and the fourth integral goes to $0$ as $\epsilon\to 0$ (can you show this?). This tells us that

$$\int_0^\infty x^\mu e^{(i-a)x}dx=e^{i(\mu+1)\varphi}\int_0^\infty x^\mu e^{-\sqrt{a^2+1}x}dx$$

We now take $a\to 0^+$ (noting that $\varphi\to\pi/2$) and get

$$\int_0^\infty x^\mu e^{ix}dx=e^{i(\mu+1)\frac{\pi}{2}}\int_0^\infty x^\mu e^{-x}dx=e^{i(\mu+1)\frac{\pi}{2}}\Gamma(\mu+1)$$

where $\Gamma$ is the Euler Gamma function. The imaginary portion of this is

$$\int_0^\infty x^\mu\sin(x)dx=\sin\left[(\mu+1)\frac{\pi}{2}\right]\Gamma(\mu+1)$$

This relates back to the integral in question of

$$J=\int_0^\infty\sin(x^2)dx$$

If we let $u=x^2$, then

$$J=\frac{1}{2}\int_0^\infty\frac{\sin(u)}{\sqrt{u}}du=\frac{1}{2}\sin\left(\frac{\pi}{4}\right)\Gamma\left(\frac{1}{2}\right)=\frac{\sqrt{2\pi}}{4}$$

For the cosine integral, we do the same process and instead take the real portion along the way.

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