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

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: 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 $$R\to\infty$$.

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

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.