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:

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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 $R\to\infty$.

  • 1
    $\begingroup$ 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 $\lim_{R \to \infty} \frac{\int_0^{\frac{\pi}{4}}e^{-R^2\cos(2\theta)}\,d\theta}{1/R}$? $\endgroup$ – Viktor Vaughn 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

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


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


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


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


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