1
$\begingroup$

This problem is from my textbook of complex analysis. I have attempted this as: let $$u=x^{2}$$ then $$dx=\frac{du}{2\sqrt{u}}$$ therefore $$\frac{1}{2}\int_{0}^{\infty} \frac{\sin u}{\sqrt{u}}\,du $$ Now what should I do further? or is there any other simple way to solve the problem? please anyone help.

$\endgroup$
  • $\begingroup$ Well, it's a complex analysis problem - are you familiar with contour integration? $\endgroup$ – πr8 Jul 3 '16 at 20:12
  • $\begingroup$ yes sir I know..but its solution is very complicated in my textbook $\endgroup$ – T.Noor Jul 3 '16 at 20:12
  • $\begingroup$ I think most solutions for tricky sine integrals tend to go through either contour integrals or playing around with Laplace transforms - perhaps give these a try? $\endgroup$ – πr8 Jul 3 '16 at 20:14
  • $\begingroup$ Is the problem don't have any simple solution like substitution I did above $\endgroup$ – T.Noor Jul 3 '16 at 20:21
  • 3
    $\begingroup$ Possible duplicate of Some way to integrate $\sin(x^2)$? $\endgroup$ – T. Bongers Jul 3 '16 at 20:23
6
$\begingroup$

Laplace transform it is. Since: $$ \mathcal{L}(\sin u) = \frac{1}{1+s^2},\qquad \mathcal{L}^{-1}\left(\frac{1}{\sqrt{u}}\right) = \frac{1}{\sqrt{\pi s}}\tag{1}$$ we have: $$\begin{eqnarray*} \int_{0}^{+\infty}\sin(x^2)\,dx &=& \frac{1}{2}\int_{0}^{+\infty}\frac{\sin u}{\sqrt{u}}\,du\\ &=& \frac{1}{2\sqrt{\pi}}\int_{0}^{+\infty}\frac{ds}{(1+s^2)\sqrt{s}}\\&=&\frac{1}{\sqrt{\pi}}\int_{0}^{+\infty}\frac{dt}{1+t^4}.\tag{2}\end{eqnarray*}$$ The last integral is straightforward to compute through the residue theorem or Euler's beta function: $\int_{0}^{+\infty}\frac{dt}{1+t^4}=\frac{\pi}{2\sqrt{2}}$. It follows that by $(1)$ and $(2)$: $$ \int_{0}^{+\infty}\sin(x^2)\,dx=\color{red}{\frac{1}{2}\sqrt{\frac{\pi}{2}}}.\tag{3}$$

$\endgroup$
  • $\begingroup$ gah! I wish I knew about Laplace transforms when I had to do this problem! $\endgroup$ – qbert Jul 3 '16 at 20:22
  • $\begingroup$ @qbert: it is a really useful tool, pretty much underestimated (especially with respect to the Fourier transform). $\endgroup$ – Jack D'Aurizio Jul 3 '16 at 20:23
  • 1
    $\begingroup$ yes sir..you got the right aswer..and I knew about Laplace transorm...thank you very much sir $\endgroup$ – T.Noor Jul 3 '16 at 20:24
  • $\begingroup$ I have been meaning to check it out, but I think I will have to get to it sooner than I thought $\endgroup$ – qbert Jul 3 '16 at 20:24
2
$\begingroup$

We seek to evaluate $\int_{0}^{\infty}\sin(x^2)$

Moving to the complex plane, we note that $$ \Im(e^{iz^{2}})=\sin(z^{2})\Rightarrow \int_{0}^{\infty}\sin(x^2)dx= \lim_{R\rightarrow \infty}\int_{0}^{R}\Im(e^{iz^{2}})dz $$ We take the contour, $C=[0,R]+\gamma_{R}+\gamma_{l}$, with $$\gamma_{R}(t)=Re^{i\pi/4t}$\\ 0\leq t\leq 1$$ and $\gamma_l$ with $$ \gamma_{-l}: [0,R]\rightarrow \mathbb{C}\\ t\mapsto t(1+i)/\sqrt{2} $$ Oriented counterclockwise. In other words, C is a pizza slice with angle $\pi/4$.

Then by cauchy's theorem and the absence of singularities within the contour we have \begin{align*} \int_{C}e^{iz^{2}}dz=0=\int_{0}^{R}e^{iz^{2}}dz+ \int_{\gamma_{R}}e^{iz^{2}}dz+\int_{\gamma_{l}}e^{iz^{2}}dz\\ \Rightarrow \int_{0}^{R}e^{iz^{2}}dz=-\int_{\gamma_{R}}e^{iz^{2}}dz- \int_{\gamma_{l}}e^{iz^{2}}dz \end{align*} With the LHS as the integral we want.

Let's evaluate $\int_{\gamma_{l}}e^{iz^{2}}dz$ first. We parametrize the the negative of the path (swapping orientation proves convenient later) \begin{align*} \gamma_{-l}: [0,R]\rightarrow \mathbb{C}\\ t\mapsto t(1+i)/\sqrt{2} \end{align*} Then we have: \begin{align*} \int_{\gamma_{-l}}e^{iz^2}=\int_{0}^{R}e^{\frac{it^2(1+i)^2}{2}}= \int_{0}^{R}e^{it^2i}\\ =\int_{0}^{R}e^{-t^2} \end{align*} A real integral (modified Gaussian Integral), which we evaluate as follows: \begin{align*} \text{Define}\;I=\int_{0}^{R}e^{-x^{2}}dx=\int_{0}^{R}e^{-y^{2}}dy\\ \text{then}\; I^{2}=\int_{0}^{R}e^{-x^{2}}dx\int_{0}^{R}e^{-y^{2}}dy\\ \Rightarrow I^{2}=\int_{0}^{R}\int_{0}^{R}e^{-x^{2}}e^{-y^{2}}dxdy\\ \Rightarrow I^{2}=\int_{0}^{R}\int_{0}^{R}e^{-(x^{2}+y^{2})}dxdy \end{align*} And converting to polar coordinates, with jacobian $r$ for the integral, and taking $\theta\in[0,\pi/2]$ since we are integrating on the positive reals in the first quadrant, we have: \begin{align*} \int_{0}^{\pi/2}\int_{0}^{R}e^{-(r^{2}\cos^{2}(\theta)+ r^{2}\sin^{2}(\theta ))}rdrd\theta \leq I^{2}\leq \int_{0}^{\pi/2}\int_{0}^{\sqrt{2}R}e^{-(r^{2}\cos^{2}(\theta)+ r^{2}\sin^{2}(\theta ))}rdrd\theta\\ \Rightarrow \int_{0}^{\pi/2}\int_{0}^{R}e^{-(r^{2})}(\theta ))rdrd\theta \leq I^{2}\leq \int_{0}^{\pi/2}\int_{0}^{\sqrt{2}R}e^{-(r^{2})}rdrd\theta\\ \end{align*} Sandwhiching our rectangular integral between the diagonal of the rectangle and the arc.

Which we can now perform a u substitution on: $u=r^2\Rightarrow \frac{du}{2r}=dr$, yielding \begin{align*} \frac{1}{2}\int_{0}^{\pi/2}\int_{0=r}^{R=r}e^{-u}dud\theta\leq I^{2}\leq \frac{1}{2}\int_{0}^{\pi/2}\int_{0=r}^{\sqrt{2}R=r}e^{-u}dud\theta\\ \Rightarrow \frac{\pi/2}{2}[\frac{-e^{-R^{2}}}{2}+\frac{1}{2}] \leq I^{2}\leq\frac{\pi/2}{2}[\frac{-e^{-2R^{2}}}{2}+\frac{1}{2}]\\ \Rightarrow \frac{\pi}{8}[1-e^{-R^{2}}]\leq I^{2}\leq \frac{\pi}{8}[1-e^{-2R^{2}}]\\ \sqrt{\frac{\pi}{8}}\sqrt{1-e^{-R^{2}}}\leq I\leq \sqrt{\frac{\pi}{8}}\sqrt{1-e^{-2R^{2}}} \end{align*} Then in the limit as $R\rightarrow \infty$ we have \begin{align*} \lim_{R\rightarrow \infty}\sqrt{\frac{\pi}{8}}\sqrt{1-e^{-iR^{2}}} \leq \lim_{R\rightarrow \infty}I\leq \lim_{R\rightarrow \infty}\sqrt{\frac{\pi}{8}}\sqrt{1-e^{-i2R^{2}}}\\ \Rightarrow \lim_{R\rightarrow \infty}I=\sqrt{\frac{\pi}{8}} = \sqrt{\frac{2\pi}{16}}\\ = \sqrt{\frac{2\pi}{16}}=\frac{\sqrt{2\pi}}{4} \end{align*} This then gives us, since $\int_{\gamma_{-l}}e^{iz^2}= -\int_{\gamma_{l}}e^{iz^2}$, \begin{align*} \lim_{R\rightarrow \infty}\int_{0}^{R}e^{iz^{2}}dz= \lim_{R\rightarrow \infty}-\int_{\gamma_{R}}e^{iz^{2}}dz- \lim_{R\rightarrow \infty}\int_{\gamma_{l}}e^{iz^{2}}dz\\ \Rightarrow \lim_{R\rightarrow \infty}\int_{0}^{R}e^{iz^{2}}dz= \lim_{R\rightarrow \infty}-\int_{\gamma_{R}}e^{iz^{2}}dz+ \lim_{R\rightarrow \infty}\int_{-\gamma_{l}}e^{iz^{2}}dz\\ \Rightarrow \lim_{R\rightarrow \infty}\int_{0}^{R}e^{iz^{2}}dz= \frac{\sqrt{2\pi}}{4} -\lim_{R\rightarrow \infty}\int_{\gamma_{R}}e^{iz^{2}}dz \end{align*} Showing $\lim_{R\rightarrow \infty}\int_{\gamma_{R}}e^{iz^{2}}dz=0$ will complete our evaluation. Usually this isn't too hard, but in this case, it is pretty hard to do rigourosly.

$\endgroup$
  • $\begingroup$ yes sir..thank you very much $\endgroup$ – T.Noor Jul 3 '16 at 20:28
2
$\begingroup$

We know $$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{\large-\frac{z^2}{2}}dz=1$$ set $z=\sqrt{2u}\,x$, we have $$\frac{\sqrt{u}}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-ux^2}dz=1$$ therefore

$$\color{blue} {\frac{2}{\sqrt{\pi}}\int_{0}^{\infty}e^{-u x^2}dz=\frac{1}{\sqrt{u}}\tag{1}}$$

Now apply $(1)$

$$I=\int_{0}^{\infty} {\color{blue} {\frac{1}{\sqrt{u}}}}e^{iu}du=\frac{2}{\sqrt{\pi}}\int_{0}^{+\infty}\int_{0}^{\infty}e^{u(i-x^2)}du\,dx$$ $$I=-\frac{2}{\sqrt{\pi}}\int_{0}^{+\infty}\frac{1}{i-x^2}dx=\frac{2}{\sqrt{\pi}}\int_{0}^{+\infty}\frac{i+x^2}{1+x^4}dx$$ Indeed

$$\color{blue}{\int_{0}^{\infty} {\color{blue} {\frac{1}{\sqrt{u}}}}e^{iu}du=\frac{2}{\sqrt{\pi}}\int_{0}^{+\infty}\frac{i+x^2}{1+x^4}dx\tag{2}}$$

We apply Euler's identity $${\int_{0}^{\infty} {\frac{\cos\,u}{\sqrt{u}}}}du+i{\int_{0}^{\infty} {\frac{\sin\,u}{\sqrt{u}}}}du=\frac{2}{\sqrt{\pi}}\int_{0}^{+\infty}\frac{x^2}{1+x^4}dx+\frac{2i}{\sqrt{\pi}}\int_{0}^{+\infty}\frac{1}{1+x^4}dx$$ In the other words

$$\color{blue}{{\int_{0}^{\infty} {\frac{\sin\,u}{\sqrt{u}}}}du=\frac{2}{\sqrt{\pi}}\int_{0}^{+\infty}\frac{1}{1+x^4}dx\tag{3}}$$

Now you can easily compute this integral.

$\endgroup$
  • $\begingroup$ $$\frac{1}{{{x}^{4}}+1}=\frac{1}{({{x}^{2}}+\sqrt{2}x+1)({{x}^{2}}-\sqrt{2}x+1)}$$ $\endgroup$ – Behrouz Maleki Jul 3 '16 at 22:02
1
$\begingroup$

Here, we present a straightforward way forward that begins with the representation

$$\begin{align} I&=\frac12 \int_0^\infty \frac{\sin(x)}{\sqrt{x}}\,dx \\\\ &=\text{Im}\left(\int_0^\infty \frac{e^{ix}}{\sqrt{x}}\right) \tag 1 \end{align}$$

We will evaluate the integral in $(1)$ by direct application of Cauchy's Integral Theorem. To that end, we analyze the closed-contour integral

$$\oint_C \frac{e^{iz}}{\sqrt{z}}\,dz=\int_0^R \frac{e^{ix}}{\sqrt{x}}\,dx+\int_0^{\pi/2} \frac{e^{iRe^{i\phi}}}{\sqrt{Re^{i\phi}}}\,iRe^{i\phi}\,d\phi+\int_R^0 \frac{e^{-y}}{\sqrt{iy}}\, i\,dy \tag 2$$

where $C$ is the quarter circle in the first quadrant, "centered" at the origin with radius $R$. Additionally, we choose the branch cut as the non-negative real axis.

Inasmuch as the integrand is analytic in and on $C$, Cauchy's Integral Theorem guarantees that the left-hand side of $(2)$ is zero. Moreover, the second integral on the right-hand side of $(2)$ vanishes as $R\to \infty$. Therefore, we find

$$\begin{align} \int_0^\infty \frac{e^{ix}}{\sqrt{x}}\,dx&=e^{i\pi/4}\int_0^\infty \frac{e^{-y}}{\sqrt{y}}\,dy\\\\ &=2e^{i\pi/4}\int_0^\infty e^{-t^2}\,dt\\\\ &=\sqrt{\pi}e^{i\pi/4} \tag 3 \end{align}$$

Finally, substituting $(3)$ into $(1)$ yields

$$\bbox[5px,border:2px solid #C0A000]{I=\frac12 \sqrt{\frac{\pi}{2}}}$$

as expected!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.