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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.

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  • $\begingroup$ Well, it's a complex analysis problem - are you familiar with contour integration? $\endgroup$
    – πr8
    Jul 3, 2016 at 20:12
  • $\begingroup$ yes sir I know..but its solution is very complicated in my textbook $\endgroup$
    – T.Noor
    Jul 3, 2016 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, 2016 at 20:14
  • $\begingroup$ Is the problem don't have any simple solution like substitution I did above $\endgroup$
    – T.Noor
    Jul 3, 2016 at 20:21
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    $\begingroup$ Possible duplicate of Some way to integrate $\sin(x^2)$? $\endgroup$
    – user296602
    Jul 3, 2016 at 20:23

4 Answers 4

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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 $(2)$: $$ \int_{0}^{+\infty}\sin(x^2)\,dx=\color{red}{\frac{1}{2}\sqrt{\frac{\pi}{2}}}.\tag{3}$$

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  • $\begingroup$ gah! I wish I knew about Laplace transforms when I had to do this problem! $\endgroup$ Jul 3, 2016 at 20:22
  • $\begingroup$ @qbert: it is a really useful tool, pretty much underestimated (especially with respect to the Fourier transform). $\endgroup$ Jul 3, 2016 at 20:23
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    $\begingroup$ yes sir..you got the right aswer..and I knew about Laplace transorm...thank you very much sir $\endgroup$
    – T.Noor
    Jul 3, 2016 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 $\endgroup$ Jul 3, 2016 at 20:24
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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.

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  • $\begingroup$ yes sir..thank you very much $\endgroup$
    – T.Noor
    Jul 3, 2016 at 20:28
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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.

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  • $\begingroup$ $$\frac{1}{{{x}^{4}}+1}=\frac{1}{({{x}^{2}}+\sqrt{2}x+1)({{x}^{2}}-\sqrt{2}x+1)}$$ $\endgroup$ Jul 3, 2016 at 22:02
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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!

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