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