Fresnel integral $\int\limits_0^\infty\sin(x^2) dx$ calculation I'm trying to calculate the improper Fresnel integral $\int\limits_0^\infty\sin(x^2)dx$ calculation.
It uses several substitutions. There's one substitution that is not clear for me.
I could not understand how to get the right side from the left one. What subtitution is done here? 
$$\int\limits_0^\infty\frac{v^2}{1+v^4} dv = \frac{1}{2}\int\limits_0^\infty\frac{1+u^2}{1+u^4} du.$$

Fresnel integral calculation:
In the beginning put $x^2=t$ and then: $$\int\limits_0^\infty\sin(x^2) dx = \frac{1}{2}\int\limits_0^\infty\frac{\sin t}{\sqrt{t}}dt$$
Then changing variable in Euler-Poisson integral we have: $$\frac{2}{\sqrt\pi}\int_0^\infty e^{-tu^2}du =\frac{1}{\sqrt{t} }$$
The next step is to put this integral instead of $\frac{1}{\sqrt{t}}$.
$$\int\limits_0^\infty\sin(x^2)dx = \frac{1}{\sqrt\pi}\int\limits_0^\infty\sin(t)\int_0^\infty\ e^{-tu^2}dudt = \frac{1}{\sqrt\pi}\int\limits_0^\infty\int\limits_0^\infty \sin (t) e^{-tu^2}dtdu$$ 
And the inner integral  $\int\limits_0^\infty \sin (t) e^{-tu^2}dt$ is equal to $\frac{1}{1+u^4}$.
The next calculation: $$\int\limits_0^\infty \frac{du}{1+u^4} = \int\limits_0^\infty \frac{v^2dv}{1+v^4} = \frac{1}{2}\int\limits_0^\infty\frac{1+u^2}{1+u^4} du = \frac{1}{2} \int\limits_0^\infty\frac{d(u-\frac{1}{u})}{u^2+\frac{1}{u^2}} $$
$$= \frac{1}{2} \int\limits_{-\infty}^{\infty}\frac{ds}{2+s^2}=\frac{1}{\sqrt2}\arctan\frac{s}{\sqrt2} \Big|_{-\infty}^\infty = \frac{\pi}{2\sqrt2} $$
In this calculation the Dirichle's test is needed to check the integral $\int_0^\infty\frac{\sin t}{\sqrt{t}}dt$ convergence. It's needed also to substantiate the reversing the order of integration ($dudt = dtdu$). All these integrals exist in a Lebesgue sense, and Tonelli theorem justifies reversing the order of integration. 
The final result is $$\frac{1}{\sqrt\pi}\frac{\pi}{2\sqrt2}=\frac{1}{2}\sqrt\frac{\pi}{2}$$
 A: There's a neat trick to evaluate the integral
$$S_n(t)=\int_0^\infty \sin(tx^n)dx.$$
First, take the Laplace transform:
$$\begin{align}
\mathcal{L}\{S_n(t)\}(s)&=\int_0^\infty e^{-st}S_n(t)dt\\
&=\int_0^\infty\int_0^\infty \sin(tx^n)e^{-st}dxdt\\
&=\int_0^\infty\int_0^\infty \sin(tx^n)e^{-st}dtdx\\
&=\int_0^\infty \frac{x^n}{x^{2n}+s^2}dx\tag1\\
&=s^{1/n-1}\int_0^\infty \frac{x^n}{x^{2n}+1}dx\\
&=\frac{s^{1/n-1}}{n}\int_0^\infty \frac{x^{1/n}}{x^2+1}dx\\
&=\frac{s^{1/n-1}}{n}\int_0^{\pi/2} \tan(x)^{1/n}dx\\
&=\frac{s^{1/n-1}}{n}\int_0^{\pi/2} \sin(x)^{1/n}\cos(x)^{-1/n}dx\\
&=\frac{s^{1/n-1}}{2n}\Gamma\left(\frac12(1+1/n)\right)\Gamma\left(\frac12(1-1/n)\right)\tag2\\
&=\frac{s^{1/n-1}\pi}{2n\cos\frac{\pi}{2n}}\tag3\\
&=\frac{\pi \sec\frac{\pi}{2n}}{2n\Gamma(1-1/n)}\mathcal{L}\{t^{-1/n}\}(s).
\end{align}$$
Thus, taking the inverse Laplace transform on both sides, 
$$S_n(t)=\frac{\pi \sec\frac{\pi}{2n}}{2nt^{1/n}\Gamma(1-1/n)}.$$
Choose $n=2$ and $t=1$ to get your integral:
$$S_2(1)=\frac12\sqrt{\frac\pi2}\ .$$
Explanation:
$(1)$: for real $q$ and $s$,
$$\begin{align}
\int_0^\infty \sin(qt)e^{-st}dt&= \text{Im}\int_0^\infty e^{iqt}e^{-st}dt\\
&=\text{Im}\int_0^\infty e^{-(s-iq)t}dt\\
&=\text{Im}\left[\frac{1}{s-iq}\right]\\
&=\frac{1}{s^2+q^2}\text{Im}\left[s+iq\right]\\
&=\frac{q}{s^2+q^2}
\end{align}$$
$(2)$: See here.
$(3)$: See here.
A: Well, if one puts $v=\frac{1}{u}$ then:
$$I=\int_0^\infty\frac{v^2}{1+v^4} dv =\int_0^\infty\frac{1}{1+u^4} du$$
So summing up the two integrals from above gives:
$$2I=\int_0^\infty\frac{1+u^2}{1+u^4} du$$
