Fresnel integral $\int\limits_0^\infty\sin(x^2) dx$ calculation

Question

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?

$$\int\limits_0^\infty\frac{v^2}{1+v^4} dv = \frac{1}{2}\int\limits_0^\infty\frac{1+u^2}{1+u^4} du.$$

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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 $$\frac{1}{\sqrt\pi}\frac{\pi}{2\sqrt2}=\frac{1}{2}\sqrt\frac{\pi}{2}$$

Answer 1

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: $$2I=\int_0^\infty\frac{1+u^2}{1+u^4} du$$

Answer 2

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.