5
$\begingroup$

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

$\endgroup$
8
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ thank you very much $\endgroup$ – Metso Jun 24 at 19:38
  • 1
    $\begingroup$ I'm glad that I could help! Btw, can you show how the arrived at that integral starting from the Fresnel Integral? $\endgroup$ – Nyssa Jun 24 at 19:39
  • 1
    $\begingroup$ I added info. Let me know if something is wrong. I'll check tomorrow. Now I'm leaving. $\endgroup$ – Metso Jun 24 at 20:25
  • 2
    $\begingroup$ @Zacky: More on Fresnel integrals, I.E Leonard AMM 95 number 5: jstor.org/stable/2322478 $\endgroup$ – FDP Jun 25 at 13:39
4
$\begingroup$

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.

$\endgroup$

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.