7
$\begingroup$

If you take the fresnel integrals to be $$S(x) = \int_{0}^{x}\sin \left(\frac { \pi \cdot t^2}{2} \right) dt$$ How do you find the asymptotic expansion? I know it begins with a $1/2$ but how?

$\endgroup$
4
$\begingroup$

I suppose you want an asymptotic expansion as $x\to \infty$. We start with

$$S(x) = \int_0^x \sin \left(\frac{\pi\cdot t^2}{2}\right)\,dt = \frac{1}{2} - \int_x^\infty \sin \left(\frac{\pi\cdot t^2}{2}\right)\,dt.$$

Now, to get a handle on that integral, we substitute $u = \frac{\pi t^2}{2}$ and obtain

$$\int_x^\infty \sin \left(\frac{\pi\cdot t^2}{2}\right)\,dt = \frac{1}{\sqrt{2\pi}} \int_{\pi x^2/2}^\infty \frac{\sin u}{\sqrt{u}}\,du.$$

Then we want an asymptotic expansion of

$$\int_y^\infty \frac{\sin u}{\sqrt{u}}\,du$$

which we get via integration by parts:

\begin{align} \int_y^\infty \frac{\sin u}{\sqrt{u}}\,du &= \left[-\frac{\cos u}{\sqrt{u}}\right]_y^\infty - \frac{1}{2}\int_y^\infty \frac{\cos u}{u^{3/2}}\,du\\ &= \frac{\cos y}{\sqrt{y}} - \frac{1}{2}\int_y^\infty \frac{\cos u}{u^{3/2}}\,du\\ &= \frac{\cos y}{\sqrt{y}} - \frac{1}{2}\left(\left[\frac{\sin u}{u^{3/2}}\right]_y^\infty + \frac{3}{2}\int_y^\infty \frac{\sin u}{u^{5/2}}\,du\right)\\ &= \frac{\cos y}{\sqrt{y}} + \frac{\sin y}{2y^{3/2}} - \frac{3}{4}\int_y^\infty \frac{\sin u}{u^{5/2}}\\ &= \frac{\cos y}{\sqrt{y}} + \frac{\sin y}{2y^{3/2}} - \frac{3\cos y}{4y^{5/2}} - \frac{15}{8} \int_y^\infty \frac{\cos u}{u^{7/2}}\,du. \end{align}

An elementary estimate shows that the last integral is $O(y^{-5/2})$ [actually, it is $O(y^{-7/2})$, as one sees when thinking about what a further integration by parts yields], so we get the asymptotic expansion

$$\int_y^\infty \frac{\sin u}{\sqrt{u}}\,du = \frac{\cos y}{\sqrt{y}} + \frac{\sin y}{2y^{3/2}} + O(y^{-5/2}),$$

and hence, inserting $y = \frac{\pi x^2}{2}$,

$$S(x) = \frac{1}{2} - \frac{\cos \frac{\pi x^2}{2}}{\pi x} - \frac{\sin \frac{\pi x^2}{2}}{\pi^2 x^3} + O(x^{-5}).$$

To get higher-order asymptotic expansions, integrate by parts more often.

$\endgroup$
  • $\begingroup$ To future readers, derivation of the $\int_0^\infty=1/2$ can be found here: en.wikipedia.org/wiki/…. It assumes you understand the guassian integral, and at the end to get 1/2 you must substitute $t^2=u^2\pi/2$ into the original and solve. $\endgroup$ – user5389726598465 May 10 '17 at 1:59
0
$\begingroup$

Using the taylor series of sine:

$$\sin\frac{\pi t^2}2=\frac{\pi t^2}2-\frac{\pi^3t^6}6+\ldots\implies$$

$$\int\limits_0^x\sin\frac{\pi t^2}2dt=\int\limits_0^x\left(\frac{\pi t^2}2-\frac{\pi^3t^6}6+\ldots\right)dt=\frac\pi6x^3-\frac{\pi^3}{42}x^7+\ldots$$

I can't see any $\;\frac12\;$ there unless $\;x=\sqrt[3]\frac3\pi\;$

$\endgroup$
  • 2
    $\begingroup$ I think the OP wants the asymptotic expansion as $x\to\infty$. The Taylor series of $\sin$ doesn't help with that. $\endgroup$ – Daniel Fischer Oct 26 '14 at 18:25

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.