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?


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.

  • $\begingroup$ 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

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 $\;x=\sqrt[3]\frac3\pi\;$

  • 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

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