I need to find the infinite series for the asymptotic expansions of the fresnel integrals as $x\rightarrow \infty$ and $x\rightarrow 0$.

Now I have computed the asyptotic expansions to be as follows

$$C(x)=\sqrt{\frac{\pi}{8}}+ \frac{\sin(x^2)}{2x} - \frac{\cos(x^2)}{4x^3} - \frac{3\sin(t^2)}{8t^5} + \mathcal{O}(x^{-7}) $$ and $$S(x)=\sqrt{\frac{\pi}{8}}- \frac{\cos(x^2)}{2x} - \frac{\sin(x^2)}{4x^3} + \frac{3\cos(t^2)}{8t^5} + \mathcal{O}(x^{-7}).$$

Not sure where to go from here, any ideas?

  • $\begingroup$ Please do not deface your posts. $\endgroup$ May 6 '14 at 16:50

I'll only cover the expansion as $x \to +\infty$ here, the expansion as $x \to 0$ is trivial. It will be easier to present the expansion if we group the real and imaginary parts together. $$C(x) + iS(x) \asymp \sqrt{\frac{\pi}{8}}(1+i) + \frac{e^{ix^2}}{2}\sum_{n=0}^\infty \frac{(-1)^{n+1} \left(\frac12\right)_n}{x^{2n+1}}\tag{*1}$$ where $\left(\lambda\right)_n = \lambda(\lambda+1)\cdots(\lambda+n-1)$ is the $n^{th}$ rising Pochhammer symbol.

It is known that $\displaystyle\;C(+\infty) = S(+\infty) = \sqrt{\frac{\pi}{8}},\;$ we have

$$\sqrt{\frac{\pi}{8}}(1+i) - (C(x) + iS(x)) = \int_x^\infty e^{i z^2} dz$$

We can view the last integral as a contour integral from $x$ to $+\infty$. Since the $e^{it^2}$ factor decays to zero rapidly as $|z| \to \infty$ in the $1^{st}$ quadrant, we can deform the contour to one from $x$ to $e^{i\pi/4} \infty$ without changing its value.

Introduce parametrization $z = x \sqrt{1 + it}$, we find

$$\int_x^\infty e^{iz^2} dz = \frac{i x e^{ix^2}}{2} \int_0^\infty e^{-x^2 t} \frac{dt}{\sqrt{1+it}} = \frac{i x e^{ix^2}}{2} \int_0^\infty e^{-x^2 t}\underbrace{\sum_{n=0}^\infty \frac{(-it)^n \left(\frac12\right)_n}{n!}}_{\text{expansion of }1/\sqrt{1+it}} dt$$

Even though the expansion inside the rightmost integral is only valid for $t < 1$, the whole thing is in a form which we can apply Watson's Lemma. We can integrate the expansion term by term and deduce the asymptotic expansion in $(*1)$.

  • $\begingroup$ does that not give the expansion rather than the infinite series? $\endgroup$
    – James
    May 2 '14 at 14:38
  • $\begingroup$ I don't understand what you are referring to. In any event, since the Taylor expansion of $\frac{1}{\sqrt{1+it}}$ has a finite radius of convergence, after we integrate it term by term, the expansion we get in $(*1)$ is only an asymptotic expansion. It need not be a Taylor expansion in $\frac{1}{x}$ i.e the expansion in $(*1)$ need not converge for any finite $x$. $\endgroup$ May 2 '14 at 14:56
  • $\begingroup$ I was just meaning the question asks for the infinite series for the asymptotic expansion, and not another expansion so to speak. so was just basically checking that it answered the required question $\endgroup$
    – James
    May 2 '14 at 15:16
  • $\begingroup$ After you split out the real and imaginary part for $C(x)$ and $S(x)$, the expansion in $(*1)$ is the one you want. $\endgroup$ May 2 '14 at 15:20

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