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What is the asymptotic solution (for $s\gg 1$) of the following integral equation $$z(s)=1+\gamma\int\limits_{-\infty}^s ds_1\int\limits_{-\infty}^{s_1}ds_2 \cos{(s_1^2-s_2^2)}z(s_2)\;?$$ In fact I need to show that $$\lim_{s\to\infty} z(s)=2\exp{\left(\frac{\pi\gamma}{4}\right)}-1.$$ The integral equation is equivalent to the following third order differential equation $$sz^{\prime\prime\prime}-z^{\prime\prime}-s(\gamma-4s^2)z^{\prime}+ \gamma z=0,$$ and the initial conditions $z(-\infty)=1,\,z^\prime(-\infty)=0,\,z^{\prime\prime}(-\infty)=\gamma$.

The question arose in the context of remarkable connection between the Landau-Zener problem and the ball rolling along the Cornu spiral (that's how I do now what the $\lim_{s\to\infty} z(s)$ should be) established by Bloch and Rojo in http://ajp.aapt.org/resource/1/ajpias/v78/i10/p1014_s1

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The paper cited by OP can be freely downloaded at dept.math.lsa.umich.edu/~abloch/RollingSpheres17.pdf. –  Jon Sep 24 '13 at 12:59
    
Try substituting $ z = \exp( f(s) ) $ in the differential equation and then look for terms that balance as $ s \rightarrow \infty $. This often allows for an asymptotic solution to be found. –  Tom Dickens Sep 25 '13 at 3:28
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The idea is to iterate on the integral equation starting with $z(s)=1$. This will give the integrals in a series $$ I_{n}=\int_{-\infty }^{\infty }ds_{1}\int_{-\infty }^{s_{1}}ds_{2}\cdots \int_{-\infty }^{s_{2n-1}}ds_{2n}\;\cos { (s_{1}^{2}-s_{2}^{2})}\;\cdots \cos {(s_{2n-1}^{2}-s_{2n}^{2})}. $$ The paper discussing them is this one. The general formula is $$ I_n=\frac{2}{n!}\left(\frac{\pi}{4}\right)^n $$ and you should take into account also powers of $\gamma$. These are the terms of the power series of the exponential multiplied by a factor 2.

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My goal was to calculate $I_n$ in a different way (without using the known Landau-Zener formula) than discussed in the paper. I hope some method exists to get the asymptotic behaviour of $z(s)$ from the integral equation quoted (or from the equivalent differential equation). Then this will give $I_n$ and therefore another derivation of the Landau-Zener formula, which was the final goal. –  Zurab Silagadze Sep 24 '13 at 15:48
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