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For $k\in\mathbb{N}_{0}$ and $x\in\mathbb{R}$, define $$I_{k}(x):=\int_{0}^{\pi/2}\cos(xg(\theta))\sin^{2k}\theta\,\mathrm{d}\theta$$ where $$g(\theta)=\int_{\sin\theta}^{1}\frac{\mathrm{d}t}{\sqrt{(1-t^{2})(1-\alpha^{2}t^{2})}}$$ and $0<\alpha<1$ is a fixed parameter.

I arrived at the expression $I_{k}(x)$ while working on certain asymptotic analysis of Jacobi elliptic functions. I doubt that the integral $I_{k}(x)$ could be evaluated somehow (even in terms of some known special functions). Nevertheless, for my purpose, it would be sufficient to know the asymptotic behavior of $I_{k}(x)$, as $k\to\infty$. It seems, however, that for my calculations, I need to know not only the leading term of the asymptotic expansion, but the second term as well.

Does anybody know how to derive the asymptotic expansion for $I_{k}(x)$, as $k\to\infty$? Many thanks.

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With the aid of Mathematica I found that $$g(\theta)=\textrm{EllipticK}(a^2)-\textrm{EllipticF}(t,a^2).$$ I get the first terms of the asymptotic expansion $$\frac{\sqrt{\pi}}{2\sqrt{k}}-\frac{\sqrt{\pi} x^2}{8(1-a^2)k^{3/2}}+ \Bigl(-\frac{k}{6}+\frac{a^2 x^2}{6(1-a^2)^2}+\frac{x^4}{24(1-a^2)^4}\Bigr) \frac{3\sqrt{\pi}}{8k^{5/2}}$$ For example with $a=2/3$, $x=1/5$ and $k=1000000$ Mathematica evaluates $$I_k(x)=0.00088622679872235827901$$ The first, second and third approximations are equal to $$0.00088622692545275801365,\quad 0.00088622690950067335550, \quad 0.00088622679872231419831.$$ I used the standard Laplace method. I expect my computations are correct. I do not make adequate checks.

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  • $\begingroup$ Thanks for the clue, the Laplace method seems to be indeed efficient here. $\endgroup$ – Twi Sep 14 '15 at 18:41

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