# Arnold's Trivium problem 52

Calculate the first term of the asymptotic expression as $k \to \infty$ of the integral

$$\int_{-\infty}^{+\infty}\frac{e^{ikx}}{\sqrt{1+x^{2n}}}dx$$

May I bother you to explain what the problem is asking and what is the intuition behind in this problem?

It is from the problem 52 of the article A mathematical trivium by Arnol'd.

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Perhaps it is asking for a power series expansion? –  Potato Dec 7 '11 at 5:04
The intuition is that $k$ is playing the rôle of the frequency of a sinusoid and when $k\to\infty$ the values of $1/\sqrt{1+x^{2n}}$ get averaged to zero. The question is to find a simple equivalent of the integral, for example, something like $C/k^4$ or $Ce^{-k^2}$ or... –  Did Dec 7 '11 at 7:20
I think that can help the result for $n=1$ as this integral, being equivalent to $2\int_0^\infty dx\frac{\cos(kx)}{\sqrt{1+x^{2n}}}$, is just $2K_0(k)$, a Bessel function with a first asymptotic term $\sqrt{\frac{\pi}{2k}}e^{-k}$. As said above, in agreement with Riemann-Lebesgue lemma, this integral is expected to go to zero for increasing $k$. –  Jon Dec 7 '11 at 14:51
Note that the OP asks to explain what the problem is asking and what is the intuition behind in this problem, not a full solution nor even the answer to the problem. –  Did Dec 7 '11 at 19:11
@Didier: You're quite right. I think I noticed that at first, then started thinking about the problem itself anyway, and then forgot about it when I'd solved the problem. By the way, considering that, your comment could have been an answer. –  joriki Dec 8 '11 at 20:20

I'm not sure this is the best way to do this, but here's one way using contour integrals:

We can complete the integral by a half-circle at infinity in the upper half plane. However, this contour would have to cross branch cuts that emanate from the zeros of the radicand. To avoid this, we can choose the branch cuts as vertical lines beginning at the zeros and leading away from the real axis, and whenever we hit one at infinity, we follow it down to the zero, go around the zero on an infinitesimal circle, and then follow the branch cut back up to infinity. There's no contribution from the infinitesimal circle since the perimeter goes with the radius and the function value goes with the inverse square root of the radius. The function value changes sign as we go around the zero, and the two integrals along the branch cut have opposite orientations, so they both yield the same contribution. There are no poles within this contour, so the integral we want is simply the sum of all the integrals along the branch cuts.

Since the integrand decays exponentially, to get the leading term in the asymptotic expansion we only have to consider the zeros closest to the real line. The first of these is $z_0=\exp(\mathrm i\pi/2n)$; the contribution of the other one at $\exp(\mathrm i\pi-\mathrm i\pi/2n)$ is the complex conjugate. Thus, we're looking for

$$\begin{eqnarray} I=4\Re\int_0^{\;\mathrm i\infty}\frac{\mathrm e^{\mathrm ik(z_0+z)}}{\sqrt{1+(z_0+z)^{2n}}}\mathrm dz\;. \end{eqnarray}$$

For large $k$, only the immediate neighbourhood of $z_0$ contributes to the integral, so we can approximate the radicand linearly:

$$\begin{eqnarray} I &\approx& 4\Re\int_0^{\;\mathrm i\infty}\frac{\mathrm e^{\mathrm ik(z_0+z)}}{\sqrt{2nz_0^{2n-1}z}}\mathrm dz \\ &=& 4\Re\frac{\mathrm e^{\mathrm ikz_0}}{\sqrt{2nz_0^{2n-1}}}\int_0^{\;\mathrm i\infty}\frac{\mathrm e^{\mathrm ikz}}{\sqrt{z}}\mathrm dz \\ &=& 4\Re\mathrm e^{\mathrm ikz_0}\sqrt{-\frac{\mathrm i\pi z_0}{2nk}}\;, \end{eqnarray}$$

where the square root takes the principal value. This leading term decays with $\exp(-k\sin(\pi/2n))$.

In the case $n=1$ (which Jon addressed in a comment), there is only a single zero in the upper half plane at $z_0=\mathrm i$, so the above result has to be divided by $2$.

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Nice solution! I was wondering how one was meant to do this. –  David Speyer Dec 8 '11 at 17:08