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I'm looking at the integral

$$I(\alpha,f)=\int_{-\pi}^\pi K_0 \left ( \alpha \sqrt{1+f(x)^2-2f(x)\cos(x)} \right ) dx.$$

Here:

  • $\alpha \gg 1$.
  • $f$ is a smooth, $2\pi$-periodic function. $f(0)=1$. The range of $f$ is $[m,M]$ with $0<m \leq 1 \leq M<\infty$.
  • $K_0$ is a modified Bessel function of the second kind.

I'd like to derive more than one term of an asymptotic expansion of $I$ for large $\alpha$ and fixed $f$.

Because of exponential decay of $K_0(z)$ for large $z$, a leading order asymptotic comes about by simply setting $f \equiv 1$, giving

$$I \sim \int_{-\pi}^\pi K_0 \left ( \alpha \sqrt{2-2\cos(x)} \right ) dx = 2 \pi K_0(\alpha) I_0(\alpha)$$

where $I_0$ is a modified Bessel function of the first kind. This result follows from the Graf addition formula:

$$K_0 \left ( \alpha \sqrt{2-2\cos(x)} \right ) = \sum_{n \in \mathbb{Z}} K_n(\alpha) I_n(\alpha) e^{inx}.$$

The Graf formula may also be applied directly to the integrand of $I$, resulting in

$$K_0 \left ( \alpha \sqrt{1+f(x)^2-2f(x)\cos(x)} \right ) = \sum_{n \in \mathbb{Z}} K_n(\alpha \max \{ 1,f(x) \}) I_n(\alpha \min \{ 1,f(x) \}) e^{inx}.$$

The problem is twofold. First, directly integrating the RHS term-by-term is nontrivial due to the "switching" at the points where $f(x)=1$. Second, although we could try to approximate the integrals of the individual terms on the RHS, the RHS is also delocalized in $n$. Specifically, the smallness of the LHS for large $x$ is achieved only through cancellation on the RHS, not smallness of the individual terms on the RHS. This means that accurately estimating the sum by estimating the individual terms requires extremely accurate estimation of a large number of individual terms. This isn't really a surprise: it is more or less the uncertainty principle in action (even though strictly speaking the RHS is not a Fourier series except when $f \equiv 1$).

Overall, as far as I can tell the Graf formula is not useful in this situation, except perhaps for numerics.

Generally the type of result I would expect would use only local features of $f$ near $x=0$, for example $f^{(k)}(0)$ for a few values of $k$.

Can anything else be done here?

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  • $\begingroup$ This looks a little similar (though, I admit, the case is different because the other Bessel function has exponential growth instead of decay) math.stackexchange.com/q/1154372/269624. $$ $$ Still, could the integrals be related to the same application? $\endgroup$ – Yuriy S Oct 21 '18 at 20:14
  • $\begingroup$ @YuriyS I mean just that, the range of $f$, not the domain of $f$. Also, the general motif here and in your link is Bessel functions evaluated at secant lengths between two points given in polar coordinates. In my setting this comes up in PDE, where $G(\mathbf{x},\mathbf{y})=K_0(|\mathbf{x}-\mathbf{y}|)$ is the fundamental solution of a certain elliptic PDE in two dimensions. $\endgroup$ – Ian Oct 21 '18 at 20:26
  • $\begingroup$ @ Ian, sorry, English not my first language, so I confused the two terms. $\endgroup$ – Yuriy S Oct 21 '18 at 20:27
  • $\begingroup$ @YuriyS No problem, I appreciate the consideration. $\endgroup$ – Ian Oct 21 '18 at 20:29
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This is just an idea and whether it works or not, heavily depends on the exact function $f(x)$.

$$K_0(a)=\int_0^\infty e^{-a \cosh t} dt$$

So we can write:

$$I(\alpha,f)=\int_0^\infty \int_{-\pi}^\pi e^{-\alpha ~g(x) \cosh t} dx~ dt$$

Where:

$$g(x)=\sqrt{1+f(x)^2-2f(x)\cos(x)}=g(2 \pi+x)$$

$$g(0)=0$$

Let's separate the integral into two parts and deal only with the first one for now:

$$I_1=\int_0^\infty \int_0^\pi e^{-\alpha ~g(x) \cosh t} dx~ dt$$

Let's assume that $g(x)$ is monotone on the interval $[0,\pi]$. Then, considering the properties of $f(x)$, there exists the inverse function $g^{-1} (y)=h(y)$ such that:

$$y=g(x)$$

Then we can make a substitution to get for the inner integral:

$$\int_0^\pi e^{-\alpha ~g(x) \cosh t} dx=\int_0^{g(\pi)} e^{-\alpha ~\cosh t~y}~ h'(y) ~dy$$

Now assume that:

$$\int_0^{g(\pi)} |h'(y)| ~dy < \infty $$

or:

$$|h'(y)| < A e^{a y}, \qquad \text{for all } y >0$$

And it can be represented as:

$$h'(y) = y^{\lambda} \sum_{n=0}^\infty \frac{c_n}{n!} y^n, \qquad c_0 \neq 0, \qquad \lambda > -1$$

Then by Watson's lemma, for $\alpha \to \infty$ we have:

$$\int_0^{g(\pi)} e^{-\alpha ~\cosh t~y}~ h'(y) ~dy \asymp \sum_{n=0}^\infty \frac{c_n \Gamma(\lambda+n+1)}{n!~ \alpha^{\lambda+n+1} ~\cosh^{\lambda+n+1} t}$$

Then we obtain for $I_1$:

$$I_1 \asymp \sum_{n=0}^\infty \frac{c_n \Gamma(\lambda+n+1)}{n!~ \alpha^{\lambda+n+1} } \int_0^\infty \frac{dt}{\cosh^{\lambda+n+1} t}$$

The integrals all converge.

The part of the original integral from $-\pi$ to $0$ in principle can be treated the same way. If $f(x)$ is even, then $I_1=I_2$, so we just multiply the asymptotic expression by $2$.


While this gives us as many asymptotic terms as we want, there's still a few challenges:

(1) Finding $h(y)$, or actually, only $\lambda$ and $c_n$ from the power series representation of $h'(y)$, if it exists.

(2) Making sure the conditions for Watson's lemma are satisfied.


Let's try an example with $f(x)=1$. We have:

$$g(x)=\sqrt{2} \sqrt{1-\cos x}$$

$$\sqrt{1-\cos x}=\frac{y}{\sqrt{2}}$$

$$\cos x=1-\frac{y^2}{2}$$

$$x=\arccos \left(1-\frac{y^2}{2} \right)=h(y)$$

$$h'(y)=\frac{2}{\sqrt{4-y^2}}=1 + \frac{y^2}{8} + \frac{3y^4}{128} + \frac{5y^6}{1024} + \cdots$$

$$g( \pi)=2$$

$$\lambda = 0$$

So we have:

$$I_1 \asymp \sum_{n=0}^\infty \frac{c_{2n}}{\alpha^{2n+1} } \int_0^\infty \frac{dt}{\cosh^{2n+1} t}$$

$$c_{2n}=1, \frac{2!}{8}, \frac{3 \cdot 4!}{128}, \frac{5 \cdot 6!}{1024}, \ldots$$

Numerically, for this case, Mathematica gives $I_1=I_2$, and gives a very good agreement with the asymptotic even for the first order:

enter image description here

Blue line is:

$$1/\left( \int_0^\pi K_0 (\alpha \sqrt{2-2 \cos x} )dx \right)$$

Orange line is:

$$\frac{2 \alpha}{ \pi}$$

as:

$$\int_0^\infty \frac{dt}{\cosh t}=\frac{\pi}{2}$$


I hope this might be helpful.

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  • $\begingroup$ The idea of changing variables to $g(x)$ on each of $[0,\pi]$ and $[-\pi,0]$ separately and then using this approach (which is essentially the corrected version of the Laplace method) is something that came to mind for me. It's unfortunately a somewhat awkward constraint: the sign of $g'$ is the sign of $f(x) f'(x) - f'(x) \cos(x) + f(x) \sin(x)$. Do you have an idea for a nice sufficient condition for this? $\endgroup$ – Ian Oct 21 '18 at 22:11
  • $\begingroup$ The approach requires you to remove the complexity from the argument of $K_0$ by changing variables, moving that complexity to "outside" where Watson's lemma can manage it. As written it breaks down immediately if $g$ isn't injective. Now in principle you could split the integration into subdomains where $g$ is injective and argue that only the one containing $x=0$ contributes appreciably for large $\alpha$. But I haven't been able to figure out how to carry out that approach. $\endgroup$ – Ian Oct 21 '18 at 22:17
  • $\begingroup$ Note that the examples where this comes up are somewhat odd geometries, for instance $f(x)=1+0.75\sin(x)$. $\endgroup$ – Ian Oct 21 '18 at 22:19
  • $\begingroup$ I gave this a more sincere thought and I realized that unless there is a minimum other than $x=0$ with $g(x)=0$, eventually an integral between a pair of critical points neither of which are $x=0$ will be exponentially small in $\alpha$. $\endgroup$ – Ian Oct 22 '18 at 19:06
  • $\begingroup$ @Ian, if $g(x) > 0$ for $x \neq 0$ then almost surely yes $\endgroup$ – Yuriy S Oct 22 '18 at 19:37
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We have $$g(x) = 1 + f^2(x) - 2 f(x) \cos x = (1 - f(x))^2 + 2 f(x) (1 - \cos x) = 0 \supset x = 0.$$ It is sufficient to approximate the integrand near $x = 0^+$: $$\sqrt {g(x)} \sim c_1 x + c_2 x^2 + c_3 x^3 = \\ \sqrt {1 + f'(0)^2} \,x + \frac {f'(0) + f'(0) f''(0)} {2 \sqrt {1 + f'(0)^2}} x^2 + \\ \frac {-1 - 4 f'(0)^2 + 6 f''(0) + 3 f''(0)^2 + 4 f'(0) f^{(3)}(0) + 4 f'(0)^3 f^{(3)}(0)} {24 (1 + f'(0)^2)^{3/2}} x^3, \\ \int_{-\pi}^\pi K_0 \!\left( \alpha \sqrt {g(x)} \right) dx \sim 2 \int_0^\infty \left( \frac 1 {c_1} + \frac {3 (2 c_2^2 - c_1 c_3) \xi^2} {c_1^5} \right) K_0(\alpha \xi) d\xi = \\ \frac \pi {c_1 \alpha} + \frac {3 \pi (2 c_2^2 - c_1 c_3)} {c_1^5 \alpha^3}.$$

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  • $\begingroup$ What happened to the $\alpha^{-2}$ term? Are you arguing it cancels out by symmetry considerations? $\endgroup$ – Ian Oct 23 '18 at 16:26
  • $\begingroup$ Correct, the expansion of $\sqrt {g(x)}$ at $x = 0^-$ is $-c_1 x - c_2 x^2 - c_3 x^3$; the $\xi^1$ terms in the expansion of the derivative of the inverse function at zero will have opposite signs. $\endgroup$ – Maxim Oct 23 '18 at 18:31

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