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?
Best Answer
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 $I_1$:
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:
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.