TITLE: Study :$ \int_0^\infty e^{-t}\left[\frac{1}{\pi}\int_0^\pi e^{\frac{t}{n}\cos(\theta)}\cos\left(a\frac{t}{n}\sin(\theta)\right)d\theta\right]^ndt$
QUESTION [5 upvotes]: I have an integral defined as follows
$$K_n(a)=\int_0^\infty e^{-t}\left[\frac{1}{\pi}\int_0^\pi e^{\frac{t}{n}\cos(\theta)}\cos\left(a\frac{t}{n}\sin(\theta)\right)d\theta\right]^ndt$$
where $a\in[-1,1]$ and I want to find the behaviour of $K_n(a)$ around $a=0$. At $a=0$ it converges for $n\geq 3$ and is infinite otherwise, but converges for all $n\geq 1$ if $a\neq 0$.
The first derivative of $K_n(a)$ vanishes at $a=0$ and the second derivative at $a=0$ can be shown to be
$$\frac{d^2K_n}{da^2}\Big|_{a=0}=\int_0^\infty e^{-t}t\left[I_0\left(\frac{t}{n}\right)\right]^{n-1}I_1\left(\frac{t}{n}\right)dt$$
where $I_b$ is the $b$th modified Bessel function of the first kind.
This integral can be shown to converge for $n\geq 5$.
So here I have two questions:

$K_n(a)$ appears to be non-analytic (or maybe just not infinitely differentiable?) at $a=0$. Indeed we need greater and greater values of $n$ for higher derivatives to not be infinite/undefined. When the second derivative does exist am I justified in approximating $$K_n(a)\simeq K_n(0)+\frac{1}{2}\frac{d^2K_n}{da^2}\Big|_{a=0}a^2$$ in any meaningful way?
Would it be valid to consider


$$K_n(a>0)\simeq\lim_{a_0\searrow 0}K_n(a_0)+\frac{1}{2}\frac{d^2K_n}{da^2}\Big|_{a=a_0}(a-a_0)^2$$
to avoid the non-analytic point $a=0$?

How would I approximate the small $a$ behaviour when I cannot calculate any derivatives? i.e. for $d<5$. For instance I know (from another result) that $K_1(a)\sim 1/a$. How can I demonstrate this using this integral?

REPLY [2 votes]: A partial answer and several remarks. In the expression
\begin{equation}
K_n(a)=\int_0^\infty e^{-t}\left[\frac{1}{\pi}\int_0^\pi e^{\frac{t}{n}\cos(\theta)}\cos\left(a\frac{t}{n}\sin(\theta)\right)d\theta\right]^ndt
\end{equation} 
the inner integral can be written as
\begin{align}
J&=\frac{1}{\pi}\int_0^\pi e^{\frac{t}{n}\cos(\theta)}\cos\left(a\frac{t}{n}\sin(\theta)\right)d\theta\\
&=\frac{1}{2\pi}\int_0^{\pi}e^{\frac{t}{n}\cos(\theta)}\left[ e^{ia\frac{t}{n}\sin(\theta)}+ e^{-ia\frac{t}{n}\sin(\theta)}\right] d\theta\\
&=\frac{1}{2\pi}\int_0^{2\pi} e^{\frac{t}{n}\left(\cos(\theta)+ia\sin(\theta)\right)} d\theta
\end{align} Defining a real number $\varphi$ such that
\begin{equation}
\cos i\varphi=\frac{1}{\sqrt{1-a^2}}\quad;\quad\sin i\varphi=\frac{ia}{\sqrt{1-a^2}}
\end{equation} 
one may write
\begin{equation}
J=\frac{1}{2\pi}\int_0^{2\pi}e^{\frac{t}{n}\sqrt{1-a^2}(\cos(\theta-i\varphi)} d\theta
\end{equation} 
It can be seen that it does not depend on $\varphi$ as may be seen from the rectangular contour (integrals over vertical sides are opposite). Then, 
\begin{align}
J&=\frac{1}{2\pi}\int_0^{2\pi}e^{\frac{t}{n}\sqrt{1-a^2}\cos(\theta)} d\theta\\
&=I_0\left( \frac{t}{n}\sqrt{1-a^2} \right)
\end{align}
where we have used an integral representation of the modified Bessel function. It remains to evaluate
\begin{align}
K_n(a)&=\int_0^\infty e^{-t}\left[I_0\left( \frac{t}{n}\sqrt{1-a^2} \right)\right]^ndt\\
&=\frac{\sqrt{1-a^2}}{n}\int_0^\infty e^{-x\frac{n}{\sqrt{1-a^2}}}\left[I_0\left( x\right)\right]^ndx
\end{align}
In order to simplify the asymptotic analysis, this expression can be written as
\begin{equation}
K_n(a)=\frac{n}{\sqrt{1-a^2}}\int_0^\infty e^{-nx\left(\frac{1}{\sqrt{1-a^2}} -1 \right)}\left[e^{-x}I_0\left( x\right)\right]^ndx
\end{equation} 
For $x\to\infty$,
$$f(x)=\left[e^{-x}I_0\left( x\right)\right]^n\sim (2\pi x)^{-\tfrac{n}{2}}$$
As claimed in the question, it is clear that the integral is finite when $a\ne0$ or, if $a=0$, when $n>2$.
Denoting $\lambda=n\left(\frac{1}{\sqrt{1-a^2}} -1 \right)$, finding the behaviour of the integral near $a=0$ is equivalent to describe the Laplace transform of $f(x)$ near the origin, as $\lambda\sim na^2/2$. This problem can be tackled using the Mellin transform method (see for example Asymptotic expansions of integrals, N. Bleistein and R.A. Handelsman ex. 4.6.1). The result is recalled here:
if, for $t\to \infty$,
\begin{equation}
f(t)\sim\sum_{m=0}^\infty c_mt^{-r_m}
\end{equation} 
where no $r_m$ is a positive integer,
\begin{equation}
\int_0^\infty e^{-\lambda t}f(t)dt\sim \sum_{m=0}^\infty c_m\Gamma(1-r_m) \lambda^{r_m-1}+\sum_{m=0}^\infty \lambda^m\frac{(-1)^m}{m!}M[f;m+1]
\end{equation} 
for $\lambda\to 0$. Here, $M[f;m+1]$ is the Mellin transform of $f$.
The leading term for $0<n<2$, is $\lambda^{r_m-1}$:
\begin{equation}
K_n(a)\sim \frac{n}{\sqrt{1-a^2}}\left( 2\pi \right)^{-\frac{n}{2}}\Gamma\left( 1-\frac{n}{2} \right)\lambda^{\frac{n}{2}-1}
\end{equation} 
For $n=1$
\begin{align}
K_1(a)&\sim \frac{1}{a}\left( \frac{1+\sqrt{1-a^2}}{2\sqrt{1-a^2}} \right)^{\frac{1}{2}}\\
&\sim \frac{1}{a}
\end{align} 
as noted in the question. The asymptotic expansion contains  integer and non-integer powers of $\lambda$. The latter can be easily calculated, however to obtain the former, the method  requires to calculate the Mellin transform of $f$, which seems to be very involved. This result is also required to obtain the cases $n>2$ where the leading term is $\lambda^0$ (the integral converges). 
Non-integer exponents are responsible for the divergence of higher derivatives. The smaller non-integer exponent being $\frac{n}{2}-1$, derivative of even order larger than $\frac{n}{2}-1$ diverge. To answer question 1, it is legitimate to use the first terms of the expansion as a truncated  series. One may also remark that when $n$ is an even integer, the expansion involves logarithmic terms.
Edit:
The case $n=2$ can be exactly worked out. Using the integral representation DLMF
\begin{equation}
\left[I_{0}\left(z\right)\right]^2=\frac{2}{\pi}\int_{0}^{\frac{1}{2}
\pi}I_{0}\left(2z\cos\theta\right)\,d\theta
\end{equation} 
one may express 
\begin{align}
K_2(a)&=\frac{4}{\pi\sqrt{1-a^2}}\int_0^\infty e^{-x\frac{2}{\sqrt{1-a^2}} }\int_{0}^{\frac{1}{2} \pi}I_{0}\left(2x\cos\theta\right)\,d\theta\,dx\\
&=\frac{4}{\pi\sqrt{1-a^2}}\int_{0}^{\frac{1}{2} \pi}\,d\theta\int_0^\infty e^{-x\frac{2}{\sqrt{1-a^2}} }I_{0}\left(2x\cos\theta\right)\,dx\\
&=\frac{2}{\pi\sqrt{1-a^2}}\int_{0}^{\frac{1}{2} \pi}\frac{d\theta}{\cos\theta}
\int_0^\infty e^{-\frac{t}{\cos\theta\sqrt{1-a^2}} }I_{0}\left(t\right)\,dt
\end{align} 
We use the Laplace transform of $I_0(t)$ which is $\left( p^2-1 \right)^{-1/2}$ (here $p=\frac{1}{\cos\theta\sqrt{1-a^2}}>1$) to obtain
\begin{align}
K_2(a)&=\frac{2}{\pi\sqrt{1-a^2}}\int_{0}^{\frac{1}{2} \pi}\frac{d\theta}{\sqrt{\frac{1}{1-a^2}-\cos^2\theta}}\\
&=\frac{2}{\pi}\int_{0}^{\frac{1}{2} \pi}\frac{d\theta}{\sqrt{1-\left( 1-a^2 \right)\sin^2\theta}}\\
&=\frac{2}{\pi}K\left( \sqrt{1-a^2} \right)\\
&=\frac{2}{\pi}K'\left( a \right)
\end{align}
where $K(.)$ and $K'(.)$ are complete elliptic integrals. Near $a=0$,
\begin{equation}
K_2(a)=\frac{2}{\pi}\left(2\ln2-\ln a^2+a^2(\frac{1}{2}\ln2-\frac{1}{8}\ln a^2-\frac{1}{4})+O\left( a^4 \right)  \right)
\end{equation} 
As expected, a logarithmic term is present. Alternatively, the Laplace transform of $\left[I_0(x)\right]^2$ is tabulated or obtained with a CAS.
For $n=1$, the result is even simpler:
\begin{align}
K_1(a)&=\frac{1}{\sqrt{1-a^2}}\int_0^\infty e^{-\frac{x}{\sqrt{1-a^2}}}I_0(x)\,dx\\
&=\frac{1}{\sqrt{1-a^2}}\frac{1}{\sqrt{\frac{1}{1-a^2}-1}}\\
&=\frac{1}{\left|a\right|}
\end{align}