I need to compute the mean value of this function:
${\displaystyle\int_0^\infty \exp\left(-\frac{a^2}{x^2}\right) I_0\left(\frac{a^2}{x^2}\right) \textrm{d}x}$
where $a$ is a constant and $I_0$ is the modified Bessel function of the first kind, but I have two main problems: I have no idea how to solve the integral and for the mean value I need to divide for the size of the integration interval which is infinite.
Now $x = k_1 + k_2 \tan \frac{\varphi}{2}$, where $k_1,k_2$ are constants, so I can solve the second problem by substituting and the mean value is given as:
$\frac{1}{\pi}{\displaystyle\int_0^\pi \exp\left[-\frac{a^2}{\left(k_1 + k_2 \tan \frac{\varphi}{2}\right)^2}\right] I_0\left[\frac{a^2}{\left(k_1 + k_2 \tan \frac{\varphi}{2}\right)^2}\right] \textrm{d}\varphi}$
Nevertheless this leads to a much more complicate integral and I have no clue on how to solve it.
If it can help the function inside the integral goes from $0$ to $1$ and its derivative should always be greater than $0$.
Wolfram Alpha gives the antiderivative (have a look here) $$I=\int\exp\left(-\frac{a^2}{x^2}\right) I_0\left(\frac{a^2}{x^2}\right) \textrm{d}x$$ $$I=\frac 1 x \exp\left(-\frac{a^2}{x^2}\right)\left(2 a^2 I_1\left(\frac{a^2}{x^2}\right)+\left(2 a^2+x^2\right) I_0\left(\frac{a^2}{x^2}\right)\right)$$ So, for $$J=\int_0^p\exp\left(-\frac{a^2}{x^2}\right) I_0\left(\frac{a^2}{x^2}\right) \textrm{d}x$$ $$J(p)=\frac 1 p \exp\left(-\frac{a^2}{p^2}\right)\left(2 a^2 I_1\left(\frac{a^2}{p^2}\right)+\left(2 a^2+p^2\right) I_0\left(\frac{a^2}{p^2}\right)\right)-2 \sqrt{\frac{2}{\pi }} a$$
Using the asymptotics of Bessel functions
$$J(p)=p+\frac{a^2}{p}-\frac{a^4}{4 p^3}+O\left(\frac{1}{p^5}\right)-2 \sqrt{\frac{2}{\pi }} a$$ and, assuming that I properly understood, $$\frac {J(p)} p=1-\frac{2a \sqrt 2 } {p\sqrt \pi }+\frac{a^2}{p^2}+O\left(\frac{1}{p^4}\right)$$
Using $a=\sqrt 2$ and $p=10^6$, the "exact" value is $$0.9999977432436658089748512$$ while the above truncated expansion gives $$\frac{500000000001}{500000000000}-\frac{1}{250000 \sqrt{\pi }}=0.9999977432436658089748522$$