From the answer of this question, it has been shown that the infinite integral
$$
f_v(x,y) = \int_0^\infty e^{-xu-y \sqrt{u^2+v^2}} \, \mathrm{d} u
$$
can conveniently be expressed in terms of infinite series involving modified Bessel functions of the second kind of the form
$$
f_v(x,y) = \sum_{n = 0}^\infty v
\left( \frac{x^2 v}{y} \right)^n
\left( \frac{2^{-n}}{n!}\, K_{n+1}(vy) – \frac{x}{2n+1} \frac{2^n n!}{(2n)!} \left( \frac{2v}{\pi y} \right)^\frac{1}{2} \, K_{n+\frac{3}{2}} (vy) \right) ,
$$
where $x \in \mathbb{C}$ and $y,v \in \mathbb{R}_+^*$.
In particular, for $x=0$, it follows readily that
$$
f_v (0,y) = \int_0^\infty e^{-y \sqrt{u^2+v^2}} \, \mathrm{d}u = v K_1(vy) \, .
$$
I would like to know under which circumstances the series converges to the above function.
For instance, for $x=3/2$, $y=4/5$, and $v=1/2$, it can be checked that the series is diverging.
What I know is the asymptotic expansion of the factorial $n! \sim \sqrt{2\pi n} (n/e)^n$ as $n \to \infty$.
However, modified Bessel functions of the second kind do not have asymptotic expansions as $n \to \infty$ apparently.
Any thoughts on how to proceed with this?
Best Answer
Before we begin, it will be useful to write the second term using the Gamma function instead of the factorial function through the Gamma duplication formula: $$ \Gamma\left(n+\frac{1}{2}\right) = 2^{1-2n}\sqrt{\pi}\frac{\Gamma(2n)}{\Gamma(n)} = 2^{-2n}\sqrt{\pi}\frac{(2n)!}{n!}, $$ which gives $$ \frac{x}{2n+1} \frac{2^n n!}{(2n)!}\sqrt{\frac{2v}{\pi y}} = \frac{x}{2n+1} \frac{\sqrt{\pi}2^{-n}}{\Gamma(n+1/2)}\sqrt{\frac{2v}{\pi y}} = \frac{2^{-n-1/2}}{\Gamma(n+3/2)}x\sqrt{\frac{v}{y}}. $$ Thus the terms of the sum can now be written in the more symmetric form $$ v \left( \frac{x^2 v}{y} \right)^n \left[ \frac{2^{-n}}{\Gamma(n+1)}\, K_{n+1}(vy) - \frac{x}{y}\frac{2^{-n-1/2}\sqrt{vy}}{\Gamma(n+3/2)} \, K_{n+\frac{3}{2}} (vy) \right], $$
which is particularly convenient because Bessel $K_n(z)$ has the asymptotic form $K_n(z) =\Gamma(n)2^{n-1}z^{-n}$ when $z\ll \sqrt{n+1}$. Since the sum goes to infinity, for any fixed value of $x,y,v$ the sum will eventually reach the regime where $vy\ll \sqrt{n+1}$, so we should be able to use the asymptotic form to do a limit comparison test. Replacing $K$ with its asymptotic form greatly simplifies the terms of the sum to $$ v \left( \frac{x^2 v}{y} \right)^n \left[ \frac{1}{(vy)^{n+1}}- \frac{x}{y}\frac{1}{(vy)^{n+1}} \right] = \left( \frac{x}{y} \right)^{2n}\frac{y-x}{y^2} , $$ which means the series can be shown to converge whenever $y > x$ and diverge whenever $y < x$ by a limit comparison test against $(x/y)^{2n}$. The $y = x$ case would require a more detailed analysis due to the cancellation between the asymptotic forms, which I can do later if you really want to know that case.