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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?

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    $\begingroup$ If $z \ll \sqrt{n+1}$, then $K_n(z)\sim (n-1)!2^{n-1} z^{-n}$. For any fixed $x,y,v$, it will eventually be true that $vy \ll \sqrt{n+1}$, so this should be enough to determine pointwise convergence. $\endgroup$ Jun 16, 2023 at 13:59
  • $\begingroup$ @eyeballfrog Thanks. A canonical answer would be highly appreciated and rewarded. $\endgroup$
    – keynes
    Jun 16, 2023 at 14:01

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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.

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  • $\begingroup$ Thank you very much. The case $x=y$ would also be of interest in case you have some spare time. It is surprising that the series is converging only when $x<y$ even though the original integral is convergent for $x>0$ or $y>0$. In reality, $x \in \mathbb{C}$ so perhaps the condition for convergence here would be $\Re(x)<y$. A nice proof from a plasma physicist by the way ;) $\endgroup$
    – keynes
    Jun 16, 2023 at 15:11
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    $\begingroup$ @Staufenberg Should be $|x| < y$ I would think, because that's still the convergence condition of $(x/y)^{2n}$ when $x$ can be complex. The $|x| = y$ case would still require further analysis, except worse because the result might depend on the phase of $x$. $\endgroup$ Jun 16, 2023 at 15:15
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    $\begingroup$ @Staufenberg Some numerical calculations suggest the $|x| = y$ case is still convergent, with terms going as $n^{-2}$. Proving this is more troublesome. $\endgroup$ Jun 16, 2023 at 15:54

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