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The integral I'm looking to solve is of the form

$$\int_x^{\infty} K_{0}(a\sqrt{r^2 + b^2})\,dr$$

where $K_{0}$ is the order 0 modified Bessel function of the second kind. I'm assuming $x > 0$.

I've been poring over Section 6.5 of Gradshteyn and Ryzhik's "Table of Integrals, Series, and Products" for a while now, trying to make a substitution that will translate the above integral into one of the formulas they have listed. In particular, I got close with Section 6.592(12) (in the 7th edition), a special case of which is:

$$\int_1^{\infty} (x - 1)^{-1/2}K_{0}(a\sqrt{x})\,dx = \Gamma(-1/2)*2^{1/2}*a^{-1/2}*K_{-1/2}(a)$$

The substitution $u = r^2+b^2$ into the original integral almost yielded something I could apply 6.592(12) to, but I wasn't quite able to make it work. I tried some other substitutions, but this one was the closest I got to something on the pages of GR.

For context, the integrand I'm working with is a probability density (I've omitted the normalizing constant above). The above integral is a key part of my attempt at getting a closed form CDF. I've already managed to find all the moments of the distribution using 6.596(3) of GR. But that was much easier for me, since the bounds of those integrals were all $(0, \infty)$.

From my attempts at solving this, I've noticed that a major hurdle here seems to be that $x \ne 0$. A vast majority of the GR formulas in this chapter have integration limits $(0, \infty)$. But I haven't been able to find a way to use any of these due to $x$. How would you approach this?

Edit: took a look through this thread. However, the arguments in the comments only seem to apply when the lower bound is 0.

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1 Answer 1

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It seems that formula $6.677.5$ works if $\gamma=0$.

Assuming that $a$ and $b$ are positive, this would give $$\int_0^{\infty} K_{0}(a\sqrt{r^2 + b^2})\,dr=\frac{\pi }{2 a} \,e^{-a b}$$

So, what would remain it $$\int_0^{x} K_{0}(a\sqrt{r^2 + b^2})\,dr$$ and, hoping that $x$ is small, we could try a series solution (have a look here and simplify).

Using, for a quick test, $a=\pi$, $b=e$,and $x=1$ with the linked formula to $O(r^8)$, this would give for the last integral $6.8691\times 10^{-5}$ while numerical integration gives $6.8780\times 10^{-5}$.

Hoping and wishing that it could help.

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  • $\begingroup$ Thanks for the reply. However, x could be arbitrarily large. The reason I'm trying to evaluate this integral is to obtain a cumulative distribution function. The integrand in my post is a PDF (where I've omitted the normalizing constant) $\endgroup$
    – user62348
    Commented Jun 7, 2023 at 12:39

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