# $\int_0^T t^{2a}K_a(t)^2\;dt=\text{ ?}$ where $K_a$ is modified Bessel function of second kind?

Let $K_a$ be the modified Bessel function of second kind, with $a>0$ a real number.

Is there a nice expression for $$\int_0^T t^{2a} K_a(t)^2\;dt,$$ where $T < \infty$?

The expression for $K_a$ is a horrible looking infinite sum, so I was hoping for a reference. I tried various books containing tables of such integrals but wasn't able to find the one I wanted.

• What are your thoughts? – tired Oct 29 '15 at 18:20
• @tired It looks too difficult to do by hand because $K_a$ has a nasty expression. So I was hoping to find a reference for this info. – TagCharacters Oct 29 '15 at 18:21
• $$\frac{\pi \Gamma \left(a+\frac{1}{2}\right) \Gamma \left(2 a+\frac{1}{2}\right)}{4 \Gamma (a+1)}$$ is the result given by mathematica if $T=\infty$, looks not so bad – tired Oct 29 '15 at 18:28
• @tired: that result can be recovered from Parseval's theorem and the orthogonality of Bessel's polynomials (en.wikipedia.org/wiki/Bessel_polynomials) – Jack D'Aurizio Oct 29 '15 at 19:45
• @JackD'Aurizio thanks! i figured it a few minutes ago.. :) – tired Oct 29 '15 at 21:55

$t^a K_a(t)$ is a multiple of the Fourier transform of $\frac{1}{(1+s^2)^{a+1/2}}$, hence by Parseval's identity the integral for $T=+\infty$ just depends on the integral: $$\int_{0}^{+\infty}\frac{ds}{(1+s^2)^{2a+1}}=\frac{\sqrt{\pi}\,\Gamma\left(2a+\frac{1}{2}\right)}{2\,\Gamma(2a+1)}$$ that can be computed by setting $s=\tan(\theta)$ then recalling the properties of Euler's beta function. For large $a$s we have:
$$\int_{0}^{+\infty}t^{2a}K_a(t)^2\,dt \approx \left(\frac{2a}{e}\right)^{2a}\sqrt{\frac{\pi^3}{8a}}.$$
• Thanks but what about if $T$ is infinite? – TagCharacters Oct 29 '15 at 22:55
• @TagCharacters: it is the only case I dealt with, $T$ being plus infinity. – Jack D'Aurizio Oct 29 '15 at 22:58
• Ok. Do you know if how to calculate $\int_0^T tK_a(t)^2$? It's simpler than the other one – TagCharacters Oct 30 '15 at 18:10