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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.

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  • $\begingroup$ What are your thoughts? $\endgroup$ – tired Oct 29 '15 at 18:20
  • $\begingroup$ @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. $\endgroup$ – TagCharacters Oct 29 '15 at 18:21
  • $\begingroup$ $$\frac{\pi \Gamma \left(a+\frac{1}{2}\right) \Gamma \left(2 a+\frac{1}{2}\right)}{4 \Gamma (a+1)}$$ is the result given by mathematica if $T=\infty$, looks not so bad $\endgroup$ – tired Oct 29 '15 at 18:28
  • $\begingroup$ @tired: that result can be recovered from Parseval's theorem and the orthogonality of Bessel's polynomials (en.wikipedia.org/wiki/Bessel_polynomials) $\endgroup$ – Jack D'Aurizio Oct 29 '15 at 19:45
  • $\begingroup$ @JackD'Aurizio thanks! i figured it a few minutes ago.. :) $\endgroup$ – tired Oct 29 '15 at 21:55
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$t^a K_a(t)$ 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:

$$ \int_{0}^{+\infty}t^{2a}K_a(t)^2\,dt \approx \left(\frac{2a}{e}\right)^{2a}\sqrt{\frac{\pi^3}{8a}}.$$

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  • $\begingroup$ Thanks but what about if $T$ is infinite? $\endgroup$ – TagCharacters Oct 29 '15 at 22:55
  • $\begingroup$ @TagCharacters: it is the only case I dealt with, $T$ being plus infinity. $\endgroup$ – Jack D'Aurizio Oct 29 '15 at 22:58
  • $\begingroup$ Ok. Do you know if how to calculate $\int_0^T tK_a(t)^2$? It's simpler than the other one $\endgroup$ – TagCharacters Oct 30 '15 at 18:10
  • $\begingroup$ @TagCharacters: it has a non-integrable singularity in the origin. $\endgroup$ – Jack D'Aurizio Oct 30 '15 at 19:47

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