Convergence of an integral involving Bessel functions

Question

I am interested to prove the convergence of the following improper integral: $$ \int_0^{\infty}t(K_s(t))^2dt, $$ where $K_s(t)$ is the modified Bessel function of the second kind and $s\in(0,1)$. I am trying to make use of the following property for $K_s(t)$, i.e. when $t\to0^{+}$: $K_s(t)\approx \frac{1}{2}\Gamma(s)(\frac{t}{2})^{-s}$ and when $t\to\infty$: $K_s(t)\approx \sqrt{\frac{\pi}{2t}}e^{-t}$. However I cannot work out a rigorous proof for this. Would be happy if anyone could work out the convergence/an upper bound for this improper integral. Thanks!

Answer 1

Concerning the convergence, split the interval $$\int_0^\infty =\int_0^1+\int_1^\infty$$ and use the two approximations you wrote in the post. $$\int_0^1 2^{s-1} t^{-s} \Gamma (s)\,dt=\frac{2^{2 s-3} \Gamma (s)^2}{1-s}$$ $$\int_1^\infty \frac{1}{2} \pi e^{-2 t}\,dt=\frac{\pi }{4 e^2}$$

Povided that $\Re(s)<1$ and $\Re(a+b)>0$

$$\int_0^\infty x\, K_s(a x)\, K_s(b x)\,dx=\frac{\pi \left(a^{2 s}-b^{2 s}\right)}{2 \left(a^2-b^2\right)\,(a b)^s} \csc (\pi s) $$ that you could find in the "Table of Integrals, Series and Products" (seventh edition) by I.S. Gradshteyn and I.M. Ryzhik. (formula 6.521.3)

$$\int_0^\infty x\, K_s(a x)\, K_s(a x)\,dx=\frac{\pi s}{2 a^2} \csc (\pi s)$$ $$\int_0^\infty x\, K_s( x)\, K_s( x)\,dx=\frac{\pi s}{2 } \csc (\pi s)$$