Integral of modified Bessel function of second kind How to prove $\int_1^\infty 2x\sqrt{x^2-1}K_1(x)dx = K_1^2(1/2)$, where $K_1$ is the modified Bessel function of second kind? Mathematica can do this calculation but I don't understand how...
 A: I'll prove the following generalization, which proves the OP's eq. for $a=1:$
$$ (1) \quad I(a):=2\int_a^\infty x\sqrt{x^2-a^2} \  K_1(x) dx = a^2 \big(K_1(a/2)\big)^2 $$
Use the specialization of a well-known formula
$$ \int_0^\infty \frac{dt}{t^2} \exp{(-t-x^2/(4t))} = \frac{4}{x} \ K_1(x) .$$
Put this into (1), justify change of $\int,$ and you'll get
$$I(a)=\frac{1}{2}\int_0^\infty \frac{dt}{t^2}e^{-t} \int_a^\infty x^2\sqrt{x^2-a^2} \ \exp{(-x^2/(4t))} \ dx$$
The inner integral can be performed explicitly.  Put the answer in and shift $t \to t/2.$
$$ I(a)=\frac{a^2}{2} \int_0^\infty \frac{dt}{t}\ \exp{\Big(-\frac{t}{2} - \frac{(a/2)^2}{t}\Big)} K_1\Big(\frac{(a/2)^2}{t}\Big) $$
Now from Gradshteyn and Rhyzik, 6.653.2
$$(2)\quad \frac{1}{2} \int_0^\infty \frac{dt}{t}\ \exp{\Big(-\frac{t}{2} - \frac{z^2+w^2}{2t}\Big)} K_s\Big(\frac{z \ w}{t}\Big)= K_s(w) \ K_s(w) $$
Set $z=w=a/2$ and $s=1$ in (2) to complete the proof.
I don't believe that I've used any formulas that can't be found in Gradshteyn and Rhyzik. That reference book has citations, so in principal this proof doesn't need a Mathematica.  My guess is that Mathematica did not do it this way, but saw it as a Meijer G-function, and then used a lookup table to get the final result.
