So, I am trying to find the regularized value of the divergent integral $I=int_1^infty sqrt{x^2-1}dx$. Since the area of $int_0^1 sqrt{1-x^2}dx=fracpi4$, I wonder whether the area under hyperbola would be interesting as well.

So, I applied my method that uses Laplace transform $mathcal{L}_t(t f(t))(x)$ so to convert divergent integrals into equivalent ones.

This way I obtained a family of equivalent integrals:

$int_1^infty sqrt{x^2-1}dx=int_0^infty frac{K_2(x)}{x}dx=int_0^infty left(x-frac{1}{2 x}right) dx =int_0^infty left(frac{2}{x^3}-frac{1}{2 x}right)dx$

Now, we already know that $int_0^infty frac1xdx$ regularizes to $gamma$ while $int_0^infty frac1{x^3}dx$ and $int_0^infty xdx$ regularize to zero, so integral $I$ regularizes to $-fracgamma2$.

This is a very interesting result.

That said, I wonder whether I could somehow find the regularized value of the area under conjugare hyperbola? The Laplace transform gives a very complicated function….