integral of ((x-sqrt(x^2-1))^2 dx (1--> inf)

[dwbent]

I have been struggling with the following integral for days, which I can integrate numerically to a simple 1/3:

int_1^inf ((x-sqrt(x^2-1))^2 dx:

Latex:

\int_{1}^{\infty} (x-\sqrt{x^2-1})^2dx. Numerically it appears to be exactly 1/3, but try as I may I cannot integrate it to prove that. The function equals 1 at x=1 and behaves as 1/(2x^2) for large x (thus converging to zero at x= infinity). I'm probably missing something stupid, but any help would be appreciated.

 

[stapel]

int_1^inf ((x-sqrt(x^2-1))^2 dx:

A good first step might be to multiply out the square:

. . . . .$\displaystyle (x\, -\, \sqrt{x^2\, -\, 1})^2$

. . . . .$\displaystyle x^2\, -\, 2x\sqrt{x^2\, -\, 1}\, +\, (x^2\, -\, 1)$

. . . . .$\displaystyle x^2\, +\, x^2\, -\, 1\, -\, 2x\sqrt{x^2\, -\, 1}$

. . . . .$\displaystyle 2x^2\, -\, 1\, -\, 2x\sqrt{x^2\, -\, 1}$

The first two terms integrate easily, so break them off and integrate separately:

. . . . .$\displaystyle \int\, 2x^2\, -\, 1\, dx\, -\, \int\, 2x\sqrt{x^2\, -\, 1}\, dx$

This leaves you with the square-root integral to puzzle with. Have you tried a u-substitution?

. . . . .$\displaystyle u\, =\, x^2\, -\, 1$

. . . . .$\displaystyle du\, =\, 2x\, dx$

Where might that lead...?

Note: You can check your work using The Integrator on the Wolfram web site. :wink:

Eliz.