Evaluate the improper integral of 1/(x + sqrt(x^2 - 1))^2 with respect to x from 1 to infinity, and explain why the lower limit cannot be 0.

Step-by-step solution

1. 1

   Fix the domain before integrating. For $0\le x<1$ the radicand $x^2-1$ is negative, so $\sqrt{x^2-1}$ is not a real number and the integrand is undefined there. The largest interval on which this is a genuine real integral starts at $x=1$ — writing the lower limit as $0$ is a statement error, not a convergence issue.

2. 2

   Choose the substitution that linearises the radical. For $x\ge 1$ put

   $x=\cosh t,\qquad t\ge 0$

   Then $\sqrt{x^2-1}=\sqrt{\cosh^2t-1}=\sinh t$ (non-negative for $t\ge0$) and $dx=\sinh t\,dt$.

3. 3

   Simplify the bracket — this is why cosh is the right choice.

   $x+\sqrt{x^2-1}=\cosh t+\sinh t=e^{t}$

   so the squared bracket is just $e^{2t}$. The limits transform as $x=1\Rightarrow t=0$ and $x\to\infty\Rightarrow t\to\infty$.

4. 4

   Rewrite the integral.

   $\int_{1}^{\infty}\frac{dx}{\left(x+\sqrt{x^2-1}\right)^{2}}=\int_{0}^{\infty}\frac{\sinh t}{e^{2t}}\,dt=\int_0^\infty e^{-2t}\sinh t\,dt$

5. 5

   Expand the hyperbolic sine into exponentials. With $\sinh t=\tfrac{e^t-e^{-t}}{2}$,

   $e^{-2t}\sinh t=\frac{e^{-t}-e^{-3t}}{2}$

   which is a difference of two decaying exponentials — both integrable on $[0,\infty)$.

6. 6

   Integrate and evaluate. Since $\int_0^\infty e^{-kt}dt=\tfrac1k$,

   $\frac12\left(\int_0^\infty e^{-t}dt-\int_0^\infty e^{-3t}dt\right)=\frac12\left(1-\frac13\right)=\frac12\cdot\frac23=\frac13$

   Numerical quadrature of $\int_0^{40}e^{-2t}\sinh t\,dt$ gives $0.3333333333$, confirming the exact value $\tfrac13$.