Evaluate the integral $$\int _0^{\infty }\frac{dx}{\Big(x+\sqrt{x\ +\ \sqrt{x\ +\ \sqrt{…}}}\Big)^2}$$
How can we evaluate this integral? I've no idea what to do with this. At first, I thought of simplifying the denominator but that seems not working here.
Can anyone give me some hints?
Edit:
I've simplified the denominator and got:
$$\int \dfrac{4}{(2x + \sqrt{4x+1} + 1)^2}dx$$
What to do now?
Best Answer
Let, \begin{align} \ x+ \sqrt {x+ {\sqrt {x + {\sqrt {...}}}}} = z^2 \Rightarrow {\frac{d}{dx}}(\ x +\sqrt {z^2})={\frac{d}{dx}}(\ z^2) \Rightarrow \ (1+{\frac{dz}{dx}})=\ 2z{\frac{dz}{dx}}
\Rightarrow {dz}= {\frac {1}{2z-1}}{\ dx} \end{align}
Now by substituting, \begin{align} \int_{0}^{\infty} {\frac {dx}{\Big(x+\sqrt{x\ +\ \sqrt{x\ +\ \sqrt{...}}}\Big)^2}} = \int_{1}^{\infty} {\frac {2z-1}{z^4}}{\ dz} = {\frac {2}{3}}\end{align}