Evaluate: $\int_0^1 \sqrt{x+\sqrt{x^2+\sqrt{x^3+\cdots}}}\, dx. $

Is there a way to evaluate the integral:$$\int_0^1\sqrt{x+\sqrt{x^2+\sqrt{x^3+\sqrt{x^4+\cdots}}}}\,dx,$$ without using numerical methods?

The integrand doesn't seem to converge to anything for any arbitrary positive real $x$. I could be wrong also. Please suggest something..

Edit. Thank you everyone for your kind responses. These help a lot.. New ideas, techniques etc etc.. Thanks..

Solution 1:

Obligatory "not an answer but too long for comments"

Let $f(x)=\sqrt{x+\sqrt{x^2+\sqrt{x^3+\dots}}}$

Surprisingly, from what I've gathered about the function from this qustion, not much is known even for the convergence of $f$ besides a few cases.

However it can be approximated extremely well.

It can easily be shown that $f(x)>\sqrt{2x}$, In fact it seems that $\lambda=\lim_{x\rightarrow\infty}(f(x)-\sqrt{2x})\approx0.1767766$.

$\lambda$ is so exceptionally close to $\frac{1}{\sqrt{32}}$, I have yet to find a digit that does not match. However, my intuition tells me that it's only a coincidence. Update: As @Uwe points out in the comments, it is true that $\lambda=\frac{1}{\sqrt{32}}$

Hence $\sqrt{2x}+\lambda$ is an extremely good approximation for $f$. However, $\int_0^{\infty}(f(x)-(\sqrt{2x}+\lambda))$ does not converge (see comments for refferences).

Also for small values of $x$, $f(x)\approx1+\frac{x}{2}$