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

Is there a way to evaluate the integral:$$int_0^1sqrt{x+sqrt{x^2+sqrt{x^3+sqrt{x^4+cdots}}}},dx,$$ 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..

Mathematics Asked on November 21, 2021

1 Answers

One Answer

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 $lambda=lim_{xrightarrowinfty}(f(x)-sqrt{2x})approx0.1767766$.

$lambda$ 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 $x$, $f(x)approx1+frac{x}{2}$

Answered by Graviton on November 21, 2021

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