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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..

Dhrubajyoti Bhattacharjee
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    See this or this. Does this help? – V.G Jul 22 '20 at 08:31
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    I upvoted the OP and the ABCD comment. My (unproven) intuition is that $\sqrt{x + \sqrt{x^2 + \sqrt{x^3 + \cdots}}}$ is strictly increasing as $x$ goes from 0 to 1. Therefore, based on the first link given in ABCD's comment, the function being integrated is bounded on $[0,1].$ That's as far as my thinking takes me. – user2661923 Jul 22 '20 at 09:25
  • Yes intuitively or based on the links provided it seems that the integrand is monotone increasing on $[0, 1]$ and also bounded above. So by monotone convergence theorem it must converge to its supremum. But I'm not actually able to find the supremum in $[0, 1]$. If it's $1$ then the integrand converges to $\varphi.$ Thank you for your comment. – Dhrubajyoti Bhattacharjee Jul 22 '20 at 10:29

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

Graviton
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    As described in this partial answer the beginning of the asymptotic expansion for $x\to\infty$ is $f(x) = \sqrt{2x}(1+\frac{1}{8\sqrt{x}}-\frac{5}{128x}+\frac{85}{1024\sqrt{x^3}}+\mathcal{O}(x^{-2}))$. Thus indeed, $\lambda$ is exactly $\frac{1}{\sqrt{32}}$, but $\int_0^\infty(f(x)-(\sqrt{2x}+\lambda))\mathrm{d}x$ does not converge at infinity. The above link gives also higher order approximations of $f(x)$ for small $x$. – Uwe Aug 02 '20 at 09:41
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    Here is another reference, giving the asymptotic expansion for $f(x)$ at $x\to\infty$ in terms of $y=\sqrt{2x}$. – Uwe Aug 02 '20 at 10:20
  • @Uwe Cheers! Fantastic update on your behalf. – Graviton Aug 02 '20 at 10:25