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We have :

$$\int_{-1}^{0}\sqrt{1+\sqrt{1+\sqrt{1+x}}}dx=\frac{8}{315}\sqrt{2}\Big(16+\sqrt{233+317\sqrt{2}}\Big)$$

We are lucky because this integral have an anti-derivative like here.

More seriously I have tried the following substitution : $x=((t-1)^2-1)^2-1)^2-1$

We get a big polynomial easily integrable .

But what about the following integral : $$\int_{-1}^{0}\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{\cdots\sqrt{1+x}}}}}dx=$$

Have you other technics to evaluate this ?

Thanks in advance for your contributions !

Update :

As it seems to be unclear I ask for a finite number of nested radical.Furthermore it's not an innocent question because I know that it converges to the golden ratio .The imlpicit question is : How to get nested radicals near from the value of the golden ratio ?

Thanks again.

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  • $\begingroup$ Why can't you use the exact same kind of substitution? $\endgroup$
    – Lee Mosher
    Commented Apr 16, 2020 at 11:56
  • $\begingroup$ @LeeMosher I ask if there is other method more practical than mine. $\endgroup$
    – Ranger-of-trente-deux-glands
    Commented Apr 16, 2020 at 13:14
  • $\begingroup$ Don't see how there would be any approaches nicer than the provided approach. To some extend, the substitution can be simplified, but not by much. $\endgroup$
    – Simply Beautiful Art
    Commented Apr 16, 2020 at 14:58
  • $\begingroup$ So if $$f_0(x)= \sqrt{1+x}$$ and $$f_{n+1}(x)=\sqrt{1+f_n(x)}$$then you want to evaluate $$\int_{-1}^0 f_n(x) \operatorname d x$$ or $$\lim_{n \to +\infty} \int_{-1}^0 f_n(x) \operatorname d x ? $$ $\endgroup$
    – Sewer Keeper
    Commented Apr 21, 2020 at 10:22

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