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Yesterday I wrote some simple integrals involving nested radicals, and/or continued fractions. This was an example with nested radicals, let $$\int_0^1\frac{dx}{\sqrt{1+(\log x)g(x)}},\tag{1}$$ where $g(x)=\sqrt{1+(\log(x)\cdot g(x))}$, that is $$g(x)=\sqrt{1+\log x^{g(x)}}.\tag{2}$$

Question. Calculate a good approximation of previous integral. Thanks in advance.

You approach can be using analysis or a numerical method.

One has that $(g(x))^2=1+g(x)(\log x)$ and thus we need to calculate, if there are no mistakes, $$\sqrt{2}\int_0^1\frac{dx}{\sqrt{2+\log^2(x)+\sqrt{\log ^2(x)+4}}}.$$

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  • $\begingroup$ I'm sorry there was a great mistake in my last formula @ClaudeLeibovici, the right integral should be (since the solution $g(x)$ of $g^2(x)-g(x)\cdot(\log x)-1=0$ is $g(x)=\frac{\log x+\sqrt{\log^2( x)+4}}{2}$) $$\int_0^1\frac{dx}{1+\sqrt{\log (x^{g(x)})}}=\sqrt{2}\int_0^1\frac{dx}{\sqrt{2+(\log(x))\color{red}{\cdot}\left(\log(x)+\sqrt{\log^2(x)+4}\right)}}.$$ $\endgroup$
    – user243301
    Commented Apr 25, 2018 at 19:21

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For $$I=\int_0^1\frac{dx}{\sqrt{2+\log^2(x)+\sqrt{\log ^2(x)+4}}}$$ we can develop the integrand as a Taylor series built around $x=1$. This should give for the integrand $$\frac{1}{2}-\frac{5}{64} (x-1)^2+\frac{5}{64} (x-1)^3-\frac{643 (x-1)^4}{12288}+\frac{163 (x-1)^5}{6144}-\frac{11989 (x-1)^6}{1179648}+O\left((x-1)^7\right)$$ Integrating terwise, we get $$\frac{x-1}{2}-\frac{5}{192} (x-1)^3+\frac{5}{256} (x-1)^4-\frac{643 (x-1)^5}{61440}+\frac{163 (x-1)^6}{36864}-\frac{11989 (x-1)^7}{8257536}+O\left((x-1)^8\right)$$ then the approximation $$I=\frac{6029213}{13762560}\approx 0.438088$$ while the exact value should be $\approx 0.433262$.

Edit

Just for the fun of it, I used the above approach (using a CAS, for sure !) up to $O\left((x-1)^n\right)$. Let $I_n$ to be the value of the integral.

As shown below, the convergence is really slow $$\left( \begin{array}{cc} n & I_n \\ 50 & 0.4337137504 \\ 100 & 0.4334521397 \\ 150 & 0.4333768933 \\ 200 & 0.4333424597 \\ 250 & 0.4333230704 \\ 300 & 0.4333107672 \\ 350 & 0.4333023266 \\ 400 & 0.4332962080 \\ 450 & 0.4332915870 \\ 500 & 0.4332879845 \end{array} \right)$$

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  • $\begingroup$ Many thanks for your calculations and approach. $\endgroup$
    – user243301
    Commented Nov 2, 2017 at 10:52
  • $\begingroup$ @user243301. You are very welcome ! In fact, I am not so happy since increasing up to $O((x-1)^k$ for $k>8$ shows a very slow convergence. I shall try something else. $\endgroup$
    – Claude Leibovici
    Commented Nov 2, 2017 at 10:55

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