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I'd like to find out a simple way for calculating the value of:

$\int_{0}^{1}\sqrt{1+\sqrt{1 + {\sqrt{1+ \sqrt{x}}}}}\,dx .$

Of course, I thought of some variable change, but it seems pretty complicated. On the other hand, I wonder if there can be made a generalization when having to deal with the expression with $k$ radicals, $k>1$.

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    My answer was for infinite radicals.2012-05-30

2 Answers 2

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  1. Let $\begin{eqnarray*} u &=&\sqrt{1+\sqrt{1+\sqrt{x}}} \Leftrightarrow &x=\left( \left( u^{2}-1\right) ^{2}-1\right) ^{2}=u^{8}-4u^{6}+4u^{4}. \end{eqnarray*}$ Since $\begin{equation*} dx=\left( 8u^{7}-24u^{5}+16u^{3}\right) du \end{equation*}$ we have $I :=\int_{0}^{1}\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{x}}}}dx\\=\int_{\sqrt{2}}^{\sqrt{1+\sqrt{2}}}\sqrt{1+u}\left(8u^{7}-24u^{5}+16u^{3}\right) du.\quad\textit{(computation below)}^† $ Each term can be integrated using the substitution $t=\sqrt{1+u}$ $\begin{equation*} \int_{a}^{b}\sqrt{1+u}u^{n}du=2\int_{\sqrt{1+a}}^{\sqrt{1+b}}t^{2}\left( t^{2}-1\right) ^{n}\,dt,\quad a=\sqrt{2},b=\sqrt{1+\sqrt{2}}. \end{equation*}$
  2. Generalization to $k=5$ radicals $\begin{equation*} J:=\int_{0}^{1}\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{x}}}}}dx. \end{equation*}$ Similarly to above the substitution is now
    $\begin{eqnarray*} v &=&\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{x}}}}\Leftrightarrow x=\left( \left( \left( v^{2}-1\right) ^{2}-1\right) ^{2}-1\right) ^{2} \\ x &=& v^{16}-8v^{14}+24v^{12}-32v^{10}+14v^{8}+8v^{6}-8v^{4}-1, \end{eqnarray*}$ and $\begin{equation*} dx=\left( 16v^{15}-112v^{13}+288v^{11}-320v^{9}+112v^{7}+48v^{5}-32v^{3}\right) dv. \end{equation*}$

Hence $\begin{eqnarray*} J &=&\int_{\alpha }^{\beta }\sqrt{1+v}\left( 16v^{15}-112v^{13}+288v^{11}-320v^{9}+112v^{7}+48v^{5}-32v^{3}\right) dv \\ \alpha &=&\sqrt{1+\sqrt{2}},\beta =\sqrt{1+\sqrt{1+\sqrt{2}}}. \end{eqnarray*}$

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In SWP I obtained

$\begin{eqnarray*} I &=&-\frac{26\,704}{765\,765}\sqrt{1+\sqrt{\sqrt{2}+1}}\sqrt{\sqrt{2}+1} \sqrt{2} \\&&+\frac{83\,584}{765\,765}\sqrt{1+\sqrt{\sqrt{2}+1}}\sqrt{\sqrt{2}+1} \\ &&+\frac{344\,096}{765\,765}\sqrt{1+\sqrt{\sqrt{2}+1}} \\ &&+\frac{67\,328}{109\,395}\sqrt{\sqrt{2}+1} \\ &&-\frac{256}{3003}\sqrt{\sqrt{2}+1}\sqrt{2} \\ &&-\frac{17\,168}{765\,765}\sqrt{1+\sqrt{\sqrt{2}+1}}\sqrt{2} \\ &\approx &1.584\,9. \end{eqnarray*}$

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Hints:

1) First substitute $\,t=\sqrt{1+\sqrt{x}}\Longrightarrow dt=\frac{dx}{4\sqrt{x}\sqrt{1+\sqrt{x}}}\Longrightarrow dx=4(t^2-1)tdt$ , and now change the limits to $\,1\,,\,\sqrt{2}$

2) Next, you have $4\int_1^{\sqrt{2}}\,t(t^2-1)\sqrt{1+\sqrt{1+t}}\,dt$ , and now substitute $y=\sqrt{1+\sqrt{1+t}}$and etc. You end up with a not-so-terrible polynomial function.