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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, $k>1$.

  • 0
    as $k \to \infty$ it seems like the integral goes to $\phi$2012-05-30
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    surprisingly, this is Wolfram integrable.2012-05-30
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    Nothing surprising - just substitute $t=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{x}}}}$.2012-05-30
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    (with 3 ones) After that substitution, Maple gets $$\frac{16 t^{17}}{17} - \frac{112 t^{15}}{15} + \frac{288 t^{13}}{13} - \frac{320 t^{11}}{11} + \frac{112 t^{9}}{9} + \frac{48 t^{7}}{7} - \frac{32 t^{5}}{5}$$2012-05-30
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    @ GEdgar: i suppose that things get worse when dealing with more radicals in place.2012-05-30
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    @picakhu It *does* converges to $\phi$ as $k \rightarrow \infty$. It is not hard to prove it since $\phi = \sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{\cdots}}}}}$ and the integral is from $0$ to $1$ and the integrand can be bound between $\phi - \epsilon(k)$ and $\phi + \epsilon(k)$ for a given $k$ and $\epsilon(k) \rightarrow 0$ as $k \rightarrow \infty$.2012-05-30
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    And the definite integral is $-{\frac {26704}{765765}}\,\sqrt {1+\sqrt {1+\sqrt {2}}}\sqrt {1+\sqrt {2}}\sqrt {2}+{\frac {83584}{765765}}\,\sqrt {1+\sqrt {1+\sqrt {2}}} \sqrt {1+\sqrt {2}}-{\frac {17168}{765765}}\,\sqrt {1+\sqrt {1+\sqrt { 2}}}\sqrt {2}+{\frac {344096}{765765}}\,\sqrt {1+\sqrt {1+\sqrt {2}}}- {\frac {256}{3003}}\,\sqrt {1+\sqrt {2}}\sqrt {2}+{\frac {67328}{ 109395}}\,\sqrt {1+\sqrt {2}} $2012-05-30
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    @Robert Israel: how did you get this result?2012-05-30
  • 0
    Using the Maple commands: > f:=sqrt(1+sqrt(1+sqrt(1+sqrt(x)))); > j:= int(f,x=0..1); > IntegrationTools[Change](j,t=f); > simplify(%,radical);2012-05-30
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    @Robert Israel: OK. Maybe things would work better with some trigonometric substitution. In any case, it doesn't seem an easy case, at all.2012-05-30
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    @Chris: The answer is what it is. I don't know what "better" result you expect to get with a trigonometric substitution.2012-05-30
  • 0
    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.