Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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, $k>1$.

share|cite|improve this question
as $k \to \infty$ it seems like the integral goes to $\phi$ – picakhu May 30 '12 at 17:36
surprisingly, this is Wolfram integrable. – user20266 May 30 '12 at 17:39
Nothing surprising - just substitute $t=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{x}}}}$. – Norbert May 30 '12 at 17:42
(with 3 ones) After that substitution, Maple gets $$\frac{16 t^{17}}{17} - \frac{112 t^{15}}{15} + \frac{288 t^{13}}{13} - \frac{320 t^{11}}{11} + \frac{112 t^{9}}{9} + \frac{48 t^{7}}{7} - \frac{32 t^{5}}{5}$$ – GEdgar May 30 '12 at 17:50
@ GEdgar: i suppose that things get worse when dealing with more radicals in place. – user 1618033 May 30 '12 at 17:52
up vote 3 down vote accepted
  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 $$\begin{eqnarray*} J &=&\int_{\alpha }^{\beta }\sqrt{1+v}\left( 16v^{15}-112v^{13}+288v^{11}-320v^{9}+112v^{7}+48v^{5}-32v^{3}\right) dv \\ \alpha &=&\sqrt{1+\sqrt{2}},\beta =\sqrt{1+\sqrt{1+\sqrt{2}}}. \end{eqnarray*}$$


In SWP I obtained

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

share|cite|improve this answer


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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.