# Integral with 4 radicals-hat

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, $k>1$.

-
as $k \to \infty$ it seems like the integral goes to $\phi$ – picakhu May 30 '12 at 17:36

1
surprisingly, this is Wolfram integrable. – user20266 May 30 '12 at 17:39

1
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. – Chris's wise sister May 30 '12 at 17:52
show 7 more comments

## 2 Answers

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

--

†In SWP I obtained

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

-

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.

-
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 8, "mathjax_display_tex": 14, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8987991213798523, "perplexity_flag": "middle"}