Integrals of $\sqrt{x+\sqrt{\phantom|\dots+\sqrt{x+1}}}$ in elementary functions

Question

Let $f_n(x)$ be recursively defined as $$f_0(x)=1,\ \ \ f_{n+1}(x)=\sqrt{x+f_n(x)},\tag1$$ i.e. $f_n(x)$ contains $n$ radicals and $n$ occurences of $x$: $$f_1(x)=\sqrt{x+1},\ \ \ f_2(x)=\sqrt{x+\sqrt{x+1}},\ \ \ f_3(x)=\sqrt{x+\sqrt{x+\sqrt{x+1}}},\ \dots\tag2$$

The functions $f_0(x)$, $f_1(x)$ and $f_2(x)$ are integrable in elementary functions, e.g.: $$\int\sqrt{x+\sqrt{x+1}}\,dx=\left(\frac{2\,x}3+\frac{\sqrt{x+1}}6-\frac14\right)\sqrt{x+\sqrt{x+1}}+\frac58\ln\left(2\,\sqrt{x+1}+2\,\sqrt{x+\sqrt{x+1}}+1\right).\tag3$$

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Question: Is there an integer $n>2$ such that $f_n(x)$ is integrable in elementary functions?

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Update: The question is reposted at MathOverflow as was suggested by moderator.

Answer 1

I think there is highly have no chance to have an integer $n>2$ such that $f_n(x)$ is integrable in elementary functions.

The reasons are mainly because of the processes of eliminating the radicals.

For $f_0(x)$ and $f_1(x)$ , they are trivially integrable in elementary functions.

For $\int f_2(x)~dx$ ,

Let $u=\sqrt{x+1}$ ,

Then $x=u^2-1$

$dx=2u~du$

$\therefore\int\sqrt{x+\sqrt{x+1}}~dx$

$=\int2u\sqrt{u^2+u-1}~du$ , which can express in elementary functions.

Starting from $\int f_3(x)~dx$ ,

Let $u=\sqrt{x+1}$ ,

Then $x=u^2-1$

$dx=2u~du$

$\therefore\int\sqrt{x+\sqrt{x+\sqrt{x+1}}}~dx$

$=\int2u\sqrt{u^2-1+\sqrt{u^2+u-1}}~du$

Introduce the Euler substitution:

Let $v=u+\sqrt{u^2+u-1}$ ,

Then $u=\dfrac{v^2+1}{2v+1}$

$du=\dfrac{2v(2v+1)-(v^2+1)2}{(2v+1)^2}dv=\dfrac{2v^2+2v-2}{(2v+1)^2}dv$

$\therefore\int2u\sqrt{u^2-1+\sqrt{u^2+u-1}}~du$

$=\int2\dfrac{v^2+1}{2v+1}\sqrt{\left(\dfrac{v^2+1}{2v+1}\right)^2-1+v-\dfrac{v^2+1}{2v+1}}\dfrac{2v^2+2v-2}{(2v+1)^2}dv$

$=\int4\dfrac{(v^2+1)(v^2+v-1)}{(2v+1)^3}\sqrt{\dfrac{(v^2+1)^2-(v^2+1)(2v+1)+(v-1)(2v+1)^2}{(2v+1)^2}}~dv$ , which is highly have no chance to express in elementary functions.

Answer 2

This is only a partial attempt at answering the question. I read this description of differential Galois theory to find out about this. I'll try to present the relevant bits of those notes---this is not original, and all mistakes are mine.

First, define what is meant by an elementary function. Let $\mathbb{C}(x,g_1,\ldots,g_n)$ be the set of all rational functions of $x$ and $g_i$, where $g_i(x)$ are functions of $x$. An elementary field $K=\mathbb{C}(x,g_1,\ldots,g_n)$ is a field of functions such that each $g_j$ is an exponential, logarithm or algebraic function of an element of $\mathbb{C}(x,g_1,\ldots,g_{j-1})$. Such an elementary field is closed under differentiation.

Liouville's theorem then says that if $f$ is a function in $K$ that has an elementary antiderivative, that antiderivative cannot be just any arbitrary function of elements of $K$, but must be equal to $$ \sum_i c_i \log g_i + h, $$ for some functions $g_i$, $h$ in $K$, and constants $c_i$.

For the given functions $f_n$, $f_n^2=x+f_{n-1}$, we can choose elementary functions to mean $K=\mathbb{C}(x,f_1,\ldots,f_n)$ (i.e., all rational expressions involving $f_i$ and $x$), and deduce that if $\int f_n$ is elementary, we must have $$ f_n = \sum_i c_i \frac{g_i'}{g_i} + h', $$ where $g_i$ and $h$ are in $K$.

While this doesn't seem like all that much, it does mean that one (rigourously) knows one doesn't ever have to look at functions like $\sqrt{x+1}\log\sqrt{x+1}$ or $e^{\sqrt{x+1}}$ when integrating $\sqrt{x+\sqrt{x+1}}$, if "elementary" functions are as defined above. As to how you can actually find those functions $g_i$, $h$ or disprove their existence, I have no idea.

I tried to integrate $f_3$ and $f_4$ by guessing a general form and solving for coefficients, but it didn't work.

Update. (This is a work in progress, so maybe the claims and expressions will be somewhat of a mystery. The question is certainly interesting.)

There is an algorithm, the Risch-Hermite-Trager algorithm (see Computer Algebra by Geddes, Czapor, Labahn, and also Integrating Algebraic Functions by Trager), for integrating arbitrary algebraic expressions in elementary form or proving their integrals are not elementary. The function $f_3(x)$ is maybe integrable in elementary form, with $$\int f_3 = \frac{-3+8x+2f_2+3 f_2 f_3/x}{12}f_3 + \int q_3, $$ where $\int q_3$ is a sum of logarithmic terms as in Liouville's theorem (but I am unable to integrate it explicitly, even the case of $f_2$ is hard for me: I asked related questions here and here, but my knowledge of algebraic geometry is lacking). $$ q_3 = \left(\left(11 x^2+23 x^3+x^4-13 x^5-x^6+x^7+\left(-6 x-13 x^2-x^3+18 x^4+6 x^5-6 x^6\right) f_2+f_1 \left(-4 x-9 x^2-9 x^3+15 x^5+x^6-4 x^7+\left(8+17 x-5 x^3-2 x^4-23 x^5+13 x^6\right) f_2\right)\right) f_3\right)/\left(32 x^2 (1+x) \left(-1-x+x^2\right) \left(-1-x+x^2-2 x^3+x^4\right)\right). $$ I am assured by the algorithm that it is integrable in elementary terms if and only if it is possible to write $$ \int q_3 = \sum c_i \log v_i, $$ where $c_i$ are numbers that belong to the field $\mathbb{Q}$ extended by the roots of the polynomial $$\begin{gathered} R(Z) = \left(-288966001+75889061888 Z^2-15925264777216 Z^4+88562225643520 Z^6+439804651110400 Z^8\right)^2 \times\\ \left(-14508481-784902408192 Z^2-757665491845120 Z^4-103142060671792840704 Z^6+635008673359146778624 Z^8\right)^2 \times\\ \left(46732663382478474361-89569724082204845664256 Z^2+904900289526729208095571968 Z^4+1144371243549912082526161076224 Z^6+3200188454940133648592935688601600 Z^8+11158139832684515154655834451715555328 Z^{10}+9545270929062413812474597798367805308928 Z^{12}-124380415953981823700469862336514528116736 Z^{14}+50404501905167945967686052562499557916672 Z^{16}\right)^2. \end{gathered}, $$ and $v_i$ have certain prescribed roots and poles, but I don't know how to calculate $c_i$ and $v_i$ explicitly (in theory it is possible to either find them or establish they cannot exist).

The point is that there is a deterministic algorithm for figuring out the algebraic portion of the integral $\int f_3$, and the integral can be evaluated or proven to be non-elementary by constructing a certain polynomial $R(Z)$, whose roots form a basis over $\mathbb{Q}$ of the field to which the coefficients $c_i$ belong. If all the roots are zero (non-zero integrand that has only zero logarithmic terms), the integral is known to be non-elementary. Otherwise, the logarithmic term is given by $$\sum_{\beta,R(\beta)=0} c_\beta \log v_\beta, $$ where (roughly) $v_\beta$ is a function of $(x,y)$ that has a root/pole wherever $q_3\,dx$ (viewed as a function of both $x$ and $y$, where $y$ is a multivalued algebraic function of $x$) has a pole with residue $\beta$, and no other roots or poles anywhere else. This is complicated by a change of variables $x=z^{-2}$, $\hat y=z y$ necessary earlier, and the fact that the curve $(x,y)$ has singular points that make some things ill-defined. Such functions $v_\beta$ (they are generators of specific principal ideals) may not exist for some other reasons, but nothing about the function $q_3$ prevents their existence.

I also calculated $R(Z)$ for $f_4$, and it has non-zero roots, meaning integrability of $f_4$ in elementary terms is equivalent to finding a set of those functions $v_\beta$; I don't know if they exist or not.