I asked a question at Math.SE last year and later offered a bounty for it, but it remains unsolved even in the simplest case. So I finally decided to repost this case here:

Is it possible to express the following indefinite integral in elementary functions? $${\large\int}\sqrt{x+\sqrt{x+\sqrt{x+1}}}\ dx$$

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    $\begingroup$ The simpler integral ${\displaystyle\int_0^1\sqrt{x+\sqrt{x+1}}\ dx}=\frac{1}{12}+\frac{1}{3}\sqrt{\frac{1+\sqrt{2}}{2}}+\frac{5\sqrt{1+\sqrt{2}}}{12}-\frac{5\ln5}{8}+\frac{5}{8}\ln\left(1+2 \sqrt{2}+2 \sqrt{1+\sqrt{2}}\right)\approx1.3015195268882140719830471603491475...$ is not recognized either, although is expressible in elementary functions. $\endgroup$ Jun 13, 2014 at 5:24
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    $\begingroup$ The reason that the simpler case works is that, if you write the integrand as $y(x) dx$, then you can rationalize it by the substitution $$x = \frac{(t-1)(t-5)(t^2+2t+5)}{16t^2}\qquad y = \frac{(5-t^2)}{4t}$$ then the integral becomes elementary. You should check whether the actual problem you are trying to solve can be rationally parametrized, which is an algebra problem. If so, then the integral should be doable. $\endgroup$ Jun 13, 2014 at 8:00
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    $\begingroup$ @Robert Bryant: setting $y=\sqrt{x+\sqrt{x+\sqrt{x+1}}}$, looking for a rational parametrization of the integral is equivalent to ask whether the algebraic curve of equation $$((y^2-x)^2-x)^2-x-1=0$$ has genus $0$ (as in the simpler case, where the parametrization do exist). A quick computation with Maple shows that the genus is $1$, so no rational parametrization of the integral can exist (at least if my calculations are correct :-) ). $\endgroup$ Jun 13, 2014 at 8:17
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    $\begingroup$ Actually, the fact that the genus of the corresponding algebraic curve is $1$ means that the integral should admit a parametrization via elliptic functions, adding further evidence to the conjecture that no expression of the primitive in elementary functions can exist. $\endgroup$ Jun 13, 2014 at 8:33
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    $\begingroup$ @FrancescoPolizzi: I only meant the rational parametrization test as the easiest way it could work out. Just because the algebraic curve associated to the problem has genus 1 doesn't necessarily mean that the integral can't be evaluated in elementary terms. For example, the curve $y^2-x^3-1=0$ has genus $1$, but the indefinite integral $$\int\frac{3x^2\ \mathrm{d} x}{\sqrt{x^3+1}}$$ can still be integrated in elementary terms. In the OP's problem, since the genus is $1$ instead of $0$, the next step is to figure out what multiple the integrand is of the holomorphic differential. $\endgroup$ Jun 13, 2014 at 20:02

2 Answers 2


The answer is 'no'. Making the substitution $$ x = \frac{(t-1)(t-5)(t^2+2t+5)}{16t^2}, $$ one finds $$ {\textstyle\sqrt{x+\sqrt{x+\sqrt{x+1}}}\,\mathrm{d}x} = \frac{(t^2-2t+5)(t^2-5)\sqrt{t^4{-}2t^2{-}40t+25}\ \mathrm{d}t}{32t^4}. $$ Denote the right hand side of the above equation by $\beta$. Now, setting $$ Q(t) = \frac{(25-25t-14t^2-3t^3+t^4)}{96t^3}, $$ one finds that $$ \beta = \mathrm{d}\left(Q(t)\sqrt{t^4{-}2t^2{-}40t+25}\right) + \frac{(10+15t-2t^2+t^3)\,\mathrm{d}t}{8t \sqrt{t^4-2t^2-40t+25}}\, $$ The second term on the right hand side is in normal form for differentials on the elliptic curve $s^2 = t^4-2t^2-40t+25$ that are odd with respect to the involution $(t,s)\mapsto(t,-s)$. Since this term represents a differential that has two poles of order $2$ (over the two points where $t=\infty$) and two poles of order $1$ (over the two points where $t=0$), an application of Liouville's Theorem (on integration in elementary terms, with the differential field taken to be the field of meromorphic functions on the elliptic curve) shows that this term is not integrable in elementary terms. Hence $\beta$ is not, and hence the original integrand is not.

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    $\begingroup$ This is extremely slick. How did you find this substitution? $\endgroup$ Jun 15, 2014 at 8:11
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    $\begingroup$ @RobertBryant Thanks! Do you think this approach can be used to resolve the general case posted at math.stackexchange.com/questions/500589/… ? $\endgroup$ Jun 15, 2014 at 15:09
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    $\begingroup$ @VladimirReshetnikov: Certainly it can be used in each individual case of $n$, but the computations will grow increasingly more elaborate as $n$ increases. I don't see any clever way to do one calculation that would settle the cases for all $n$, but there might be such a way. I expect that it's not an elementary integral for any $n>2$; the cancellations that would have to happen for large $n$ to make the integral of $f_n$ elementary would border on the miraculous. Why are you interested in this particular sequence? $\endgroup$ Jun 15, 2014 at 21:47
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    $\begingroup$ @PeterMueller: In fact, I found that substitution when I was looking for the explanation that the simpler integrand $f_2$ could be integrated in elementary terms. It was only later that I looked at the mathstackexchange version and saw Harry Peter's substitution. My reasoning for that case was: Make $x+1=u^2$, so that the integrand becomes $\sqrt{u^2-1+u}$ and then make another substitution to rationalize the conic $y^2=u^2+u-1$ in terms of a parameter $t$. That leads directly to the substitution I used. $\endgroup$ Jun 15, 2014 at 21:51
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    $\begingroup$ @VladimirReshetnikov: OK. Actually, it turns out that there is an argument that works for general $n>2$. I thought of it this morning and have decided to enter it as a separate 'answer' (in quotes because it's not really just an answer to your MO question, but to your original SE question). It turns out to be a pretty exercise in Riemann surface theory. $\endgroup$ Jun 17, 2014 at 18:29

I'm adding a separate answer for the general question that the OP asked, which settles the question in the negative for all $n>2$ (and gives an alternate proof for $n=3$ to the one I gave above).

Recall that the OP defined a sequence of algebraic functions $f_n$ by the rule $f_0(x) = 1$, $f_1(x) = \sqrt{x+1}$, and $f_{n+1}(x) = \sqrt{x + f_n(x)}$ for all $n\ge 1$. It was observed that $f_n$ has an elementary antiderivative for $n=0$, $1$, and $2$, and the problem was to determine whether $f_n$ has an elementary antiderivative for some $n>2$.

I am going to show that there is no elementary antiderivative of $f_n$ when $n>2$.

Assume $n>2$ (NB: This is important, because the argument below will not work for $n\le2$; the reader may enjoy finding where it breaks down), and let $K_n = {\mathbb C}\bigl(x,f_n(x)\bigr)$ be the elementary differential field generated by $x$ and $f_n(x)$. Then $K_n$ is the field of meromorphic functions on the normalization $\hat C_n$ of the algebraic curve $C_n$ defined by the minimal degree $y$-monic polynomial $P_n(x,y)$ that satisfies $P_n\bigl(x,f_n(x)\bigr) \equiv 0$. This minimal degree is $2^n$; for example, $P_2(x,y) = (y^2-x)^2-x-1$ and $P_3(x,y) = \bigl((y^2-x)^2-x\bigr)^2-x-1$, etc.

Since $P_{n+1}(x,y) = (P_n(x,y)+1)^2-x-1$ for $n\ge 1$ with $P_1(x,y)=y^2-x-1$, one sees, by applying the Eisenstein Criterion to $P_n(x,y)$ regarded as an element of $D[y]$ with $D$ being the integral domain ${\mathbb C}[x]$, that $P_n(x,y)$ is irreducible for all $n\ge 1$. Hence, $\hat C_n$ is connected.

It will be important in what follows to observe that $K_n$ has an involution $\iota$ that fixes $x$ and sends $f_n(x)$ to $-f_n(x)$; this is because $P_n(x,y)$ is an even polynomial in $y$. The fixed field of $\iota$ is ${\mathbb C}\bigl(x,\,f_n(x)^2\bigr)$, and the $(-1)$-eigenspace of $\iota$ is ${\mathbb C}\bigl(x,\,f_n(x)^2\bigr)f_n(x) = K_{n-1}{\cdot}f_n(x)$.

Now, the curve $C_n\subset \mathbb{CP}^2$ has only one point on the line at infinity, namely $[1,0,0]$, but the normalization $\hat C_n$ has $2^{n-1}$ points lying over this point. They can be parametrized as follows: First, establish the convention that $\sqrt{u}$ means the unique analytic function on the complex $u$-plane minus its negative axis and $0$ that satisfies $\sqrt1 = 1$ and $\bigl(\sqrt{u}\bigr)^2 = u$. Let $\epsilon = (\epsilon_1,\ldots,\epsilon_{n-1})$ be any sequence with ${\epsilon_k}^2=1$ and consider the sequence of functions $g^\epsilon_k(t)$ defined by the criteria $g^\epsilon_1(t) = \sqrt{1+t^2}$ and $g^\epsilon_{k+1}(t) = \sqrt{1+\epsilon_{n-k}t g^\epsilon_k(t)}$ for $1\le k < n$. Choose, as one may, a $\delta_n>0$ sufficiently small so that, when $t$ is complex and satisfies $|t|<\delta_n$, all of the functions $g^\epsilon_k$ are analytic when $|t|<\delta_n$. In particular, one finds an expansion $$ g^\epsilon_n(t) = 1+\tfrac12\epsilon_1\,t + \tfrac18(2\epsilon_1\epsilon_2-1)t^2 + O(t^3). $$

Also, it is easy to verify that the disk in $\mathbb{CP}^2$ defined by $$ [x,y,1] = [1,\ t g^\epsilon_n(t),\ t^2]\qquad\text{for}\quad |t|<\delta_n $$ is a nonsingular parametrization of a branch of $C_n$ in a neighborhood of the point $[1,0,0]$. In the normalization $\hat C_n$, this is then a local parametrization of a neighborhood of a point $p_\epsilon\in \hat C_n$. Obviously, this describes $2^{n-1}$ distinct points on $\hat C_n$.

When $x$ and $f_n$ are regarded as meromorphic functions on $\hat C_n$, it follows that there is a unique local coordinate chart $t_\epsilon:D_\epsilon\to D(0,\delta_n)\subset \mathbb{C}$ of an open disk $D_\epsilon\subset \hat C_n$ about $p_\epsilon$ such that $t_\epsilon(p_\epsilon)=0$ and on which one has formulae $$ x = \frac1{{t_\epsilon}^2} \quad\text{and}\quad f_n(x) = \frac{g^\epsilon_n(t_\epsilon)}{t_\epsilon} = \frac{1+\tfrac12\epsilon_1\ t_\epsilon +\tfrac18(2\epsilon_1\epsilon_2-1)\ {t_\epsilon}^2} {t_\epsilon} + O({t_\epsilon}^2). $$ In particular, it follows that $f_n(x)$, as a meromorphic function on $\hat C_n$, has polar divisor equal to the sum of the $p_\epsilon$ and hence has degree $2^{n-1}$. Of course, this implies that the zero divisor of $f_n(x)$ on $\hat C_n$ must be of degree $2^{n-1}$ as well.

Note that the functions $g^\epsilon_k$ satisfy $g^{-\epsilon}_k(-t) = g^{\epsilon}_k(t)$, where $-\epsilon = (-\epsilon_1,\ldots,-\epsilon_{n-1})$. This implies that $\iota(p_\epsilon) = p_{-\epsilon}$ and that $t_\epsilon\circ\iota = -t_{-\epsilon}$.

Now, the $2^{n-1}$ zeroes of $f_n(x)$ on $\hat C_n$ are distinct, for they are the zeros of the polynomial $q_n(x) = P_n(x,0) = (q_{n-1}+1)^2-x-1$, and the discriminant of $q_n$, being the resultant of $q_n$ and $q_n'$, is clearly an odd integer, and hence is not zero. Thus, $C_n$ is a branched double cover of $C_{n-1}$, branched exactly where $f_{n}$ has its zeros. This induces a branched cover $\pi_n:\hat C_n\to \hat C_{n-1}$ that is exactly the quotient of $\hat C_n$ by the involution $\iota$ (whose fixed points are where $f_n$ has its zeros). Since one then has the Riemann-Hurwitz formula $$ \chi(\hat C_n) = 2\chi(\hat C_{n-1}) - B_n = 2\chi(\hat C_{n-1}) - 2^{n-1}, $$ and $\chi(\hat C_1) = \chi(\hat C_2) = 2$, induction gives $\chi(\hat C_n) = (3{-}n)2^{n-1}$, so the genus of $\hat C_n$ is $(n{-}3) 2^{n-2} + 1$. (This won't actually be needed below, but it is interesting.)

The only poles of $x$ and $f_n(x)$ on $\hat C_n$ are the points $p_\epsilon$, and computation using the above expansions shows that, in a neighborhood of $p_\epsilon$, one has an expansion of the form $$ f_n(x)\,\mathrm{d} x - \mathrm{d}\left(f_n(x)\bigl(\tfrac12\ x + \tfrac16\ f_n(x)^2\bigr) \right) = \left(\frac{ (1-\epsilon_1\epsilon_2) } {4{t_\epsilon}^2} + O({t_\epsilon}^{-1})\right)\ \mathrm{d} t_\epsilon\ . $$ Thus, the meromorphic differential $\eta$ on $\hat C_n$ defined by the left hand side of this equation has, at worst, double poles at the points $p_\epsilon$ and no other poles.

Now, by Liouville's Theorem, $f_n$ has an elementary antiderivative if and only if $f_n(x)\ \mathrm{d} x$ and, hence, the form $\eta$ are expressible as finite linear combinations of exact differentials and log-exact differentials. Thus, $f_n(x)$ has an elementary antiderivative if and only if $\eta$ is expressible in the form $$ \eta = \mathrm{d} h + \sum_{i=1}^m c_i\,\frac{\mathrm{d} g_i}{g_i} $$ for some $h,g_1,\cdots g_m\in K_n$ and some constants $c_1,\ldots,c_m$. Suppose that these exist. Since $\eta$ has, at worst, double poles at the $p_\epsilon$ and no other poles, it follows that $h$ must have, at worst, simple poles at the points $p_\epsilon$ and no other poles; in fact, $h$ is uniquely determined up to an additive constant because its expansion at $p_\epsilon$ in terms of $t_\epsilon$ must be of the form $$ h = \frac{\epsilon_1\epsilon_2-1}{4t_\epsilon} + O(1). $$
Moreover, because $\eta$ is odd with respect to $\iota$, it follows that $h$ (after adding a suitable constant if necessary) must also be odd with respect to $\iota$. This implies, in particular, that $h$ vanishes at each of the zeros of $f_n$ (which, by the argument above, are simple zeros). This implies that $h = r\,f_n$ for some $r\in K_{n-1}$ that has no poles and satisfies $r(p_\epsilon) = (\epsilon_1\epsilon_2-1)/4$ for each $\epsilon$. However, since $r$ has no poles and $\hat C_n$ is connected, it follows that $r$ is constant. Thus, it cannot take the two distinct values $0$ and $-1/2$, as the equation $r(p_\epsilon) = (\epsilon_1\epsilon_2-1)/4$ implies.

Thus, the desired $h$ does not exist, and $f_n$ cannot be integrated in elementary terms for any $n>2$.

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    $\begingroup$ Thanks! You could repost this answer (or just a link to it) at M.SE, there is still an active bounty on that question. $\endgroup$ Jun 17, 2014 at 19:14

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