Let fn(x) be recursively defined as

f0(x)=1, fn+1(x)=√x+fn(x),

i.e. fn(x) contains n radicals and n occurences of x:

f1(x)=√x+1, f2(x)=√x+√x+1, f3(x)=√x+√x+√x+1, …The functions f0(x), f1(x) and f2(x) are integrable in elementary functions, e.g.:

∫√x+√x+1dx=(2x3+√x+16−14)√x+√x+1+58ln(2√x+1+2√x+√x+1+1).

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

Update: The question is reposted atMathOverflowas was suggested by moderator.

**Answer**

**I posted an answer to this question as an answer to the related question on MathOverflow. See my (second) answer at https://mathoverflow.net/questions/171733. However, as Krokop pointed out, it would be better to have a copy here on Math.SE to the original question. Thanks to Krokop for copying my answer over to Math.SE.**

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

Assume n>2 (NB: This is important, because the argument below will *not* work for n≤2; the reader may enjoy finding where it breaks down), and let Kn=C(x,fn(x)) be the elementary differential field generated by x and fn(x). Then Kn is the field of meromorphic functions on the normalization ˆCn of the algebraic curve Cn defined by the minimal degree y-monic polynomial Pn(x,y) that satisfies Pn(x,fn(x))≡0. This minimal degree is 2n; for example, P2(x,y)=(y2−x)2−x−1 and P3(x,y)=((y2−x)2−x)2−x−1, etc.

Since Pn+1(x,y)=(Pn(x,y)+1)2−x−1 for n≥1 with P1(x,y)=y2−x−1, one sees, by applying the Eisenstein Criterion to Pn(x,y) regarded as an element of D[y] with D being the integral domain C[x], that Pn(x,y) is irreducible for all n≥1. Hence, ˆCn is connected.

It will be important in what follows to observe that Kn has an involution ι that fixes x and sends fn(x) to −fn(x); this is because Pn(x,y) is an even polynomial in y. The fixed field of ι is C(x,fn(x)2), and the (−1)-eigenspace of ι is C(x,fn(x)2)fn(x)=Kn−1⋅fn(x).

Now, the curve Cn⊂CP2 has only one point on the line at infinity, namely [1,0,0], but the normalization ˆCn has 2n−1 points lying over this point. They can be parametrized as follows: First, establish the convention that √u means the unique analytic function on the complex u-plane minus its negative axis and 0 that satisfies √1=1 and (√u)2=u. Let ϵ=(ϵ1,…,ϵn−1) be any sequence with ϵk2=1 and consider the sequence of functions gϵk(t) defined by the criteria gϵ1(t)=√1+t2 and gϵk+1(t)=√1+ϵn−ktgϵk(t) for 1≤k<n. Choose, as one may, a δn>0 sufficiently small so that, when t is complex and satisfies |t|<δn, all of the functions gϵk are analytic when |t|<δn. In particular, one finds an expansion

gϵn(t)=1+12ϵ1t+18(2ϵ1ϵ2−1)t2+O(t3).

Also, it is easy to verify that the disk in CP2 defined by

[x,y,1]=[1, tgϵn(t), t2]for|t|<δn

is a nonsingular parametrization of a branch of Cn in a neighborhood of the point [1,0,0]. In the normalization ˆCn, this is then a local parametrization of a neighborhood of a point pϵ∈ˆCn.

Obviously, this describes 2n−1 distinct points on ˆCn.

When x and fn are regarded as meromorphic functions on ˆCn,

it follows that there is a unique local coordinate chart tϵ:Dϵ→D(0,δn)⊂C of an open disk Dϵ⊂ˆCn about pϵ such that tϵ(pϵ)=0 and on which one

has formulae

x=1tϵ2andfn(x)=gϵn(tϵ)tϵ=1+12ϵ1 tϵ+18(2ϵ1ϵ2−1) tϵ2tϵ+O(tϵ2).

In particular, it follows that fn(x), as a meromorphic function on ˆCn,

has polar divisor equal to the sum of the pϵ and hence has degree 2n−1. Of course, this implies that the zero divisor of fn(x) on ˆCn must be of degree 2n−1 as well.

Note that the functions gϵk satisfy g−ϵk(−t)=gϵk(t), where −ϵ=(−ϵ1,…,−ϵn−1).

This implies that ι(pϵ)=p−ϵ and that

tϵ∘ι=−t−ϵ.

Now, the 2n−1 zeroes of fn(x) on ˆCn are distinct, for they are the zeros of the polynomial qn(x)=Pn(x,0)=(qn−1+1)2−x−1, and the discriminant of qn, being the resultant of qn and q′n, is clearly an odd integer, and hence is not zero. Thus, Cn is a branched double cover of Cn−1, branched exactly where fn has its zeros. This induces a branched cover πn:ˆCn→ˆCn−1 that is exactly the quotient of ˆCn by the involution ι (whose fixed points are where fn has its zeros). Since one then has the Riemann-Hurwitz formula

χ(ˆCn)=2χ(ˆCn−1)−Bn=2χ(ˆCn−1)−2n−1,

and χ(ˆC1)=χ(ˆC2)=2, induction gives χ(ˆCn)=(3−n)2n−1, so the genus of ˆCn is (n−3)2n−2+1. (This won't actually be needed below, but it is interesting.)

The only poles of x and fn(x) on ˆCn are the points pϵ,

and computation using the above expansions shows that,

in a neighborhood of pϵ, one has an expansion of the form

fn(x)dx−d(fn(x)(12 x+16 fn(x)2))=((1−ϵ1ϵ2)4tϵ2+O(tϵ−1)) dtϵ .

Thus, the meromorphic differential η on ˆCn

defined by the left hand side of this equation has, at worst, double poles

at the points pϵ and no other poles.

Now, by Liouville's Theorem, fn has an elementary antiderivative if and only if fn(x) dx and, hence, the form η are expressible as finite linear combinations of exact differentials and log-exact differentials.

Thus, fn(x) has an elementary antiderivative if and only if η is expressible in the form

η=dh+m∑i=1cidgigi

for some h,g1,⋯gm∈Kn and some constants c1,…,cm. Suppose that these exist. Since η has, at worst, double poles at the pϵ and no other poles, it follows that h must have, at worst, simple poles at the points pϵ and no other poles; in fact, h is uniquely determined up to an additive constant because its expansion at pϵ in terms of tϵ must be of the form

h=ϵ1ϵ2−14tϵ+O(1).

Moreover, because η is odd with respect to ι, it follows that h (after adding a suitable constant if necessary) must also be odd with respect to ι. This implies, in particular, that h vanishes at each of the zeros of fn (which, by the argument above, are simple zeros). This implies that h=rfn for some r∈Kn−1 that has no poles and satisfies r(pϵ)=(ϵ1ϵ2−1)/4 for each ϵ. However, since r has no poles and ˆCn is connected, it follows that r is constant. Thus, it cannot take the two distinct values 0 and −1/2, as the equation r(pϵ)=(ϵ1ϵ2−1)/4 implies.

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

**Attribution***Source : Link , Question Author : Vladimir Reshetnikov , Answer Author : Community*