Given an expression such as $f(x) = x^x$, is it possible to provide a thorough and rigorous proof that there is no function $F(x)$ (expressible in terms of known algebraic and transcendental functions) such that $ \frac{d}{dx}F(x) = f(x)$? In other words, how can you rigorously prove that $f(x)$ does not have an elementary antiderivative?

Jack M
  • 26,283
  • 6
  • 57
  • 113
  • 1,984
  • 3
  • 16
  • 19
  • 1
    Differential Galois theory, I think. – Ethan Dec 27 '12 at 06:15
  • 8
    http://en.wikipedia.org/wiki/Liouville%27s_theorem_(differential_algebra) – Qiaochu Yuan Dec 27 '12 at 06:16
  • 3
    Please look at the [Risch Algorithm.](http://en.wikipedia.org/wiki/Risch_algorithm) This is not quite an algorithm, but does the job well, in principle, for many elementary functions. – André Nicolas Dec 27 '12 at 06:19
  • @QiaochuYuan Thanks. Unfortunately, I don't have the mathematical background to thoroughly understand that page. Does the theorem simply place a restriction on which functions can be integrated or does it show with certainty whether or not any given function is integrable? The page seems to indicate the former. – hesson Dec 27 '12 at 06:22
  • 2
    @hesson: the former. The latter is, as far as I know, an open problem. By the way, the way you're using "integrable" is nonstandard; you should say "has an elementary antiderivative" or something like that. – Qiaochu Yuan Dec 27 '12 at 06:46
  • You want the theorem of Liouville above that Qiaochu cited. If it's mathematically beyond you, you could try using e.g. wolframalpha to find an antiderivative. If it can't, chances are there is no antiderivative in terms of standard functions. – Benjamin Dickman Dec 27 '12 at 08:58
  • @AndréNicolas: How is the Risch algorithm not an algorithm? It is a decision procedure that in finite time either proves no antiderivative exists in the class considered, or it hands you one. What more could one want? (Well that it were simpler and faster,of course.) – Marc van Leeuwen Dec 27 '12 at 09:16
  • 3
    @AndréNicolas: OK reading the WP article I see there is a contradiction between its intro and what it says in the section on decidability: you need to, but cannot in general, decide whether an expression is identically $0$. I never realised that (normally something called an algorithm or decision procedure is exactly that). I guess though that running into such an expression that is not obviously $0$ is extremely unlikely. – Marc van Leeuwen Dec 27 '12 at 09:31

1 Answers1


To get some background and references you may start with this SE thread.

Concerning your specific question about $z^z$ here is an extract from a sci.math answer by Matthew P Wiener :

"Finally, we consider the case $I(z^z)$.

So this time, let $F=C(z,l)(t)$, the field of rational functions in $z,l,t$, where $l=\log z\,$ and $t=\exp(z\,l)=z^z$. Note that $z,l,t$ are algebraically independent. (Choose some appropriate domain of definition.) Then $t'=(1+l)t$, so for $a=t$ in the above situation, the partial fraction analysis (of the sort done in the previous posts) shows that the only possibility is for $v=wt+\cdots$ to be the source of the $t$ term on the left, with $w$ in $C(z,l)$.

So this means, equating $t$ coefficients, $1=w'+(l+1)w$. This is a first order ODE, whose solution is $w=\frac{I(z^z)}{z^z}$. So we must prove that no such $w$ exists in $C(z,l)$.
So suppose (as in one of Ray Steiner's posts) $w=\frac PQ$, with $P,Q$ in $C[z,l]$ and having no common factors. Then $z^z= \left(z^z\cdot \frac PQ\right)'=z^z\cdot\frac{[(1+l)PQ+P'Q-PQ']}{Q^2}$, or $Q^2=(1+l)PQ+P'Q-PQ'$.

So $Q|Q'$, meaning $Q$ is a constant, which we may assume to be one. So we have it down to $P'+P+lP=1$.

Let $P=\sum_{i=0}^n [P_i l^i]$, with $P_i, i=0\cdots n \in C[z]$. But then in our equation, there's a dangling $P_n l^{n+1}$ term, a contradiction."


For future references here is a complete re-transcript of Matthew P Wiener's $1997$ sci.math article (converted in $\LaTeX$ by myself : feel free to fix it!).
A neat translation in french by Denis Feldmann is available at his homepage.

What's the antiderivative of $\ e^{-x^2}\ $? of $\ \frac{\sin(x)}x\ $? of $\ x^x\ $?

These, and some similar problems, can't be done.

More precisely, consider the notion of "elementary function". These are the functions that can be expressed in terms of exponentals and logarithms, via the usual algebraic processes, including the solving (with or without radicals) of polynomials. Since the trigonometric functions and their inverses can be expressed in terms of exponentials and logarithms using the complex numbers $\mathbb{C}$, these too are elementary.

The elementary functions are, so to speak, the "precalculus functions".

Then there is a theorem that says certain elementary functions do not have an elementary antiderivative. They still have antiderivatives, but "they can't be done". The more common ones get their own names. Up to some scaling factors, "$\mathrm{erf}$" is the antiderivative of $e^{-x^2}$ and "$\mathrm{Si}$" is the antiderivative of $\frac{\sin(x)}x$, and so on.

For those with a little bit of undergraduate algebra, we sketch a proof of these, and a few others, using the notion of a differential field. These are fields $(F,+,\cdot,1,0)$ equipped with a derivation, that is, a unary operator $'$ satisfying $(a+b)'=a'+b'$ and $(a.b)'=a.b'+a'.b$. Given a differential field $F$, there is a subfield $\mathrm{Con}(F)=\{a:a'=0\}$, called the constants of $F$. We let $I(f)$ denote an antiderivative. We ignore $+C$s.

Most examples in practice are subfields of $M$, the meromorphic functions on $\mathbb{C}$ (or some domain). Because of uniqueness of analytic extensions, one rarely has to specify the precise domain.

Given differential fields $F$ and $G$, with $F$ a subfield of $G$, one calls $G$ an algebraic extension of $F$ if $G$ is a finite field extension of $F$.

One calls $G$ a logarithmic extension of $F$ if $G=F(t)$ for some transcendental $t$ that satisfies $t'=\dfrac{s'}s$, some $s$ in $F$. We may think of $t$ as $\;\log s$, but note that we are not actually talking about a logarithm function on $F$. We simply have a new element with the right derivative. Other "logarithms" would have to be adjoined as needed.

Similarly, one calls $G$ an exponential extension of $F$ if $G=F(t)$ for some transcendental $t$ that satisfies $t'=t.s'$, some $s$ in $F$. Again, we may think of $t$ as $\;\exp s$, but there is no actual exponential function on $F$.

Finally, we call $G$ an elementary differential extension of $F$ if there is a finite chain of subfields from $F$ to $G$, each an algebraic, logarithmic, or exponential extension of the next smaller field.

The following theorem, in the special case of $M$, is due to Liouville. The algebraic generality is due to Rosenlicht. More powerful theorems have been proven by Risch, Davenport, and others, and are at the heart of symbolic integration packages.

A short proof, accessible to those with a solid background in undergraduate algebra, can be found in Rosenlicht's AMM paper (see references). It is probably easier to master its applications first, which often use similar techniques, and then learn the proof.

MAIN THEOREM. Let $F,G$ be differential fields, let $a$ be in $F$, let $y$ be in $G$, and suppose $y'=a$ and $G$ is an elementary differential extension field of $F$, and $\mathrm{Con}(F)=\mathrm{Con}(G)$. Then there exist $c_1,...,c_n \in \mathrm{Con}(F), u_1,\cdots,u_n, v\in F$ such that

$$a = c_1\frac{u_1'}{u_1}+ ... + c_n\frac{u_n'}{u_n}+ v'$$

That is, the only functions that have elementary antiderivatives are the ones that have this very specific form. In words, elementary integrals always consist of a function at the same algebraic "complexity" level as the starting function (the $v$), along with the logarithms of functions at the same algebraic "complexity" level (the $u_i$'s).

This is a very useful theorem for proving non-integrability. Because this topic is of interest, but it is only written up in bits and pieces, I give numerous examples. (Since the original version of this FAQ from way back when, two how-to-work-it write-ups have appeared. See Fitt & Hoare and Marchisotto & Zakeri in the references.)

In the usual case, $F,G$ are subfields of $M$, so $\mathrm{Con}(F)=\mathrm{Con}(G)$ always holds, both being $\mathbb{C}$. As a side comment, we remark that this equality is necessary. Over $\mathbb{R}(x)$, $\frac 1{1+x^2}$ has an elementary antiderivative, but none of the above form.

We first apply this theorem to the case of integrating $f\cdot\exp(g)$, with $f$ and $g$ rational functions. If $g=0$, this is just $f$, which can be integrated via partial fractions. So assume $g$ is nonzero. Let $t=\exp(g)$, so $t'=g't$. Since $g$ is not zero, it has a pole somewhere (perhaps out at infinity), so $\exp(g)$ has an essential singularity, and thus $t$ is transcendental over $C(z)$. Let $F=C(z)(t)$, and let $G$ be an elementary differential extension containing an antiderivative for $f\cdot t$.

Then Liouville's theorem applies, so we can write

$$f\cdot t = c_1\frac{u_1'}{u_1} + \cdots + c_n \frac{u_n'}{u_n} + v'$$

with the $c_i$ constants and the $u_i$ and $v$ in $F$. Each $u_i$ is a ratio of two $C(z)[t]$ polynomials, $\dfrac UV$ say. But $\dfrac {(U/V)'}{U/V}=\dfrac {U'}U-\dfrac{V'}V$ (quotient rule), so we may rewrite the above and assume each $u_i$ is in $C(z)[t]$. And if any $u_i=U\cdot V$ factors, then $\dfrac{(U\cdot V)'}{(U\cdot V)}=\dfrac {U'}U+\dfrac {V'}V$ and so we can further assume each $u_i$ is irreducible over $C(z)$.

What does a typical $\frac {u'}u$ look like? For example, consider the case of $u$ quadratic in $t$. If $A,B,C$ are rational functions in $C(z)$, then $A',B',C'$ are also rational functions in $C(z)$ and

\begin{align} \frac {(A.t^2+B.t+C)'}{A.t^2+B.t+C} &= \frac{A'.t^2 + 2At(gt) + B'.t + B.(gt) + C'}{A.t^2 + B.t + C}\\ &= \frac{(A'+2Ag).t^2 + (B'+Bg).t + C'}{A.t^2 + B.t + C}\\ \end{align}

(Note that contrary to the usual situation, the degree of a polynomial in $t$ stays the same after differentiation. That is because we are taking derivatives with respect to $z$, not $t$. If we write this out explicitly, we get $(t^n)' = \exp(ng)' = ng'\cdot \exp(ng) = ng'\cdot t^n$.)

In general, each $\frac {u'}u$ is a ratio of polynomials of the same degree. We can, by doing one step of a long division, also write it as $D+\frac Ru$, for some $D \in C(z)$ and $R \in C(z)[t]$, with $\deg(R)<\deg(u)$.

By taking partial fractions, we can write $v$ as a sum of a $C(z)[t]$ polynomial and some fractions $\frac P{Q^n}$ with $\deg(P)<\deg(Q)$, $Q$ irreducible, with each $P,Q \in C(z)[t]$. $v'$ will thus be a polynomial plus partial fraction like terms.

Somehow, this is supposed to come out to just $f\cdot t$. By the uniqueness of partial fraction decompositions, all terms other than multiples of $t$ add up to $0$. Only the polynomial part of $v$ can contribute to $f\cdot t$, and this must be a monomial over $C(z)$. So $f\cdot t=(h\cdot t)'$, for some rational $h$. (The temptation to assert $v=h\cdot t$ here is incorrect, as there could be some $C(z)$ term, cancelled by $\frac {u'}u$ terms. We only need to identify the terms in $v$ that contribute to $f\cdot t$, so this does not matter.)

Summarizing, if $f\cdot \exp(g)$ has an elementary antiderivative, with $f$ and $g$ rational functions, $g$ nonzero, then it is of the form $h\cdot \exp(g)$, with $h$ rational.

We work out particular examples, of this and related applications. A bracketed function can be reduced to the specified example by a change of variables.

$\quad\boxed{\displaystyle\exp\bigl(z^2\bigr)}$ $\quad\left[\sqrt{z}\cdot\exp(z),\frac{\exp(z)}{\sqrt{z}}\right]$

Let $h\cdot \exp\bigl(z^2\bigr)$ be its antiderivative. Then $h'+2zh=1$.
Solving this ODE gives $h=\exp(-z^2)\cdot I\left(\exp\bigl(z^2\bigr)\right)$, which has no pole (except perhaps at infinity), so $h$, if rational, must be a polynomial. But the derivative of $h$ cannot cancel the leading power of $2zh$, contradiction.

$\quad\boxed{\displaystyle\frac{\exp(z)}z}$ $\quad\left[\exp(\exp(z)),\frac 1{\log(z)}\right]$

Let $h\cdot \exp(z)$ be an antiderivative. Then $h'+h=\frac 1z$. I know of two quick ways to prove that $h$ is not rational.

One can explicitly solve the first order ODE (getting $\exp(-z)\cdot I\left(\frac{\exp(z)}z\right))$, and then notice that the solution has a logarithmic singularity at zero. For example, $h(z)\to\infty$ but $\sqrt{z}\cdot h(z)\to 0$ as $z\to 0$. No rational function does this.

Or one can assume $h$ has a partial fraction decomposition. Obviously no $h'$ term will give $\frac 1z$, so $\frac 1z$ must be present in $h$ already. But $\left(\frac 1z\right)'=-\frac 1{z^2}$, and this is part of $h'$. So there is a $\frac 1{z^2}$ in $h$ to cancel this. But $\left(\frac 1{z^2}\right)'$ is $-\frac 2{z^3}$, and this is again part of $h'$. And again, something in $h$ cancels this, etc etc etc. This infinite regression is impossible.

$\quad\boxed{\displaystyle\frac {\sin(z)}z}$ $\quad[\sin(\exp(z))]$

$\quad\boxed{\displaystyle\sin\bigl(z^2\bigr)}$ $\quad\left[\sqrt{z}\sin(z),\frac{\sin(z)}{\sqrt{z}}\right]$

Since $\sin(z)=\frac 1{2i}[\exp(iz)-\exp(-iz)]$, we merely rework the above $f\cdot \exp(g)$ result. Let $f$ be rational, let $t=\exp(iz)$ (so $\frac {t'}t=i$) and let $T=\exp(iz^2)$ (so $\frac{T'}T=2iz$) and we want an antiderivative of either $\frac 1{2i}f\cdot\left(t-\frac 1t\right)$ or $\frac 1{2i}f\cdot(T-\frac 1T)$. For the former, the same partial fraction results still apply in identifying $\frac 1{2i}f\cdot t=(h\cdot t)'=(h'+ih)\cdot t$, which can't happen for $f=\frac 1z$, as above. In the case of $f\cdot\sin\bigl(z^2\bigr)$, we want $\frac 1{2i}f\cdot T=(h\cdot T)'=(h'+2izh)\cdot T$, and again, this can't happen for $f=1$, as above.

Although done, we push this analysis further in the $f\cdot \sin(z)$ case, as there are extra terms hanging around. This time around, the conclusion gives an additional $\frac kt$ term inside $v$, so we have $-\frac 1{2i}\frac ft=\left(\frac kt\right)'=\frac{k'-ik}t$. So the antiderivative of $\frac 1{2i}f\cdot\left(t-\frac 1t\right)$ is $h\cdot t+\frac kt$.

If $f$ is even and real, then $h$ and $k$ (like $t=\exp(iz)$ and $\frac 1t=\exp(-iz)$) are parity flips of each other, so (as expected) the antiderivative is even.
Letting $C=\cos(z), S=\sin(z), h=H+iF$ and $k=K+iG$, the real (and only) part of the antiderivative of $f$ is $(HC-FS)+(KC+GS)=(H+K)C+(G-F)S$.
So over the reals, we find that the antiderivative of (rational even).$\sin(x)$ is of the form (rational even).$\cos(x)+$ (rational odd).$\sin(x)$.

A similar result holds for (odd)$\cdot\sin(x)$, (even)$\cdot\cos(x)$, (odd)$\cdot\cos(x)$. And since a rational function is the sum of its (rational) even and odd parts, (rational)$\cdot\sin$ integrates to (rational)$\cdot\sin$ + (rational)$\cdot\cos$, or not at all.

Let's backtrack, and apply this to $\dfrac {\sin(x)}x$ directly, using reals only. If it has an elementary antiderivative, it must be of the form $E\cdot S+O\cdot C$. Taking derivatives gives $(E'-O)\cdot S+(E+O')\cdot C$. As with partial fractions, we have a unique $R(x)[S,C]$ representation here (this is a bit tricky, as $S^2=1-C^2$: this step can be proven directly or via solving for $t, \frac 1t$ coefficients over $C$). So $E'-O=\frac 1x$ and $E+O'=0$, or $O''+O=-\frac 1x$. Expressing $O$ in partial fraction form, it is clear only $(-\frac 1x)$ in $O$ can contribute a $-\frac 1x$. So there is a $-\frac 2{x^3}$ term in $O''$, so there is a $\frac 2{x^3}$ term in $O$ to cancel it, and so on, an infinite regress. Hence, there is no such rational $O$.

$\quad\boxed{\displaystyle\frac{\arcsin(z)}z}$ $\quad[z.\tan(z)]$

We consider the case where $F=C(z,Z)(t)$ as a subfield of the meromorphic functions on some domain, where $z$ is the identify function, $Z=\sqrt{1-z^2}$, and $t=\arcsin z$. Then $Z'=-\frac zZ$, and $t'=\frac 1Z$. We ask in the main theorem result if this can happen with $a=\frac tz$ and some field $G$. $t$ is transcendental over $C(z,Z)$, since it has infinite branch points.

So we consider the more general situation of $f(z)\cdot \arcsin(z)$ where $f(z)$ is rational in $z$ and $\sqrt{1-z^2}$. By letting $z=\frac {2w}{1+w^2}$, note that members of $C(z,Z)$ are always elementarily integrable.

Because $x^2+y^2-1$ is irreducible, $\frac{C[x,y]}{x^2+y^2-1}$ is an integral domain, $C(z,Z)$ is isomorphic to its field of quotients in the obvious manner, and $C(z,Z)[t]$ is a UFD whose field of quotients is amenable to partial fraction analysis in the variable $t$. What follows takes place at times in various $z$-algebraic extensions of $C(z,Z)$ (which may not have unique factorization), but the terms must combine to give something in $C(z,Z)(t)$, where partial fraction decompositions are unique, and hence the $t$ term will be as claimed.

Thus, if we can integrate $f(z)\cdot\arcsin(z)$, we have $f\cdot t$ = $\sum$ of $\frac {u'}u s$ and $v'$, by the main theorem.

The $u$ terms can, by logarithmic differentiation in the appropriate algebraic extension field (recall that roots are analytic functions of the coefficients, and $t$ is transcendental over $C(z,Z)$), be assumed to all be linear $t+r$, with $r$ algebraic over $z$. Then $\frac {u'}u=\frac {1/Z+r'}{t+r}$. When we combine such terms back in $C(z,Z)$, they don't form a $t$ term (nor any higher power of $t$, nor a constant).

Partial fraction decomposition of $v$ gives us a polynomial in $t$, with coefficients in $C(z,Z)$, plus multiples of powers of linear $t$ terms. The latter don't contribute to a $t$ term, as above.

If the polynomial is linear or quadratic, say $v=g\cdot t^2 + h\cdot t + k$, then $v'=g'\cdot t^2 + \left(\frac{2g}Z+h'\right)\cdot t + \left(\frac hZ+k'\right)$. Nothing can cancel the $g'$, so $g$ is just a constant $c$. Then $\frac {2c}Z+h'=f$ or $I(f\cdot t)=2c\cdot t+I(h'\cdot t)$. The $I(h'.t)$ can be integrated by parts. So the antiderivative works out to $c\cdot(\arcsin(z))^2 + h(z)\cdot \arcsin(z) - I\left(\frac{h(z)}{\sqrt{1-z^2}}\right)$, and as observed above, the latter is elementary.

If the polynomial is cubic or higher, let $v=A.t^n+B.t^{n-1}+\cdots$, then $v'=A'.t^n + \left(n\cdot\frac AZ+B'\right).t^{n-1} +\cdots$ $A$ must be a constant $c$. But then $\frac{nc}Z+B'=0$, so $B=-nct$, contradicting $B$ being in $C(z,Z)$.

In particular, since $\frac 1z + \frac c{\sqrt{1-z^2}}$ does not have a rational in "$z$ and/or $\sqrt{1-z^2}$" antiderivative, $\frac {\arcsin(z)}z$ does not have an elementary integral.

$\quad\boxed{\displaystyle z^z}$

In this case, let $F=C(z,l)(t)$, the field of rational functions in $z,l,t$, where $l=\log z$ and $t=\exp(z\,l)=z^z$. Note that $z,l,t$ are algebraically independent. (Choose some appropriate domain of definition.) Then $t'=(1+l)t$, so for $a=t$ in the above situation, the partial fraction analysis (of the sort done in the previous posts) shows that the only possibility is for $v=wt+\cdots$ to be the source of the $t$ term on the left, with $w$ in $C(z,l)$.

So this means, equating $t$ coefficients, $1=w'+(l+1)w$. This is a first order ODE, whose solution is $w=\frac{I(z^z)}{z^z}$. So we must prove that no such $w$ exists in $C(z,l)$. So suppose (as in one of Ray Steiner's posts) $w=P/Q$, with $P,Q$ in $C[z,l]$ and having no common factors. Then $z^z= \left(z^z\cdot \frac PQ\right)'=z^z\cdot\frac{[(1+l)PQ+P'Q-PQ']}{Q^2}$, or $Q^2=(1+l)PQ+P'Q-PQ'$.

So $Q|Q'$, meaning $Q$ is a constant, which we may assume to be one. So we have it down to $P'+P+lP=1$.

Let $P=\sum_{i=0}^n [P_i l^i]$, with $P_i, i=0\cdots n \in C[z]$. But then in our equation, there's a dangling $P_n l^{n+1}$ term, a contradiction.

On a slight tangent, this theorem of Liouville will not tell you that Bessel functions are not elementary, since they are defined by second order ODEs. This can be proven using differential Galois theory. A variant of the above theorem of Liouville, with a different normal form, does show however that $J_0$ cannot be integrated in terms of elementary methods augmented with Bessel functions.

What follows is a fairly complete sketch of the proof of the Main Theorem. First, I just state some easy (if you've had Galois Theory 101) lemmas.

Throughout the lemmas $F$ is a differential field, and $t$ is transcendental over $F$.

  • Lemma $1$: If $K$ is an algebraic extension field of $F$, then there exists a unique way to extend the derivation map from $F$ to $K$ so as to make $K$ into a differential field.
  • Lemma $2$: If $K=F(t)$ is a differential field with derivation extending $F$'s, and $t'$ is in $F$, then for any polynomial $f(t)$ in $F[t]$, $f(t)'$ is a polynomial in $F[t]$ of the same degree (if the leading coefficient is not in $\mathrm{Con}(F)$) or of degree one less (if the leading coefficient is in $\mathrm{Con}(F)$).
  • Lemma $3$: If $K=F(t)$ is a differential field with derivation extending $F$'s, and $\frac{t'}t$ is in $F$, then for any $a$ in $F$, $n$ a positive integer, there exists $h$ in $F$ such that $(a\cdot t^n)'=h\cdot t^n$. More generally, if $f(t)$ is any polynomial in $F[t]$, then $f(t)'$ is of the same degree as $f(t)$, and is a multiple of $f(t)$ iff $f(t)$ is a monomial.

These are all fairly elementary. For example, $(a\cdot t^n)'=\bigl(a'+a\frac {t'}t\bigr)\cdot t^n$ in lemma $3$. The final 'iff' in lemma $3$ is where transcendence of $t$ comes in. Lemma $1$ in the usual case of subfields of $M$ is an easy consequence of the implicit function theorem.

MAIN THEOREM. Let $F,G$ be differential fields, let $a$ be in $F$, let $y$ be in $G$, and suppose $y'=a$ and $G$ is an elementary differential extension field of $F$, and $\mathrm{Con}(F)=\mathrm{Con}(G)$. Then there exist $c_1,...,c_n \in \mathrm{Con}(F), u_1,\cdots,u_n, v\in F$ such that

$$(*)\quad a = c_1\frac{u_1'}{u_1}+ ... + c_n\frac{u_n'}{u_n}+ v'$$

In other words, the only functions that have elementary antiderivatives are the ones that have this very specific form.


By assumption there exists a finite chain of fields connecting $F$ to $G$ such that the extension from one field to the next is given by performing an algebraic, logarithmic, or exponential extension. We show that if the form $(*)$ can be satisfied with values in $F2$, and $F2$ is one of the three kinds of allowable extensions of $F1$, then the form $(*)$ can be satisfied in $F1$. The form $(*)$ is obviously satisfied in $G$: let all the $c$'s be $0$, the $u$'s be $1$, and let $v$ be the original $y$ for which $y'=a$. Thus, if the form $(*)$ can be pulled down one field, we will be able to pull it down to $F$, and the theorem holds.

So we may assume without loss of generality that $G=F(t)$.

  • Case $1$ : $t$ is algebraic over $F$. Say $t$ is of degree $k$. Then there are polynomials $U_i$ and $V$ such that $U_i(t)=u_i$ and $V(t)=v$. So we have $$a = c_1 \frac{U_1(t)'}{U_1(t)} +\cdots + c_n \frac{ U_n(t)'}{U_n(t)} + V(t)'$$ Now, by the uniqueness of extensions of derivatives in the algebraic case, we may replace $t$ by any of its conjugates $t_1,\cdots, t_k,$ and the same equation holds. In other words, because $a$ is in $F$, it is fixed under the Galois automorphisms. Summing up over the conjugates, and converting the $\frac {U'}U$ terms into products using logarithmic differentiation, we have $$k a = c_1 \frac{[U_1(t_1)\times\cdots\times U_1(t_k)]'}{U_1(t_1)\times \cdots \times U_n(t_k)}+ \cdots + [V(t_1)+\cdots +V(t_k)]'$$ But the expressions in $[\cdots]$ are symmetric polynomials in $t_i$, and as they are polynomials with coefficients in $F$, the resulting expressions are in $F$. So dividing by $k$ gives us $(*)$ holding in $F$.

  • Case $2$ : $t$ is logarithmic over $F$. Because of logarithmic differentiation we may assume that the $u$'s are monic and irreducible in $t$ and distinct. Furthermore, we may assume v has been decomposed into partial fractions. The fractions can only be of the form $\dfrac f{g^j}$, where $\deg(f)<\deg(g)$ and $g$ is monic irreducible. The fact that no terms outside of $F$ appear on the left hand side of $(*)$, namely just $a$ appears, means a lot of cancellation must be occuring.

    Let $t'=\dfrac{s'}s$, for some $s$ in $F$. If $f(t)$ is monic in $F[t]$, then $f(t)'$ is also in $F[t]$, of one less degree. Thus $f(t)$ does not divide $f(t)'$. In particular, all the $\dfrac{u'}u$ terms are in lowest terms already. In the $\dfrac f{g^j}$ terms in $v$, we have $a g^{j+1}$ denominator contribution in $v'$ of the form $-jf\dfrac{g'}{g^{j+1}}$. But $g$ doesn't divide $fg'$, so no cancellation occurs. But no $\dfrac{u'}u$ term can cancel, as the $u$'s are irreducible, and no $\dfrac{(**)}{g^{j+1}}$ term appears in $a$, because $a$ is a member of $F$. Thus no $\dfrac f{g^j}$ term occurs at all in $v$. But then none of the $u$'s can be outside of $F$, since nothing can cancel them. (Remember the $u$'s are distinct, monic, and irreducible.) Thus each of the $u$'s is in $F$ already, and $v$ is a polynomial. But $v' = a -$ expression in $u$'s, so $v'$ is in $F$ also. Thus $v = b t + c$ for some $b$ in $\mathrm{con}(F)$, $c$ in $F$, by lemma 2. Then $$a= c_1 \frac{u_1'}{u_1} +\cdots + c_n\frac{u_n'}{u_n} + b \frac{s'}s + c'$$ is the desired form. So case 2 holds.

  • Case $3$ : $t$ is exponential over $F$. So let $\dfrac {t'}t=s'$ for some $s$ in $F$. As in case 2 above, we may assume all the $u$'s are monic, irreducible, and distinct and put $v$ in partial fraction decomposition form. Indeed the argument is identical as in case 2 until we try to conclude what form $v$ is. Here lemma 3 tells us that $v$ is a finite sum of terms $b\cdot t^j$ where each coefficient is in $F$. Each of the $u$'s is also in $F$, with the possible exception that one of them may be $t$. Thus every $\dfrac {u'}u$ term is in $F$, so again we conclude $v'$ is in $F$. By lemma 3, $v$ is in $F$. So if every $u$ is in $F$, $a$ is in the desired form. Otherwise, one of the $u$'s, say $u_n$, is actually $t$, then $$a = c_1\frac{u_1'}{u_1} + \cdots + (c_n s + v)'$$ is the desired form. So case 3 holds.


A D Fitt & G T Q Hoare "The closed-form integration of arbitrary functions", Mathematical Gazette (1993), pp 227-236.
I Kaplansky An introduction to differential algebra (Hermann, 1957)
E R Kolchin Differential algebra and algebraic groups (Academic Press, 1973)
A R Magid "Lectures on differential Galois theory" (AMS, 1994)
E Marchisotto & G Zakeri "An invitation to integration in finite terms", COLLEGE MATHEMATICS JOURNAL (1994), pp 295-308.
J F Ritt Integration in finite terms (Columbia, 1948).
J F Ritt Differential algebra (AMS, 1950).
M Rosenlicht "Liouville's theorem on functions with elementary integrals", PACIFIC JOURNAL OF MATHEMATICS (1968), pp 153-161.
M Rosenlicht "Integration in finite terms", AMERICAN MATHEMATICS MONTHLY, (1972), pp 963-972.
G N Watson A treatise on the theory of Bessel functions (Cambridge, 1962).

-Matthew P Wiener

Raymond Manzoni
  • 41,072
  • 5
  • 79
  • 133
  • 154
    Raymond, I don't believe you've given enough detail. – Will Jagy Dec 27 '12 at 20:10
  • Yes @Will the proof and references were missing (or did you speak of the $x^x$ case ?). – Raymond Manzoni Dec 27 '12 at 20:21
  • 17
    (I am neglecting the possibility of people making fun of my short answers... :-)) – Raymond Manzoni Dec 27 '12 at 20:30
  • 1
    I am amazed that you managed to find more to edit in. There is a good chance Matthew Wiener is the one who started graduate school with me, shared an office, worked with Solovay, then went back to Pennsylvania to care for his ill father in the late 1980's. Oh, even in graduate school he got interested in online forums very early, spent a ton of time on those. But he knew absolutely everything. – Will Jagy Dec 27 '12 at 20:45
  • @Will: we used his excellent article rather often at sci.math. His demonstrations of non-elementary are very clear and straightforward (not sure the OP will appreciate this remark...) that's not so common (I appreciated the Geddes, Czapor and Labahn book but it's longer!). Cheers, – Raymond Manzoni Dec 27 '12 at 20:53
  • 4
    For your listening pleasure: http://meta.math.stackexchange.com/questions/6992/is-this-the-longest-answer-on-the-site – Will Jagy Dec 27 '12 at 21:26
  • 2
    Ha ha Thanks for that thread @Will ! Some hard competition I see from Joriki, yourself, Javier, Arturo and others. Another of my 'impressive' posts (from the weight point of view) is this [one](http://math.stackexchange.com/questions/138084/regularization-of-a-divergent-integral/139093#139093). Editing was really awful (it further contains pngs files!). – Raymond Manzoni Dec 27 '12 at 22:11
  • 9
    @Raymond, many thanks for putting so much effort to answer this question. – hesson Dec 27 '12 at 22:43
  • Thanks @hesson ! Non elementary proofs are often asked here perhaps that Matthew's article will help other people too (as well as latex interpreter optimizer...). – Raymond Manzoni Dec 27 '12 at 23:05
  • @RaymondManzoni: Please add a longer answer; such short answers are disapproved. ;) – P.K. Dec 28 '12 at 20:12
  • 17
    @RaymondManzoni Thank you. Thank you, thank you, thank you, thank you. Thank you for giving this superlatively comprehensive explanation, coming "free, *gratis*, and for nothing", out of the goodness of your heart. I will be returning to this again and again as I soak up the details. What an amazing addition to this site. (Also, thanks at WillJagy for tipping me off!) – Chris Jan 01 '13 at 20:29
  • Woah ! I am glad you liked that @user1296727 but all the credit is due to the author **Matthew P. Wiener** (I merely edited his excellent explications in $\LaTeX$ for more pleasant reading as did Denis Feldmann earlier for his french translation...). I'll be happy if Matthew's work may be useful for others here (this answer is CW : if you find typos/errors please edit it!). Anyway a Happy New Year ! – Raymond Manzoni Jan 01 '13 at 21:22
  • @RaymondManzoni too long didn't read –  Feb 08 '14 at 15:27
  • 2
    Thanks again @حكيم الفيلسوف الضائع (for the bounty) but please don't make me an annuitant! :-) Cheers, – Raymond Manzoni May 17 '14 at 15:38
  • 2
    Hahaha @RaymondManzoni Cheers! ;-) – Hakim May 17 '14 at 18:58
  • Thanks for the edit @Jarek Kuben ! – Raymond Manzoni May 07 '16 at 23:54