\documentclass{article}
\usepackage{graphicx}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{enumerate}

\usepackage{hyperref}

\usepackage{ifthen}
\usepackage{ulem} % for strikethrough \sout{...} command
\usepackage{color}

\usepackage{wasysym}

\setlength{\unitlength}{0.1in}

\setlength{\oddsidemargin}{-0.35in}
\setlength{\evensidemargin}{-0.35in}
\setlength{\topmargin}{-0.35in}
\setlength{\headheight}{0.0in}
\setlength{\headsep}{0.0in}
\setlength{\topskip}{0.0in}
\setlength{\textheight}{9.45in}
\setlength{\textwidth}{7.2in}

\newcommand{\bdot}{\, {\scriptstyle{\bullet}} \,}
\newcommand{\B}[1]{{\bf {#1}}}
\newcommand{\la}{\langle}
\newcommand{\ra}{\rangle}

\newcommand{\myspace}{\parbox[c][0.25in]{0.25in}{\hfill}}
\newcommand{\varspace}[2]{\mbox{\parbox[c][#1]{#2}{\hfill}}}
\newcommand{\force}{{\noindent $\mbox{\;}$}}

\newcommand{\diagquotient}[2]{\displaystyle{{#1 \atop \;}\hspace*{-0.1in} \mbox{\put(0,0){\line(2,1){2}}} \hspace*{0.1in} {\; \atop #2}}}


\newcommand{\F}{\mathbb{F}}
\newcommand{\E}{\mathbb{E}}
\newcommand{\K}{\mathbb{K}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\C}{\mathbb{C}}

\newcommand{\LEMMA}{\vspace{0.05in} \noindent \fbox{\large\sc Lemma:}  }
\newcommand{\COR}{\vspace{0.05in} \noindent \fbox{\large\sc Corollary:}  }
\newcommand{\DEF}{\vspace{0.05in} \noindent \fbox{\large\sc Definition:}  }
\newcommand{\THM}{\vspace{0.05in} \noindent \fbox{\large\sc Theorem:}  }
\newcommand{\EX}{\vspace{0.05in} \noindent \fbox{\large\sc Example:}  }
\newcommand{\bproof}{\vspace{0.1in} \noindent {\bf Proof:} }
\newcommand{\eproof}{$\blacksquare$ \vspace{0.05in}}


\begin{document}

%\pagestyle{empty}

\noindent
{\bf Math 4010/5530}
\hfill {\Large \bf Elementary Functions and Liouville's Theorem} \hfill 
{\bf April 2016}

\vspace{0.1in}

\DEF Let $R$ be a commutative ring with $1$. We call a map $\partial:R \to R$ a {\bf derivation} if $\partial(a+b)=\partial(a)+\partial(b)$ and $\partial(ab)=\partial(a)b+a\partial(b)$ for all $a,b \in R$. A ring equipped with a particular derivation is called a {\bf differential ring}. For convenience, we write $\partial(a)=a'$ (just like in calculus).
If $R$ is a differential ring and an integral domain, it is a  {\bf differential integral domain}. If $R$ is a differential ring and a field, it is a {\bf differential field}. 

\vspace{0.1in}

\noindent Let $R$ be a differential ring.
\begin{itemize} \setlength\itemsep{0pt}
\item $1' = (1\cdot 1)' = 1' \cdot 1 + 1 \cdot 1'$ so $1'=1'+1'$ and so $0=1'$.  Notice that $\partial$ is a group homomorphism (consider $R$ as an abelian group under addition), so $\partial(na)=n\partial(a)$ for all $a \in R$ and $n \in \Z$. Therefore, $\partial(n1)=n\partial(1)=n0=0$. Thus everything in the prime subring is a {\bf constant} (i.e. has derivative zero).
\item Given $a \in R$, notice that $(a^2)' = (a \cdot a)' = a'a+aa' = 2aa'$. In fact, we can prove (using induction), that 

\vspace{-0.15in}

$$(a^n)'=na^{n-1}a' \mbox{ for any }n \in \Z_{\geq 0} \qquad \mbox{(The power rule)}$$

\vspace{-0.05in}

\item Let $a \in R^\times$ ($a$ is a unit). Then $0=1' = (aa^{-1})' = a'a^{-1} + a(a^{-1})'$ so $a(a^{-1})' = -a^{-1}a'$. Dividing by $a$, we get that $(a^{-1})' = -a^{-2}a'$.  Again, using induction, we have that $(a^n)'=na^{n-1}a'$ for all $n \in \Z$ (if $a^{-1}$ exists).
\item Let $a \in R$ and $b \in R^\times$. Then $(ab^{-1})' = a'b^{-1}+a(b^{-1})' = a'b^{-1}+a(-b^{-2}b')$. Therefore,

\vspace{-0.15in}

$$ \partial\left(\dfrac{a}{b}\right) = \dfrac{a'b-ab'}{b^2} \qquad \mbox{(The quotient rule)}$$

\vspace{-0.05in}

This then implies that if $\mathbb{Q} \subseteq R$, every element of $\mathbb{Q}$ is a constant.

{\it Note:} If $R$ is a differential integral domain, then $R$'s field of fractions $\mathbb{F}$ can be turned into a differential field in a unique way. We define $\partial(a/b)$ via the quotient rule. It is a simple exercise to show that this is well defined, extends the definition of $\partial$ from $R$ to $\mathbb{F}$, and there is no other way to do this.
\end{itemize}

\DEF A {\bf differential subring} of a differential ring $R$, is a subring (containing $1$) which is closed under differentiation. Thus $S \subseteq R$ is a differential subring if $1 \in S$, $a,b \in S$ implies that $a-b$, $ab$, and $a' \in S$. 

Likewise, a {\bf differential ideal} of $R$ is an ideal that is closed under differentiation.  If $I$ is a differential ideal of $R$, $R/I$ becomes a differential ring if we define $\partial(a+I)=\partial(a)+I$ (it's easy to show that this is well defined). 

A {\bf differential homomorphism} between two differential rings $R,S$ is a map $\varphi:R \to S$ which is a ring homomorphism (i.e. $\varphi(a+b)=\varphi(a)+\varphi(b)$, $\varphi(ab)=\varphi(a)\varphi(b)$, and $\varphi(1)=1$) that preserves differentiation (i.e. $\varphi(a') = \varphi(a)'$). It's easy to show that the kernel of a differential homomorphism, $\mathrm{ker}(\varphi)=\{ x \in R \;|\; \varphi(x)=0\}$, is a differential ideal of the domain. Also, the image is a differential subring of the codomain and the isomorphism theorems all hold.

%\vspace{0.1in}

\LEMMA Let $R$ be a differential ring. The {\bf constants} of $R$ is the differential subring $C = \mathrm{ker}(\partial) = \{ x \in R \;|\; x'=0 \}$. Moreover, if $R$ is a differential field, then $C$ is a differential subfield.

\bproof  First, $1 \in C$ since $1'=0$. Next, suppose $a,b \in C$. Then $(a-b)' = a'-b'=0-0=0$ and $(ab)' = a'b+ab' = 0b+a0=0$ so $a-b, ab \in C$. Finally, for all $a \in C$, $a'=0 \in C$. So $C$ is a differential subring. Now in addition, suppose $R$ is a field. Let $a \in C$ such that $a \not=0$. Then $(a^{-1})' = -a^{-2}a' = -a^{-2}0=0$ so $a^{-1} \in C$. Thus $C$ is a field.
\eproof

\EX Any commutative ring with $1$ is a differential ring when given the derivation $\partial=0$. In particular, $\mathbb{Q}$ is a differential ring with $\partial=0$ and by some of the above discussion, this is our only choice!

The field of rational functions with real coefficients, $\mathbb{R}(x)$, is a differential field when given $\partial = \dfrac{d}{dx}$ (the usual derivative). Here the field of constants is $\mathbb{R}$ (as we would expect).

It is interesting (and strange) to consider that $\mathbb{Q}(\pi)$ (which is isomorphic to $\mathbb{Q}(x)$) is a differential field when given the derivation $\partial = ``d/d\pi$''. In other words, $\pi'=1$ and $(\pi^3)' = 3\pi^2$ etc. This indicates that $\mathbb{R}$ can be turned into a differential field in many {\it very strange} ways. 

%\vspace{0.1in}

\LEMMA Let $\F$ be a differential field of characteristic 0 and let $\K / \F$ be an algebraic extension (as regular old fields). The derivation on $\F$ can be extended to a derivative on $\K$ in exactly one way (i.e. $\K$ can be turned into a differential field and this can only be done in one way).

\bproof  [Sketch] I'll just show uniqueness. For a complete proof see \cite{crespo} Proposition 5.3.1.

Let $\alpha \in \K$ and let $f(x) \in \F[x]$ be the minimal polynomial for $\alpha$ over $\F$ (every element in an algebraic extension is {\it by definition} algebraic). Let $Df(x)$ be the formal derivative of our polynomial $f(x)$ (treating the coefficients like constants). Then $f(x)$ and $Df(x)$ are relatively prime ($f(x)$ is irreducible and $Df(x)$ has lower degree and is non-zero since $\F$ has characteristic 0). This means that $Df(\alpha) \not=0$. 

Consider $f(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_0$. Then $(f(x))' = nx^{n-1}x' +(a_{n-1}'x^{n-1}+(n-1)a_{n-1}x^{n-2}x')+\cdots+(a_1'x+a_1x')+a_0'$. Therefore, 
$(f(x))' = Df(x)x'+g(x)$ where $g(x)=a_{n-1}'x^{n-1}+\cdots+a_1'x+a_0'$. Now $\alpha$ is the root of $f(x)$, so we must have $(f(\alpha))'=0$. In other words, $0=Df(\alpha)\alpha' + g(\alpha)$. So the only consistent way to define $\alpha'$ is $\dfrac{g(\alpha)}{Df(\alpha)}$. \eproof

\vspace{-0.1in}

\EX We know that the only derivation on $\mathbb{Q}$ is zero. Consider $\sqrt{5} \in \mathbb{Q}[\sqrt{5}]$. As in the above proof, we can figure out what the derivative of $\sqrt{5}$ must be from its minimal polynomial. $(\sqrt{5})^2-5=0$ so $2(\sqrt{5})(\sqrt{5})'-0=0$ and thus $(\sqrt{5})' = 0/(2\sqrt{5}) = 0$ (as we might have guessed). 

\vspace{0.05in}

Even more can be said. Notice in the above proof, if $\alpha$ is algebraic over $\F$'s constants (i.e., $0=f(\alpha)=a_n\alpha^n+\cdots+a_0$ where $a_n,\dots,a_0$ are constant), then $g(x)=a_n'x^n+\cdots+a_0'=0$. Thus we must have $\alpha'=g(\alpha)/Df(\alpha)=0$. Therefore, any element that is algebraic over the constants, is itself a constant. In particular, the only possible derivation on the field of algebraic numbers (i.e., $\overline{\mathbb{Q}}$) is $\partial=0$.

\vspace{0.05in}

\DEF Let $\F$ be a differential field. We call $\F(t)/\F$ a {\bf logarithmic} extension if there exists some $s \in \F$ such that $t' = \dfrac{s'}{s}$. Similarly we call $\F(t)/\F$ an {\bf exponential} extension if there exists some $s \in \F$ such that $t'=ts'$. 

\vspace{0.1in}

These definitions mimic the defining properties of the natural logarithm and the exponential function. Suppose that $t'=\dfrac{s'}{s}$ and we could ``integrate''. Then $t = \int t' = \int \frac{s'}{s} =  \ln(s)$. Likewise, solving the ``differential equation'' $t'=ts'$ would give us $t=e^s$. The above definitions yield formal characterizations of these functions without requiring us to work with actual fields of functions (or dealing with actual real or complex variables).

\vspace{0.05in}

\DEF Let $\F$ be a differential field with field of constants $C$ ($= \mathrm{ker}(\partial)$). A differential field extension $\E/\F$ is an {\bf elementary} extension if there exists a tower of differential fields (where all fields have the same constants):

\vspace{-0.15in}

$$\F=\F_0 \subseteq \F_1 \subseteq \F_2 \subseteq \cdots \subseteq \F_\ell=\E$$

\vspace{-0.05in}

such that each extension $\F_{j+1}/\F_j$ is either an algebraic, logarithmic, or exponential extension.

If $\F=\C(x)$ (thought of as actual rational functions in a single complex variable), then $\mathcal{E}$ is defined to be the collection of all complex functions which lie in some elementary extension of $\C(x)$ (we allow these functions to have domains which aren't all of $\C$). $\mathcal{E}$ is the field of elementary functions.

\EX Essentially elementary functions are the functions that can be built from $\C$ and $f(x)=x$ using algebra, exponentiation, and taking logarithms. For example: $f(x) = \sqrt{\dfrac{e^{\sqrt[3]{x^5+1}}-7}{\ln(x^{15}+\sqrt{x^4-1})+\ln(e^{1/x}-6)}}$ is elementary.

Keep in mind that elementary functions also include some functions which may be unfamiliar to us. For example: The Bring radical is $\mathrm{BR}(a)$ is the unique real root of the polynomial $x^5+x+a$ (then extend $\mathrm{BR}(a)$ analytically to a function of a complex variable). Because $Y=\mathrm{BR}(x)$ is a root of the equation $Y^5+Y+x=0$, we have that $\C(x)(\mathrm{BR}(x))$ is an elementary extension of $\C(x)$ and so $\mathrm{BR}(x)$ is an elementary function. By the way, for $Y=\mathrm{BR}(x)$, $0=(Y^5+Y+x)'=5Y^4Y'+Y'+1$ so $Y' = -\dfrac{1}{5Y^4+1}$. Thus $\mathrm{BR}'(x) = -\dfrac{1}{5\cdot\mathrm{BR}(x)^4+1}$. 

%\vspace{0.1in}

\EX What about trigonometric functions? The trig functions can be built from log's and exponential's. For example: $\sin(x) = \dfrac{e^{ix}-e^{-ix}}{2i}$ and $\cos(x) = \dfrac{e^{ix}+e^{-ix}}{2}$. Then $\tan(x)$, $\sec(x)$, $\csc(x)$, and $\cot(x)$ can be built from sine and cosine. Inverse trig functions? For example: $\mathrm{arcsin}(x) = -i \ln(ix+\sqrt{1-x^2})$, $\mathrm{arccos}(x) = -i\ln(x+\sqrt{x^2-1})$, and $\mathrm{arctan}(x) = \dfrac{i}{2}\left(\ln(1-ix)-\ln(1+ix)\right)$. The other inverse functions have similar looking formulas.

This means that functions like $f(x)=e^{\tan(\sqrt{x})}\ln(\sin(x)+\sec^5(e^{\sqrt{x}}))$ are elementary.

\vspace{0.1in}

We should be careful. By definition, elementary functions include all algebraic functions. They are closed under algebraic operations (for example: addition, multiplication, taking roots). They are also closed under function composition. However, they are not closed under function inverses. For example, $f(x)=x e^x$ under proper restrictions of domain can be inverted. One such branch of $f^{-1}(x)$ is called the Lambert $W$-function $W(x)$. Even though $xe^x$ is elementary, $W(x)$ is not. See \cite{elRitt} for a characterization of when an inverse of an elementary function is still elementary.

\vspace{0.05in}

Now let's prove Liouville's theorem which characterizes which functions live in an elementary extension. We will follow the proof as presented in \cite{rosen}.

\vspace{0.05in}

\newpage

\LEMMA Let $\F$ be a differential field with differential field extension $\F(t)$. Suppose that $\F(t)$ and $\F$ have the same constants (i.e. the kernels of the derivations match) and that $t$ is transcendental over $\F$. 
\begin{enumerate}
\item Let $t' \in \F$ and $f(t) \in \F[t]$ with $\deg(f(t))>0$. Then $(f(t))' \in \F[t]$ and $\deg(f(t))=\deg((f(t))')$ iff the leading coefficient of $f(t)$ is non-constant. If the leading coefficient of $f(t)$ is constant, then $\deg((f(t))')=\deg(f(t))-1$. 
\item Let $t'/t \in \F$. Then for all $a \in \F^\times$, $n \in \Z_{\not=0}$, $(at^n)' = ht^n$ for some $h \in \F^\times$. Moreover, if $f(t) \in \F[t]$ with $\deg(f(t))>0$, then $\deg(f(t))=\deg((f(t))')$. Also, $(f(t))'=cf(t)$ for some $c \in \F$ iff $f(t)$ is a monomial.
\end{enumerate}

\bproof Let $f(t) = a_nt^n+a_{n-1}t^{n-1}+\cdots+a_0 \in \F[t]$ with $a_n \not=0$ so that $\deg(f(t))=n>0$. 

For item 1, $(f(t))' = a_n't^n+(na_nt'+a_{n-1}')t^{n-1} + ((n-1)a_{n-1}t'+a_{n-2}')t^{n-2}+\cdots+(a_1t'+a_0')$. Notice that the coefficients all lie in $\F$ since we have assumed $t' \in \F$ (if $a_j \in \F$, then $a_j' \in \F$ because $\F$ is a differential field). 

Recall that $a_n$ is non-constant iff $a_n' \not=0$ (by definition). Thus the degree does not drop iff the leading coefficient is not constant. In the case that it is, we have $a_n'=0$. Suppose that $(na_nt'+a_{n-1}')=0$ (i.e. the coefficient of $t^{n-1}$ is zero). This implies that $(na_nt+a_{n-1})' = na_n't+na_nt'+a_{n-1}' = na_nt'+a_{n-1}'=0$ (we are assuming $a_n'=0$). Therefore, $na_nt+a_{n-1}=c$ for some constant $c \in \F$. Since $na_n \not=0$ and $na_n,a_{n-1}\in \F$, we have $t$ is algebraic over $\F$ (contradicting our hypothesis that $t$ is transcendental over $\F$). Thus $(na_nt'+a_{n-1}')\not=0$ and so the degree of $(f(t))'$ is $n-1$.

For item 2, let $t'/t = b \in \F$. Let $a \in \F^\times$. Then $(at^n)' = a't^n+nat^{n-1}t' = (a'+nab)t^n$ since $t'=bt$. If $a'+nab=0$, then $at^n$ would be constant. So $t$ would satisfy the polynomial $aX^n+c$ for the constant $c=-at^n$. Since $a \not=0$ and $a,c \in \F$, this would mean that $t$ is algebraic over $\F$ (again, a contradiction). Therefore, $a'+nab \not=0$ (i.e. $(at^n)'=ht^n$ where $h=a'+nab \in \F^\times$). 

Notice that our calculation above shows that each non-zero term of $f(t)$ yields a non-zero term (of the same degree) in $(f(t))'$. Thus $\deg(f(t))=\deg((f(t))')$. Also, if $f(t)$ is a non-zero monomial, say $f(t)=at^n$, the above calculation says that $(at^n)'=ht^n$ so $(f(t))'=c f(t)$ where $c=h/a \in \F^\times$. 

Conversely, suppose that $(f(t))'=cf(t)$ for some $c \in \F$. Then the above calculations say that $c=h/a_j=(a_j'+ja_jb)/a_j$ for all $a_j \not=0$. Suppose that $a_m,a_\ell \not=0$ (where $m\not=\ell$). Then $\dfrac{a_m'+ma_mb}{a_m} = \dfrac{a_\ell'+\ell a_\ell b}{a_\ell}$. That means $(a_\ell'+\ell a_\ell b)a_m = (a_m'+ma_mb)a_\ell$. Therefore, $\left(\dfrac{a_mt^m}{a_\ell t^\ell}\right)' = \dfrac{(a_m'+ma_mb)a_\ell t^{m+\ell}-(a_\ell'+\ell a_\ell b)a_m t^{m+\ell}}{a_\ell^2 t^{2\ell}}=0$.
Thus $\dfrac{a_mt^m}{a_\ell t^\ell}=z$ for some constant $z \in \F$. Therefore, $a_mt^m-za_\ell t^\ell=0$ ($t$ satisfies a non-zero polynomial over $\F$). Thus $t$ is algebraic over $\F$ (again, a contradiction). Therefore, at most one coefficient is non-zero (i.e. $f(t)$ is a monomial). 
\eproof

\vspace{0.1in}

We can now give Liouville's theorem which characterizes which ``functions'' in $\F$ can be ``integrated'' in terms of elementary functions.

\vspace{0.1in}

\THM (Liouville) Let $\F$ be a differential field of characteristic 0. Let $\alpha \in \F$. If $y'=\alpha$ has a solution $y$ in  an elementary extension of $\F$ (with the same constants), then there exists constants $c_1,\dots,c_n$ and elements $u_1,\dots,u_n,v \in \F$ such that 

\vspace{-0.25in}

$$ \alpha = v' + c_1\dfrac{u_1'}{u_1}+c_2\dfrac{u_2'}{u_2}+\cdots+c_n\dfrac{u_n'}{u_n}$$

\bproof
Suppose that $y=\int \alpha$, a solution of $y'=\alpha$ is elementary over $\F$. This implies that there exists a tower of differential fields:

\vspace{-0.25in}

$$\F \subseteq \F(t_1) \subseteq \F(t_1,t_2) \subseteq \cdots \F(t_1,t_2,\dots,t_N)$$
such that all of these fields have the same constants, $y \in \F(t_1,\dots,t_N)$, and each $t_i$ is either algebraic, logarithmic, or exponential over $\F(t_1,\dots,t_{i-1})$.

Use proof by induction. If $N=0$, $y \in \F$ and thus $\alpha=y' \in \F$ so we're done.  Suppose the result is true for any tower of height $N-1$. Notice that the tower with $\F$ deleted is $N-1$ steps tall. Therefore, by induction, there exists constants $c_1,\dots,c_n$ and elements $u_1,\dots,u_n,v \in \F(t_1)$ such that $\alpha = v'+\sum\limits_{i=1}^n c_i \dfrac{u_i'}{u_i}$. We need to get elements in $\F$ (not $\F(t_1)$). To do this we consider several cases: $t_1$ is algebraic or transcendental (then when $t_1$ is transcendental it must be either logarithmic, or exponential). 

Case 1: Suppose $t_1=t$ is algebraic over $\F$. There exists some non-zero polynomials in $\F[t]$ such that $u_i=U_i(t)$ and $v=V(t)$. Consider the distinct conjugates of $t$ in some algebraic closure of $\F(t)$ [or some splitting field over $\F(t)$] say $t=\tau_1,\tau_2,\dots,\tau_s$. Now $\E=\F(\tau_1,\dots,\tau_s)$ is an algebraic extension of $\F$ so it extends uniquely as a differential field.  There is an automorphism of $\E$ fixing $\F$ such that $t=\tau_1$ maps to $\tau_j$ (this is the very definition of a ``conjugate''), call this map $\sigma_j$, then we have $\alpha = \sigma_j(\alpha) = \sigma_j\left(V(t)'+\sum\limits_{i=1}^n c_i \dfrac{U_i'(t)}{U_i(t)}\right) = V(\tau_j)'+\sum\limits_{i=1}^n c_i \dfrac{U_i'(\tau_j)}{U_i(\tau_j)}$ (because $\alpha$ and all of the coefficients in the polynomials $U_i$ and $V$ lie in $\F$ and so are fixed by $\sigma$).

Now we ``symmetrize''! Add up all of these conjugate expressions and divide by $s$ (note that $s^{-1}$ exists because we are working in characteristic 0). First, note that by ``logarithmic differentiation'' we have:
$$\dfrac{(A_1\cdots A_s)'}{A_1\cdots A_s} = \dfrac{A_1'(A_2\cdots A_s)+ (A_1)A_2'(A_3\cdots A_s)+ \cdots + (A_1\cdots A_{s-1})A_s'}{A_1\cdots A_s} =  \dfrac{A_1'}{A_1}+\cdots+\dfrac{A_s'}{A_s}$$
Therefore, symmetrizing yields:

\vspace{-0.1in}

$$\alpha = \dfrac{1}{s} \sum_{j=1}^s \alpha = \dfrac{1}{s} \sum_{j=1}^s \sigma_j(\alpha) = \dfrac{1}{s} \sum_{j=1}^s \left(V(\tau_j)'+\sum\limits_{i=1}^n c_i \dfrac{U_i'(\tau_j)}{U_i(\tau_j)}\right) = \dfrac{1}{s}\sum_{j=1}^s V(\tau_j)' + \sum_{i=1}^n \dfrac{c_i}{s} \dfrac{(U_i(\tau_1)\cdots U_i(\tau_s))'}{U_i(\tau_1)\cdots U_i(\tau_s)}$$

\vspace{-0.05in}

Notice both $\frac{1}{s}(V(\tau_1)'+\cdots+V(\tau_s)')$ and$\dfrac{(U_i(\tau_1)\cdots U_i(\tau_s))'}{U_i(\tau_1)\cdots U_i(\tau_s)}$ are fixed by all automorphisms because they are symmetric. Hence, each of these lies in the ground field $\F$.  Letting $v=\dfrac{1}{s}(V(\tau_1)+\cdots+V(\tau_s))$ and $u_i = U_i(\tau_1)\cdots U_i(\tau_s)$ we're done.

Case 2: Suppose that $t_1=t$ is transcendental $\F$. Again we have $v,u_1,\dots,u_n \in \F(t_1)=\F(t)$. But now, since $t$ is transcendental, we need to consider rational polynomials in $t$ with coefficients in $\F$. Such polynomials can be factored and for any $w=u_i$ we can get $w = a_1(t)^{k_1}\cdots a_\ell(t)^{k_\ell} \cdot b$ where $a_i(t) \in \F[t]$ is monic and irreducible, $k_j \in \Z_{\not=0}$ (positive and negative powers), and $b \in \F^\times$. Again, using logarithmic differentiation we have 

\vspace{-0.1in}

$$\dfrac{w'}{w} = \dfrac{(ba_1(t)^{k_1}\cdots a_\ell(t)^{k_\ell})'}{ba_1(t)^{k_1}\cdots a_\ell(t)^{k_\ell}} = \dfrac{b'}{b}+k_1\dfrac{a_1'(t)}{a_1(t)}+\cdots+k_\ell\dfrac{a_\ell'(t)}{a_\ell(t)}$$

\vspace{-0.05in}

So without loss of generality we can assume each $u_i$ is either a monic irreducible $a(t) \in \F[t]$ or an element of $\F$. 
As for $v$, decompose it into its partial fraction decomposition. After possibly a polynomial term, each term in the fractional part of this decomposition is of the form $\dfrac{g(t)}{f(t)^r}$ for some monic irreducible $f(t) \in \F[t]$ and some $g(t) \in \F[t]$ where $\deg(g(t))<\deg(f(t))$.

Case 2A: Suppose that (in addition to being transcendental) $t$ is logarithmic over $\F$. This means that there exists some $a \in \F^\times$ such that $t' = \dfrac{a'}{a}$. Thus $t' = a'/a \in \F$. By our lemma, if $f(t) \in \F[t]$, then $(f(t))' \in \F[t]$ and $(f(t))'$ has degree 1 less than that of $f(t)$. Since $f(t)$ is irreducible (and the $\F$ is separable since it has characteristic 0), $(f(t))'$ and $f(t)$ must be relatively prime. Notice that 

\vspace{-0.1in}

$$\left(\dfrac{g(t)}{f(t)^r}\right)' = \dfrac{(g(t))'(f(t))^r-g(t)r(f(t))^{r-1}(f(t))'}{(f(t))^{2r}} = \dfrac{(g(t))'}{(f(t))^r} - \dfrac{rg(t)(f(t))'}{(f(t))^{r+1}}$$

\vspace{-0.05in}

Now $\deg(g(t))$ and $\deg((f(t))')$ are both less than $\deg(f(t))$. This along with the fact that $f(t)$ is irreducible implies that $f(t)$ does not divide $g(t)$ or $(f(t))'$. Therefore, the partial fraction decomposition of $v'$ involves a term with the denominator $(f(t))^{r+1}$. Thus this appears in $\alpha$'s partial fraction decomposition. But $\alpha \in \F$ so it has no fractional part to its decomposition (it's already its own decomposition). Thus such terms cannot appear in $v$'s decomposition. Likewise, they cannot appear in the $u_i$'s either. Therefore, $u_1, \dots, u_n \in \F$ and $v=V(t) \in \F[t]$ (a polynomial in $t$ -- that is -- $v$ has no fractional part).

At this point we have, $\alpha = (V(t))'-\sum\limits_{i=1}^n c_i \dfrac{u_i'}{u_i}$ where $c_i$, $u_i$, $u_i'$, and $\alpha$ all lie in $\F$. Therefore, $(V(t))' \in \F$. Hence $v=V(t)=ct+d$ for some $c,d \in \F$. But $v'$ has degree 0, so $c$ must be a constant. Therefore, $(V(t))' = c\dfrac{a'}{a}+d'$ where $a,d \in \F$ and $c$ is a constant. We then have 

\vspace{-0.1in}

$$\alpha = d' + c\dfrac{a'}{a}+ \sum\limits_{i=1}^n c_i \dfrac{u_i'}{u_i}$$ 

\vspace{-0.1in}

\noindent just as we wanted.

Case 2B: Suppose that (in addition to being transcendental) $t$ is exponential over $\F$. This means $\dfrac{t'}{t}=b'$ for some $b \in \F$. Recall that our lemma says, $\deg(f(t))=\deg((f(t))')$ and $f(t)$ divides $(f(t))'$ only if  $f(t)$ is a monomial. So if $f(t)$ is monic irreducible and $f(t)\not=t$, then $f(t)$ is not a monomial and so $f(t)$ does not divide $(f(t))'$. As before, if $f(t) \not= t$, then $\dfrac{(f(t))'}{f(t)}$ can be written as a polynomial in $t$ plus a proper fraction with denominator $f(t)$. We are led to the same contradiction since if $\dfrac{g(t)}{(f(t))^r}$ appears in the partial fraction decomposition of $v$, then $v'$ will have a term in it's decomposition with denominator $(f(t))^{r+1}$, so this appears in the decomposition of $\alpha \in \F$ (which has no fractional part). The same applies to the $u_i$'s. Therefore, the only fractional parts that can appear must have denominators $f(t)^m=t^m$. This implies that $v=V(t)=\sum a_jt^j$ for some $a_j \in \F$ where the sum ranges over a finite set of integers (some powers can be negative). Likewise, all $u_i \in \F$ with a possible exception of some $u_i=t$, without loss of generality, say $u_1=t$. Then $\sum\limits_{i=1}^n c_i \dfrac{u_i'}{u_i} = c_1\dfrac{t'}{t} + \sum\limits_{i=2}^n c_i\dfrac{u_i'}{u_i}$. But $\dfrac{t'}{t}=b' \in \F$. Therefore, again $(V(t))' = \alpha  - c_1b' - \sum\limits_{i=2}^n c_i \dfrac{u_i'}{u_i} \in \F$. Recall that $(at^n)'=ht^n$ for some $h \in \F^\times$. But $(V(t))'$ has no $t$ terms (it belongs to $\F$), thus $V(t) = at^0 = a \in \F^\times$. Therefore,
$\displaystyle \alpha = c_1\dfrac{t'}{t} + \sum\limits_{i=2}^n c_i\dfrac{u_i'}{u_i} + v' = (c_1b+v)' + \sum\limits_{i=2}^n c_i\dfrac{u_i'}{u_i}$
as desired. \eproof

\vspace{0.1in}

Now with Liouville's theorem in hand, we can show that various functions are not elementary functions. This means that we ``cannot integrate'' them (well, we can't integrate them and get a formula for the antiderivative written in terms of elementary functions).

\vspace{0.1in}

\COR Let $f(x), g(x) \in \C(x)$ be nonzero and suppose that $g(x)$ is also non-constant. Then $$\int f(x)e^{g(x)}\,dx \mbox{\ \   is elementary \quad if and only if \quad} f(x)=a'(x)+a(x)g'(x) \mbox{\ \   for some } a(x) \in \C(x).$$

\bproof
We will suppress the variable $x$ and write $f=f(x)$, $g=g(x)$, etc. Let $\F = \C(x)$ and $t=e^{g}$ so that we have $t'/t = g'$ (i.e. $\F(t)$ is an exponential extension). Also, since $g$ is non-constant, $\F(t)$ is a pure transcendental extension of $\F$. Suppose that $\int fe^{g}\,dx = \int ft\,dx$ is elementary, by Liouville's theorem we can write 

\vspace{-0.1in}

$$ft = v' + \sum\limits_{i=1}^n c_i \dfrac{u_i'}{u_i} \quad \mbox{ for some } \quad v,u_1,\dots,u_n \in \F(t) \quad \mbox{ and } \quad c_1,\dots,c_n \in \C.$$

\vspace{-0.05in}

As in the proof of Liouville's theorem, we can factor each $u_i \not\in \F$ into a power product of irreducible elements of $\F[t]$ and then using logarithmic differentiation, we can guarantee (at the expense of a possibly longer summation) that each of $u_i \not\in \F$ is distinct, monic, and irreducible. 

Now imagine that $v$ has been expanded in its partial fraction decomposition (with respect to the ring $\F[t]$). If this decomposition of $v$ had some (nonzero) term $\dfrac{a}{b^k}$ (necessarily with $\deg(a)<\deg(b)$ and $k \in \Z_{>0}$ as this is part of what it takes to be a term in a partial fraction decomposition), then when computing $v'$ we would have a term $\dfrac{a'b^k-akb^{k-1}b'}{b^{2k}} = \dfrac{a'}{b^k}-k\dfrac{ab'}{b^{k+1}}$. Notice that $v' = ft-\sum c_i u_i'/u_i$ where the $u_i$'s of degree $>0$ are distinct, monic, and irreducible and only {\it appear in denominators raised to the \underline{first} power}. This means that either $a'b^k-akb^{k-1}b'=0$ (i.e. $(a/b)'=0$) or $k=1$ (i.e. only the first power of $b$ can possibly appear in the denominator). 

Now if $k=1$, we have $\dfrac{a'}{b}-\dfrac{ab'}{b^2}$ and thus the second term cannot be unreduced (after reducing we can only have the first power of $b$ appearing) so $b$ must divide $-ab'$ (in $\F[t]$). Now $b$ is irreducible so $b$ either divides $-a$ or $b'$. But $\deg(a)<\deg(b)$, so $b$ must divide $b'$. However, by our lemma, part 2, $\deg(b)=\deg(b')$, this means that $b'=cb$ for some $c \in \F$. Again by our lemma, $b$ must be a monomial. It's irreducible, monic, and monomial so $b=t$. 

On the other hand, $a'b^k-akb^{k-1}b'=0$ and so $a'b=kab'$. Thus $b$ divides $kab'$ so as in the paragraph above we will be led to conclude that $b=t$. Therefore, the only fractional terms in the partial fraction decomposition of $v$ have denominators $b^k=t^k$. Thus $v=\sum p_jt^j$ for some $p_j \in \F$ (the sum ranging over a finite set of integers).

Now turning back to the $u_i$'s. Notice that $\sum_{i=1}^n c_i u_i'/u_i = ft-v'=ft-\sum p_jt^j$ and $u_j$ is either in $\F$ or monic irreducible. Thus the only possible $u_i \not\in \F$ would be $u_i=t$. In this case, $u_i'/u_i = t'/t \in \F$. Thus $\sum_{i=1}^n c_i u_i'/u_i \in \F$.

At this point we have $ft=v'+\sum_{i=1}^n c_i\dfrac{u_i'}{u_i} = \sum p_j't^j+\sum p_jjt^{j-1}t' + q$ where $q = \sum_i u_i'/u_i \in \F$. Recall that $t'/t=g' \in \F$ so $ft = \sum p_j't^j+ \sum jp_jg't^j+q$. Comparing coefficients of $t^1$, we get $f=p_1'+1p_1g'$. Letting $a=p_1$ ($\in \F$), we have $f=a'+ag'$ as desired.

Conversely, suppose that $f=a'+ag'$ for some $a \in \F$. Then $\int fe^g = \int (a'+ag')e^g = ae^g$.
\eproof

\EX $\displaystyle \int e^{x^2}\,dx$ is not elementary. \quad In our corollary, this goes with $f(x)=1$ and $g(x)=x^2$. So if it were elementary, we would have a solution $a(x) \in \C(x)$ for $1=a'(x)+2xa(x)$. Consider $a(x)=p(x)/q(x)$ for some $p(x),q(x) \in \C[x]$ and $q(x) \not=0$. Without loss of generality, assume $p(x)$ and $q(x)$ are relatively prime.
Then $1=\dfrac{p'(x)q(x)-p(x)q'(x)}{q(x)^2}+2x\dfrac{p(x)}{q(x)}$ . Clearing denominators we get $q(x)^2=p'(x)q(x)-p(x)q'(x)+2xp(x)q(x)$. In other words, $q(x)(q(x)-2xp(x)-p'(x))=-p(x)q'(x)$. Notice that if $q(x)$ has a root, say $r$, then $p(r) \not=0$ (since $p(x)$ and $q(x)$ are relatively prime). But $q(x)(q(x)-2xp(x)-p'(x))=-p(x)q'(x)$ so if $q(x)$ has a factor $(x-r)^k$ then $q'(x)$ must have the same factor! This is impossible ($q'(x)$ must have one less factor). Thus $q(x)$ cannot have any roots (i.e. $q(x)=c$ is constant). But then we have $a(x)=p(x)/c$ so $a(x)$ must be a polynomial. But then looking at degrees we have $0=\deg(1)=\deg(a'(x)+2xa(x))=\deg(a(x))+1$ (contradiction). Hence, no such $a(x)$ exists. Thus our integral is not elementary. This means that the error function: $\mathrm{erf}(x) = (2/\sqrt{\pi})\int_0^x e^{t^2}\,dt$ isn't elementary.

\EX $\displaystyle \int \dfrac{e^x}{x}\,dx$ is not elementary. \quad In our corollary, this goes with $f(x)=1/x$ and $g(x)=x$. So if it were elementary, we would have a solution $a(x) \in \C(x)$ for $1/x=a'(x)+1a(x)$. In other words, $1=x(a'(x)+a(x))$. If $a(x)=p(x)/q(x)$ is again written in lowest terms, $a'(x)=(p'(x)q(x)-p(x)q'(x))/q(x)^2$ and so after clearing denominators $q(x)^2=x[p'(x)q(x)-p(x)q'(x)+p(x)q(x)]$. Suppose that $q(x)$ has some root $r \not=0$ with multiplicity $k$, then $xp(x)q'(x)=q(x)[xp'(x)+p(x)-q(x)]$ and again we reach the same kind of contradiction as in the example above. Thus the only root $q(x)$ can have is $0$. Since $q(x)^2=x[p'(x)q(x)-p(x)q'(x)+p(x)q(x)]$, $0$ is a root (and must be repeated). Thus $q(x)=cx^k$ for some $k>1$. Then, $x^k(ckp(x)) = xp(x)kcx^{k-1} = xp(x)q'(x) = q(x)[xp'(x)+p(x)-q(x)] = cx^k[xp'(x)+p(x)-cx^k]$ and so $kp(x)=xp'(x)+p(x)-cx^k$ and thus $(k-1)p(x)=xp'(x)-cx^k$ so that $p(x)$ has $0$ as a root. But then $p(x)$ and $q(x)$ aren't relatively prime (contradiction). Thus our integral is not elementary.

This also implies that the real and imaginary parts of this integral are not elementary. Therefore, the sine integral function $\mathrm{Si}(x) = \int_0^t \sin(t)/t \,dt$ isn't elementary.

\EX $\displaystyle \int \dfrac{1}{\ln(x)}\,dx$ is not elementary. \quad Let $u=\ln(x)$ (so $e^u=x$) then $du=1/x\,dx$ so that $\displaystyle \int 1/\ln(x)\,dx = \int x/(x\ln(x))\,dx = \int e^u/u\,du$. Thus if the integral of $1/\ln(x)$ were elementary, then $\displaystyle \int e^x/x\,dx$ would be too -- which it isn't!

\vspace{0.05in}

The above proof of Liouville's theorem was drawn from the article \cite{rosen}. Additional examples of non-elementary integrals as well as some more details about our examples can be found in the article \cite{mead}. The books \cite{kaplan} and \cite{ritt} are classical introductions to differential algebra whereas \cite{magid} and \cite{crespo} give more modern introductions to the subject.

\vspace{0.1in}

It turns out that there is a complete algorithm for determining when a function can be integrated in terms of elementary functions. This is the so called ``Ricsh Algorithm'' developed by Robert Risch in the late 60's. For an excellent, very readable, introduction to this as well as more details about results like Liouville's theorem, I highly recommend \cite{geddes}. 


\vspace{-0.1in}

\begin{thebibliography}{diffalg-bib}
\bibitem[Crespo \& Hajto]{crespo} T. Crespo and Z. Hajto, {\it Algebraic Groups and Differential Galois Theory}, AMS Graduate Studies in Mathematics {\bf Vol. 122} (2011). % An early version of this text can be found at \href{https://www4.ncsu.edu/~singer/papers/Crespo_Hajto.pdf}{https://www4.ncsu.edu/$\sim$singer/papers/Crespo\underline{\ }Hajto.pdf}  LINK BROKEN :(
\bibitem[Geddes]{geddes} K. Geddes, S. Czapor, and G. Labahn, {\it Algorithms for Computer Algebra},  Kluwer Acad. Pub. (1992).
\bibitem[Kaplansky]{kaplan} I. Kaplansky, {\it An Introduction to Differential Algebra}, {Pub. of De L'Inst. De Math. De L'Univ. De Nancago} (1957).\quad Web: \   \href{http://mmrc.iss.ac.cn/~weili/DifferentialAlgebra/References/Kaplansky.pdf}{http://mmrc.iss.ac.cn/$\sim$weili/DifferentialAlgebra/References/Kaplansky.pdf}
\bibitem[Magid]{magid} A. Magid, {\it Lectures on Differential Galois Theory}, AMS University Lecture Series {\bf Vol. 7} (1994).
\bibitem[Mead]{mead} D. Mead, {\it Integration}, Am. Math. Monthly {\bf Vol. 68} No. 2 (Feb. 1961) 152--156.
\bibitem[Ritt1]{ritt} J. Ritt, {\it Differential Algebra}, Dover Publications (1950).%\\ Available free online at: \quad \href{https://www4.ncsu.edu/~singer/foraachen/DA.pdf}{https://www4.ncsu.edu/$\sim$singer/foraachen/DA.pdf} BROKEN LINK :(
\bibitem[Ritt2]{elRitt} J. Ritt {\it Elementary Functions and their Inverses}, Trans. Amer. Math. Soc. {\bf 27} (1925), 68--90.\\
\force\hspace*{-0.4in} Web: \  \href{https://www.ams.org/journals/tran/1925-027-01/S0002-9947-1925-1501299-9/S0002-9947-1925-1501299-9.pdf}{\small https://www.ams.org/journals/tran/1925-027-01/S0002-9947-1925-1501299-9/S0002-9947-1925-1501299-9.pdf}
\bibitem[Rosenlicht]{rosen} M. Rosenlicht, {\it Integration in Finite Terms}, Am. Math. Monthly {\bf Vol. 79} No. 9 (Nov. 1972) 963--972.
\end{thebibliography}



\end{document}