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The title is pretty self explanatory, I'd like to prove that \begin{align*} \int\frac{1}{x}\,dx \end{align*} cannot be a rational function. I have attempted a proof by contradiction, but it doesn't seem to lead anywhere. If it is assumed that $F(x)=\frac{p(x)}{q(x)}$, where $p(x)$ and $q(x)$ are polynomials, then using logarithmic differentiation, \begin{align*} \frac{1}{x}=\frac{p(x)}{q(x)}\left(\frac{p'(x)}{p(x)}-\frac{q'(x)}{q(x)}\right) \end{align*} I don't see how this leads to a contradiction. I get similar results using the quotient rule

\begin{align*} \frac{1}{x}=\frac{p'(x)q(x)-p(x)q'(x)}{[q(x)]^2} \end{align*} Writing out the terms of $p(x)$ and $q(x)$ seems too messy. Any help or suggestions would be greatly appreciated.

José Carlos Santos
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Alex S
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5 Answers5

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As you wrote in a comment, this is the same thing as proving that $\log$ is not a rational function. Suppose it was. Then we could express $\log$ as $\frac pq$, where $p$ and $q$ are polynomial functions. Furthermore, $\deg p>\deg q$, since $\lim_{x\to+\infty}\log(x)=+\infty$. So $\frac pq=P+R$, where $P$ is a non constant polynomial function and $R$ is a rational function such that $\lim_{x\to+\infty}R(x)=0$. Besides,$$1=\frac{e^{\log x}}x=\frac{e^{P(x)}}xe^{R(x)}.$$This is impossible, since$$\lim_{x\to+\infty}\frac{e^{P(x)}}xe^{R(x)}=(+\infty)\times1=+\infty$$or$$\lim_{x\to+\infty}\frac{e^{P(x)}}xe^{R(x)}=0\times1=0.$$

José Carlos Santos
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Let $p(x),q(x)$ be polynomials without common factors such that $\ln x=p(x)/q(x)$.

$\lim_{x\to 0^+}\ln x=-\infty$ means $q(x)$ has a root at $0$, so $q(x)=x q_1(x)$ for some polynomial $q_1(x)$.

$\lim_{x\to 0^+} x \ln x=0$ means $q_1(x)$ does not have a root at $x=0$ and that $p(x)$ has a root at $0$, so $x$ divides $p(x)$, contradicting the assumption of no common factors!

Kyle Miller
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Let $f(x)=p(x)/q(x)$ in lowest terms and consider $f'(x)$. If $q(x)$ does not have the factor $x$ then the denominator of $f'(x)$ doesn't have $x$ as a factor. But if $q(x)$ has a factor of $x^r$, then the denominator of $f'(x)$ has a factor of $x^{r+1}$.

Angina Seng
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  • Okay, so $q(x)$ must have $x$ as a factor? – Alex S Oct 18 '17 at 06:09
  • Could you please explain why the degree of $q(x)$ must increase (if that's what you're arguing)? – Alex S Oct 18 '17 at 06:12
  • Still not convinced, sorry – Alex S Oct 18 '17 at 06:35
  • If $f(x)=p(x)/q(x)$ is reduced, then $f'(x)=1/x$ is equivalent to $x(q(x)p'(x)-p(x)q'(x))=q(x)^2$. Suppose $q(x)$ is divisible by $x^r$ with $r$ least (and the outer $x$ on the left implies $r\geq 1$). This means $q(x)^2$ is divisible by $x^{2r}$, and less strongly, that $q(x)p'(x)-p(x)q'(x)$ is divisible by $x^{r}$ since $r+1\leq 2r$. Since $q(x)$ is divisible by $x^{r}$, it follows $p(x)q'(x)$ is divisible by $x^r$. But $p(x)$ is not divisible by $x$, so $q'(x)$ is divisible by $x^r$. This implies $q(x)$ is divisible by $x^{r+1}$, contradicting the fact $r$ was least. – Kyle Miller Oct 18 '17 at 07:20
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Yet another way to get at it: if $\int \frac{dx}{x}$ were rational, we could write

$\displaystyle \int \dfrac{dx}{x} = \dfrac{p(x)}{q(x)}, \tag 1$

with $p(x), q(x) \in \Bbb R[x]$.

First question: how do we know we can take $p(x), q(x) \in \Bbb R[x]$, and not $p(x), q(x) \in \Bbb C[x]$? Well, if $p(x), q(x) \in \Bbb C[x]$, we could write

$\dfrac{p(x)}{q(x)} = \dfrac{\bar q(x) p(x)}{\bar q(x) q(x)}, \tag 2$

with

$\bar q(x) q(x) \in \Bbb R[x]$. We then see that

$\Im(\bar q(x) p(x)) = 0, \tag 3$

since $\int \frac{dx}{x}$ is real. Thus

$\dfrac{p(x)}{q(x)} = \dfrac{\Re(\bar q(x) p(x))}{\bar q(x) q(x)}, \tag 4$

the quotient of two real polynomials; so we might as well assume that

$p(x), q(x) \in \Bbb R[x] \tag 5$

from the beginning.

We can differentiate (1) and obtain

$\dfrac{1}{x} = \dfrac{p'(x)q(x) - p(x)q'(x)}{q^2(x)}, \tag 6$

which we may re-write as

$q^2(x) = x(p'(x)q(x) - p(x)q'(x)). \tag 7$

Now it is easy to see that

$\deg(p'(x)q(x) - p(x)q'(x)) = \deg p(x) + \deg q(x) - 1, \tag 8$

and so

$\deg x(p'(x)q(x) - p(x)q'(x)) = \deg p(x) + \deg q(x); \tag 9$

also,

$\deg q^2(x) = 2 \deg 2q(x), \tag{10}$

so we find

$2\deg q(x) = \deg p(x) + \deg q(x), \tag{11}$

whence

$\deg q(x) = \deg p(x). \tag{12}$

Now say

$p(x) =\sum_0^n p_i x^i, \tag{13}$

and

$q(x) = \sum_0^n q_i x^i; \tag{14}$

then

$\dfrac{p(x)}{q(x)} = \dfrac{\sum_0^n p_i x^i}{\sum_0^n q_i x^i} = \dfrac{x^n\sum_0^n p_i x^{i - n}}{x^n\sum_0^n q_i x^{i - n}} = \dfrac{\sum_0^n p_i x^{i - n}}{\sum_0^n q_i x^{i - n}}, \tag{15}$

whence

$\lim_{x \to \infty}\dfrac{p(x)}{q(x)} = \lim_{x \to \infty} \dfrac{\sum_0^n p_i x^{i - n}}{\sum_0^n q_i x^{i - n}} = \dfrac{p_n}{q_n} < \infty; \tag{16}$

but this contradicts

$\lim_{x \to \infty}\displaystyle \int \dfrac{dx}{x} = \lim_{x \to \infty} \ln x = \infty; \tag{17}$

thus (1) is impossible.

Robert Lewis
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If ln(x) is rational, consider what happens for large x.

Depending on the degrees of the numerator and denominator, it can go tp zero, tend tp a constant, or grow like an integral power of x.

But ln does not behave like any of these.

Therefore it can't be rational.

marty cohen
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  • So using little o notation, $o(0)<o(c)<o(\ln(x))<o(x^n)$? And if two functions are equal, they must have the same behavior at infinity? – Alex S Oct 18 '17 at 08:07
  • I suppose you could prove this rigorously using limit theorems – Alex S Oct 18 '17 at 08:09
  • Yep. One of my recent answers shows that$\ln x =o(x^a)$ for any $a>0$ from the definition of log as the integral of 1/x. – marty cohen Oct 18 '17 at 15:00