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First of all, I'm not entirely sure what to call polynomial functions that can have x raised to negative integers powers, so I will just call them polynomial type functions.

The integral of every polynomial type function is another polynomial type function, unless, of course, our polynomial type function has $\frac{1}{x}$ in it. In that case, our integral is obviously $\ln{x}$.

In the case of every other function in the form $x^{k}$ for some integer $k$, we can use power rule to find the integral. With $\frac{1}{x}$, there is a problem, since our integral will have had a constant of $0$ multiplying the term, nullifying it.

So my question is, why does it happen to be true that every single polynomial type function has a polynomial type integral, except $\frac{1}{x}$? $\frac{1}{x}$ happens to be the only function where using power rule to evaluate the integral doesn't work, and it also happens to the be the only one without a polynomial type integral.

This can't be a coincidence: the two special properties of $\frac{1}{x}$ must be related. What about $\frac{1}{x}$ makes it so special?

Maxim Enis
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    Functions involving negative (whole) powers of $x$ are called rational functions. – Alex S Oct 17 '17 at 01:50
  • This answer demonstrates the connection quite brilliantly in my opinion. – Kenny Lau Oct 17 '17 at 01:51
  • While the derivative of a rational function is always a rational function, anti-derivative of a rational function is not necessarily guaranteed to be a rational function. It is easy to prove using rules of derivatives that there is no rational function whose derivative is $1/x$ and with some more effort one can show that no algebraic function can have derivative $1/x$. So it's more a matter of the way derivatives and their rules work. – Paramanand Singh Oct 17 '17 at 09:49
  • Good question, which fortunately already has a lot of good answers elsewhere. In addition to the one listed above, see also here and here –  Oct 18 '17 at 02:38

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One way to see it is that the derivative of $\ln(x)$ is equal to $\frac{1}{x}$: $$y=\ln(x)$$ $$e^y=x$$ Differentiating both sides with respect to x, $$\frac{dy}{dx}e^y=1$$ $$\frac{dy}{dx}=\frac{1}{e^y}$$ But remember that $e^y=x$. This means that $$\frac{dy}{dx}=\frac{1}{x}$$ This means that the integral of $\frac{1}{x}$ has to be $\ln(x)$.

I'm not sure what you mean when you say that $\frac{1}{x}$ doesn't have a "polynomial type" derivative. From your definition $\frac{d}{dx}\frac{1}{x}=-\frac{1}{x^2}$ is a polynomial type function.

Dispersion
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$\frac 1x$ does have a polynomial type derivative. The derivative is $-\frac 1{x^2}$ which is polynomial type. This is a central assumption of the question which is wrong.

Ross Millikan
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  • There's a Latin phrase that summaries this answer nicely "a falsis principiis proficisci", meaning "you proceed from false principles" – Alex S Oct 17 '17 at 01:59
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The integral of $\frac{1}{x}$ is not a rational function. Several proofs of this can be found on this website. There are however power series expansions of $\ln(x)$ for some but not all of the real line. Recall the geometric series: \begin{align*} 1+x+x^2+x^3+\cdots=\frac{1}{1-x} \end{align*} This series converges for $|x|<1$. Integrating both sides yields \begin{align*} \int(1+x+x^2+x^3+\cdots)\,dx&=x+\frac{x^2}{2}+\frac{x^3}{3}+\cdots\\ &=\sum_{k=1}^\infty \frac{x^k}{k}. \end{align*} Therefore \begin{align*} \int\frac{1}{1-x}\,dx=\sum_{k=1}^\infty \frac{x^k}{k} \end{align*} This is a but unwieldy though because of the negative sign, so flipping this around gives \begin{align*} \int\frac{1}{1+x}\,dx=\sum_{k=1}^\infty (-1)^{k+1}\frac{x^k}{k} \end{align*} Using a change of variables and the definition of the natural logarithm, the power series on the RHS is equal to $\ln(1+x)$. So to summarize, the integral of $\frac{1}{x}$ can represented as a power series, but not a rational function. If you want to see some more cool stuff like this, you can follow this link.

Alex S
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