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I have recently learned that $\int\frac{1}{x}\,dx=\ln x$.

However, by the power law of integration, the integral of $1/x$ is equal to $x^0 / 0$ which is undefined. Therefore, is $x^0 / 0 = \ln x$?

Am I correct in this assumption?

Daniel W. Farlow
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Gigabit
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  • If they taught you the power law is true for $n=-1$, then you've been misled. Obviously, it is not true, since $x^0/0$ is, as you say, undefined, and it makes no sense to define it as $\ln x$. – Thomas Andrews Mar 28 '15 at 17:12
  • Actually, it's $\dfrac{x^0-1}0$ , or, better said, $\displaystyle\lim_{h\to0}\frac{x^h-1}h$ . – Lucian Mar 28 '15 at 19:03

5 Answers5

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Your observation can be properly understood using a limit. Specifically, $f(x,p)=\int_1^x y^{p} dy$ is a continuous function of $p$, even at $p=-1$. This means that

$$\lim_{p \to -1} \frac{x^{p+1}-1}{p+1} = \ln(x).$$

You could prove this directly with L'Hospital's rule, provided you already knew that $\frac{d}{dx} a^x = a^x \ln(a)$.

Ian
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  • Why the -1 in the numerator? – isarandi Mar 28 '15 at 17:20
  • @isarandi For $p \neq -1$, $\int_1^x y^p dy=\frac{x^{p+1}-1}{p+1}$. The $-1$ comes from evaluation at the lower limit. – Ian Mar 28 '15 at 17:21
  • I find it confusing. Your expression is an analogue of $(x^0 - 1)/0$, not $x^0/0$ – isarandi Mar 28 '15 at 17:24
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    @isarandi Replace "confusing" by "the correct version of the analogy". Indeed, $\log x$ is the limit of $(x^a-1)/a$ when $a\to0$, not of $x^a/a$. – Did Mar 28 '15 at 17:33
  • Okay, I see now. The "problem" is that the power rule formula gives the integral assuming that we begin at 0, while the ln formula requires starting from 1. Which of course doesn't matter when simply looking for antiderivatives (the actual purpose of these formulas). – isarandi Mar 28 '15 at 17:41
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The power rule cannot be used when integrating $\int \frac 1x\,dx = \int x^{-1}\,dx$.

I assume you are referring to the fact that for $n\neq -1$, $$\int x^{n} \,dx = \dfrac{x^{n+1}}{n+1}+ C$$ Note that this holds for all integers $n \neq -1$.

So $\dfrac {x^0}0 \neq \ln |x|$, and as you know, $\dfrac {x^0}0$ is not defined.

Jordan Glen
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  • Thank you for responding.

    Would you therefore explain how the integral of $1/x$ = $lnx$, then?

     – Gigabit Mar 28 '15 at 17:07
  • Because we know that $\dfrac{d}{dx}(\ln x) = \dfrac 1x$, and it is sometimes used as the definition of $\int \frac 1x,dx$. – Jordan Glen Mar 28 '15 at 17:10
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    @AlecR The answer varies depending on how you set up the definitions. Often $\int_1^x \frac{1}{y} dy$ is the definition of $\ln(x)$. Other times we proceed through other definitions, such as defining $\exp$ and then defining $\ln$ to be its inverse. – Ian Mar 28 '15 at 17:10
  • You should add a $C$ to your indefinite integral... – Thomas Andrews Mar 28 '15 at 17:31
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We can define $\log{x}$ explicitly as $$ \int_1^x \frac{dt}{t}; $$ but why this definition?

Note that, for general powers $x^{n-1}$, we have $$ \int_1^x t^{n-1} \, dt = \frac{x^n-1}{n}, $$ so we can't just set $n=0$ here to have a sensible answer without getting $0/0$. However, this suggests that $$ \lim_{n \to 0} \frac{x^n-1}{n} = \log{x}. \tag{1} $$ This is in fact true, even if we define the logarithm as the inverse to the exponential: recall the definition of $e^x$ as a limit, $$ y = e^x = \lim_{m \to \infty} \left( 1+\frac{x}{m}\right)^m $$ say, then we can do some algebra to find $$ x= \lim_{m \to \infty} m(y^{1/m}-1), $$ (or just substitute this in and see what happens) and if you set $m=1/n$ you recover the limit (1).

Chappers
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A slight, but important correction to make: $\int\frac{1}{x}\,dx=\ln |x|$ + c

We use the absolute value of x in order to have the same domain for both the right and left hand side.

If we only take the natural log of x, our x values are restricted to being > 0.

Whereas in $x^-1$, Our x values are only restricted to x !=0

Jordan
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all the answers so far use integral representation of $\ln x.$ i will use the fact that $$ \text{ the derivative of } a^x \text{ at } x = 0 \text{ is }\ln a.$$ written in the form $$\lim_{x \to 0} \frac{a^x - 1}{x} = \ln a. $$ switching the roles of $x$ and $a$ we have $$\lim_{a \to 0} \frac{x^a - 1} a = \ln x. $$ in a less than rigorous way, we have $$\frac{x^0 - 1} 0 = \ln x $$

abel
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