# Why can't you integrate all power functions without a log function?

You need a logarithm function to solve all power functions. That's a fact.

Power functions look like this: $f\colon x \mapsto a x^r \qquad a,r \in \mathbb{R}$

But why would you need a logarithm function to be able to solve such a function? Exponential functions do have an unknown exponent but they look different: $x \mapsto a^x$

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I don't understand your question. What exactly do you mean by "solving all power functions", and in what sense do you "need" a logarithm function? – Nate Eldredge Jun 14 '11 at 20:39
By "solve all power functions", do you mean "to define all power functions"? It is true that in order to be able to give a general definition that encompasses all possible real exponents, one usually uses logarithms. But it is possible to define rational exponents without relying on logarithms, and one can define arbitrary exponents as limits of rational exponents, so it is not quite "a fact" that you "need logarithms" to define these functions. If not what you meant, then what does it mean to "solve a function"? – Arturo Magidin Jun 14 '11 at 20:40
And how is your question related to integration, which is what your title mentions? I confess to being confused. – Arturo Magidin Jun 14 '11 at 20:41
I think the OP is asking why $\int x^{-1}\,dx$ involves a logarithm. – Myself Jun 14 '11 at 20:48
I have to admit I also don't really understand the question. Here is what I can say: an antiderivative of $x^\alpha$ is $\frac{x^{\alpha+1}-1}{\alpha+1}$. Now let $\alpha \to -1$ and watch how the miracle appears... – Fabian Jun 14 '11 at 20:50

The problem is this: you know that for $n \neq -1$, the indefinite integral of $x^n$ is equal to $\frac{x^{n+1}}{n+1}$. But this formula can't possibly be valid for $n = -1$ because the denominator vanishes. So instead you have to take the limit. It's easiest to see how this works with the definite integral

$$\int_a^b x^n \, dx = \frac{a^{n+1} - b^{n+1}}{n+1}.$$

If you want to see what happens at $n = -1$, what you do is to take the limit as $n \to -1$. By l'Hopital's rule, remembering that $x^k = e^{k \ln x}$, we find that

$$\lim_{n \to -1} \frac{a^{n+1} - b^{n+1}}{n+1} = \lim_{n \to -1} \frac{a^{n+1} \ln a - b^{n+1} \ln b}{1} = \ln a - \ln b.$$

So that's where the logarithm appears: it naturally comes out of the value of this limit, and in fact this limit can be used to define the logarithm.

More generally speaking, if you have a collection of functions closed under differentiation, you are in no way guaranteed that that collection of functions is also closed under integration. In fact given a class of functions, integration generally gives you new functions not in that class.

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There is some circularity in using $\lim_{n\to-1}(a^{n+1}-1)/(n+1)=\log a$ to define the logarithm, when you're defining $a^{n+1}$ to be $e^{(n+1)\log a}$. – Gerry Myerson Jun 14 '11 at 23:53
@Qiaochu , @Gerry, to simplify one can take a=1, then the limit will define the $\log (b)$ – Arjang Jun 15 '11 at 0:08
@Arjang, I did that (well, I took $b=1$), but my question remains: how can you use the limit to define the logarithm, when you've already used logarithms to define $a^{n+1}$? – Gerry Myerson Jun 15 '11 at 1:34
@Gerry: if you use the above limit to define the logarithm, then naturally you aren't going to apply l'Hopital's rule. I just wanted to give a reason why the limit turns out the way it does that would make sense to someone with a standard calculus background. – Qiaochu Yuan Jun 15 '11 at 8:21
@Gerry: I hope we can agree that there is more than one way to define exponentials (for example defining them for integer exponents, then rational exponents, then extending by continuity). The remark about logarithms was, as I have been saying, a way to make the limit maximally sensible to evaluate for someone with a standard calculus background. – Qiaochu Yuan Jun 16 '11 at 9:00