# Understanding the integral of $x^a$

If we integrate $x^a$ we get $\frac{x^{a+1}}{a+1}$, except when $a=-1$, then we get $\log x$. Why is $a=-1$ special, how can I understand this intuitively? (similarly, the differential equation $x' = x^a$)

• Because when $a=-1$ you can't divide by zero. Mar 8, 2015 at 17:50
• Is there actually a way to see this in a visual way? That is, by means of a graph. Mar 8, 2015 at 17:58
• Would this be of any help ? Mar 8, 2015 at 21:33

The following is more about computation than intuition. But it may throw some light on the matter. Let $x$ be positive and $t\ne -1$. Then $$\int_1^x w^t \,dw=\frac{x^{t+1}-1}{t+1}=\frac{e^{(t+1)\ln x}-1}{t+1}.$$ We find the limit of the right-hand side as $t\to -1$. There are various ways to do this. Out of laziness we will use L'Hospital's Rule. So differentiate top and bottom with respect to $t$. We get $$\frac{(\ln x)e^{(t+1)\ln x}}{1},$$ which has limit $\ln x$ as $t\to -1$.

So in the limit sense the usual formula "works" when $a=-1$.

• Thanks, that does help, but it does not completely resolve my confusion. I feel like there is something going on with the order of the limits, because if you have the limit of $3^a/a$ as $a$ goes to 0 then it's not $\log 3$. What am I missing? edit: oh, I already see my mistake, thanks! I was confused because I only looked at the indefinite integral. Mar 8, 2015 at 18:11
• I'll ask a follow-up question if you don't mind. I hope it is not too vague. Does this mean that in some sense the logarithm ought to belong to some extension of the set of polynomials with integer exponents, where the logarithm is related to $x^0$? Mar 8, 2015 at 18:24
• The problem is to make precise in what sense. Because of my background, I tend to think in terms of non-standard models of analysis. Then for $x$ positive and real, we can think of $\log x$ as (the standard part of) $\frac{x^w-1}{w}$ where $w$ is a non-zero infinitesimal. Mar 8, 2015 at 18:44
• In Taylor series and Laurent series we think of $x^0 = 1$. I wonder if there is an alternative series expansion which uses the logarithm for the 0th power. If I remember correctly there is a problem with the logarithm in conventional complex analysis, because the singularity at 0 is a strange kind of singularity. Or maybe I'm talking nonsense... Mar 8, 2015 at 19:07
• Interesting question. I don't know. As I mentioned earlier, I went instinctually to a non-standard analysis viewpoint. But that is far from being the only reasonable viewpoint! Mar 8, 2015 at 19:14

Think of derivatives instead of integrals. The derivative of $x^b$ is $b\,x^{b-1}$. In order to get $x^{-1}$ you would need $b=0$. But $x^0\equiv1$ is constant and its derivative is $0$. The moral of the story is that the derivative of a power function is never $1/x$, and consequently the integral of $1/x$ is not a power function.

If $a=-1$, then $a+1=0$, and $\frac{x^{a+1}}{a+1}$ is undefined.