While I was studying integrals by my own, I learnt these two rules for integrating $f(x) = x^k$:
- if $k \neq -1$, then $\int{x^k dx}=\frac {x^{k+1}}{k+1}+c$;
- if $k =- 1$, then $\int{x^{-1} dx} = \ln {|x|} + c$.
What I find interesting is that, for a fixed $x_0$, the function $g(x_0, k)$ (defined below) has a discontinuity at $-1$, but it is still defined.
Let $g(x_0,k)=\int_1^{x_0} {x^k dx}$ and $x_0 \in (0, +\infty)$. Notice that $\lim_{k\rightarrow -1} g(x_0, k) = \pm\infty$, but $g(x_0, -1) = \ln x_0 +c$.
If you graph$^1$ $g(x_0, k)$ (with $x_0 = e$ and $k$ represented by the $x$-axis), you get this:
My question is: why? Why is $g(x_0, -1)$ well defined?
I mean:
- it makes sense that $1/x$ should have a primitive; also I can graphically calculate the area underneath it
- I understand the proofs for $\int{x^{-1} dx} = \ln {|x|} + c$
- $\int{x^{-1} dx} = \ln {|x|} + c$ just works, so it must be correct
But it seems like this result is completely out of context when you study $x^k$.
What am I missing out? And, is there any relationship between $\int{x^k dx}$ (with $x\ne -1$) and $\int{x^{-1} dx}$ at all? If there are none, what's special about $x^{-1}$?
NOTES:
- graph$^1$: done with GeoGebra. I added the point manually, as GeoGebra was graphing $h(x) = \frac {e^{x+1}} {x + 1}$ for every $x$, instead of $h(x) = \ln e$ when $x = -1$.