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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: g(e, k): there is a discontinuity at x = -1

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 $x = -1$.
Mike
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lax48
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  • Essentially your choice of $+C=0$ was not the most natural, since both solutions have zeros at different points. Aligning the $k\neq -1$ case to have a zero exactly where $\ln x$ has a zero at $x=1$ obtains a continuous function. – Ninad Munshi Jan 19 '22 at 23:23
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    @ThomasAndrews shouldn't it be that $\frac{x^{k+1}-1}{k+1}\rightarrow \ln x,x>0$ – Golden_Ratio Jan 19 '22 at 23:23
  • Note, your definition of $g$ is not a function, since your indefinite integrals are not defined completely. You need something like $$g(x,k)=\int_1^{x} y^k,dy$$ or some other starting bound, but, when $k<-1,$ this only appears to be defined for $x>0.$ – Thomas Andrews Jan 19 '22 at 23:23
  • Deleting and reposting comment due to error: If you look at $f(y)=\frac{a^y-1}{y},$ you see that as $y\to 0,$ this converges to $\ln a.$ Letting $a=x,y=k+1,$ then as $k\to -1,$ you have $$\frac{x^{k+1}-1}{k+1}\to \ln x.$$ – Thomas Andrews Jan 19 '22 at 23:25
  • I'm sure this is asked here a zillion times already. Someone look for a dupe please... – metamorphy Jan 19 '22 at 23:59
  • Overall I do think this question is confused. A part of that is that the OP was not precise on what the limits of integration are. Certainly the graph in the OP is incorrect; for $x_0=e$ and in fact for any $x_0\in (0,\infty)$ the value of $g(x_0,k)$ is well-defined and finite for any $k$. For $k <-1$ this is bounded by some $C_k$ for all $x_0\in[1,\infty)$, and for $k>-1$ this is bounded by $D_k$ for all $x_0\in [0,1]$. As pointed out precisely in the answer below things converge as $k \rightarrow -1$ as one might expect. – Mike Jan 20 '22 at 00:48
  • But to sum up, for any fixed $x_0 \in (0,\infty)$ the function $g(x_0,k)$ is continuous in $k$ over the entire real line, even at $k=-1$. – Mike Jan 20 '22 at 00:58
  • or https://math.stackexchange.com/questions/64432/what-is-so-special-about-alpha-1-in-the-integral-of-x-alpha – Henry Jan 20 '22 at 10:04

2 Answers2

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You cannot say much about $\lim g$ when $g$ is an indefinite integral.

Make it definite over a particular interval such as $[1,x_0]$ with $x_0>1$ so $$g(x_0,k)= \int_1^{x_0} x^k\,dx = \frac{x_0^{k+1}-1}{k+1}$$

Now consider the limit, and you will find $$\lim\limits_{k \to -1} g(x_0,k) = \lim\limits_{k \to -1} \frac{x_0^{k+1}-1}{k+1} = \log_e(x_0) = \int_1^{x_0} x^{-1}\,dx$$ and all is right with the world

Henry
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  • Very nice. I do think that adding another line showing how you derived in particular the second equation would be helpful to the reader though.... – Mike Jan 19 '22 at 23:29
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    e.g., For small $|r|$ and $x$ fixed the following holds: $(x^r-1)/r =(e^{r\ln x}-1)/r \approx ((r \ln x+1)-1)/r =\ln x$ – Mike Jan 19 '22 at 23:33
  • @Mike - yes, that works (here $r=k+1$). I really want to say do not take "limits of indefinite integrals" and then added a little – Henry Jan 19 '22 at 23:37
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    Thank you! Could you point out why we cannot say much about $\lim g$ when $g$ is an indefinite integral? – lax48 Jan 19 '22 at 23:38
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    @lax48 You have that $+C$, which could be any number so not suitable for taking a limit. If you ignore it, you are in effect considering $\int_0^{x_0} x^k , dx$ which diverges for some negative $k$ including $k=-1$ – Henry Jan 19 '22 at 23:42
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    Thank you for the explanation. Your answer and @David K’s one, plus @Mike’s comments, helped me understand my mistakes. I’m accepting your answer, but could you also credit David K and Mike? – lax48 Jan 20 '22 at 12:30
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The graph of the alleged function $g(x_0,k)$ for $x_0=e$ is a composite of the graphs of three different definite integrals.

For $k > -1$ you have graphed $$ y = \int_0^e x^k \;\mathrm dx. $$

For $k = -1$ you have graphed $$ y = \int_1^e x^k \;\mathrm dx. $$

For $k < -1$ you have graphed $$ y = \int_\infty^e x^k \;\mathrm dx = -\int_e^\infty x^k \;\mathrm dx. $$

For a fixed $k,$ the indefinite integral really gives you a family of functions to evaluate at $x=x_0$ (a different function for each value of $C$), not a single function. You don't get a function $g(x_0,k)$ until you have chosen a value of $C$ to use in each integral's solution. If you insist on using indefinite integrals, to make your graph continuous simply select $C=-1/(k+1)$ for each $k\neq -1,$ but $C=0$ when $k=-1.$

David K
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