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Of course, it is easy to see, that the integral (or the antiderivative) of $f(x) = 1/x$ is $\log(|x|)$ and of course for $\alpha\neq - 1$ the antiderivative of $f(x) = x^\alpha$ is $x^{\alpha+1}/(\alpha+1)$.

I was wondering if there is an intuitive (probably geometric) explanation why the case $\alpha=-1$ is so different and why the logarithm appears?

Some answers which I thought of but which are not convincing:

  1. Taking the limit $\alpha=-1$ either from above or below lead to diverging functions.

  2. Some speciality of the case $\alpha=-1$ are that both asymptotes are non-integrable. However, the antidrivative is a local thing, and hence, shouldn't care about the behavior at infinity.

Alex M.
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Dirk
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    What is so special about $\epsilon = \frac{1}{\infty}$ in the derivative of $x^\epsilon$? – Dan Brumleve Sep 14 '11 at 09:55
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    The limit works, only the implicit $+C$ diverges as $\alpha\to-1$ and needs accounting; ameliorate with definite integration: take $$\lim_{\alpha\to-1}\int_1^x u^\alpha du=\lim_{\beta\to0}\frac{x^{\beta}-1}{\beta},$$ then write $x^\beta$ as $\exp(\beta\log x)$ and use the Taylor series expansion of the exponential function. / Honestly though I don't see any geometric intuition behind why $\alpha=-1$ is so special of a situation, and I wish I did, but it reminds me of the fact that $\zeta(s)$ can be analytically continued to all of $\mathbb{C}$ except for $s=1$ ... maybe there's a connection? – anon Sep 14 '11 at 11:01
  • 168335, I do not get the meaning of the animation. Is there an explanation somewhere? – Dirk Sep 14 '11 at 11:16
  • Related: http://math.stackexchange.com/questions/6595/how-do-i-evaluate-this-integral/6604#6604 – Hans Lundmark Sep 14 '11 at 12:39
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    It may be also worth to ask why the anti-derivative function defined by @Didier in his answer is inverse for the exponent. It's strange since for other power functions the inverses of anti-derivatives are also power functions (up to the scaling parameter). – SBF Sep 14 '11 at 19:57
  • In what sense is the antiderivative a local thing? And why aren't you convinced by your first answer? – Qiaochu Yuan Sep 14 '11 at 21:26
  • Qiaochu: If I modify a function outside of a neighborhood of a point, the antiderivative at that point will not change. However, as anon's comment showed, my first answer can be turned in something more convincing. – Dirk Sep 15 '11 at 06:01
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    I tend to accept anon's comment as an answer if it was given as an answer... – Dirk Sep 20 '11 at 07:01
  • There's not much special if one does it correctly, as in@anon's comment. Compare the plots in https://math.stackexchange.com/a/2158190/96384. – Torsten Schoeneberg Jul 23 '19 at 20:35

7 Answers7

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Assume you know that for every $\beta$ the derivative of the function $x\mapsto x^\beta$ is the function $x\mapsto\beta x^{\beta-1}$ and that you want to choose $\beta$ such that the derivative is a multiple of the function $x\mapsto x^{\alpha}$. You are led to solve the equation $\beta-1=\alpha$, which yields $\beta=\alpha+1$. If $\alpha=-1$, this gets you $\beta=0$, but then the derivative you obtain is the function $x\mapsto 0x^{-1}=0$, which is not a nonzero multiple of $x\mapsto x^{-1}$. For every other $\alpha$, this procedure gets you an antiderivative but for $\alpha=-1$, you failed. Or rather, you proved that no power function is an antiderivative of $x\mapsto x^{-1}$. Your next step might be (as mathematicians often do when they want to transform one of their failures into a success) to introduce a new function defined as the antiderivative of $x\mapsto x^{-1}$ which is zero at $x=1$, and maybe to give it a cute name like logarithm, and then, who knows, to start studying its properties...

Edit (Second version, maybe more geometric.)

Fix $s>t>0$ and $c>1$ and consider the area under the curve $x\mapsto x^\alpha$ between the abscissæ $x=t$ and $x=s$ on the one hand and between the abscissæ $x=ct$ and $x=cs$ on the other hand. Replacing $x$ by $cx$ multiplies the function by a factor $c^\alpha$. The length of the interval of integration is multiplied by $c$ hence the integral itself is multiplied by $c^{\alpha+1}$.

On the other hand, if an antiderivative $F$ of $x\mapsto x^\alpha$ is a multiple of $x\mapsto x^\beta$ for a given $\beta$, then $F(ct)=c^\beta F(t)$ and $F(cs)=c^\beta F(s)$ hence $F(ct)-F(cs)=c^\beta (F(t)-F(s))$. Note that this last relation holds even if one assumes only that $F$ is the sum of a constant and a multiple of $x\mapsto x^\beta$.

Putting the two parts together yields $c^{\alpha+1}=c^\beta$. Once again, if $\alpha=-1$, this would yield $\beta=0$, hence $F$ would be constant and the area $F(t)-F(s)$ under the curve $x\mapsto x^\alpha$ from $s$ to $t\ge s$ would be zero for every such $s$ and $t$, which is impossible since the function $x\mapsto x^\alpha$ is not zero. (And for every $\alpha\ne1$, this scaling argument yields the correct exponent $\beta$.)

Community
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Did
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    The problem is to explain the choice of $x^{a+1}/(a+1)$ as "the" antiderivatives (out of the whole universe of solutions to $f'(x) = x^a$, with its freedom to choose a different constant of integration for each value of $a$). If the choice is made by requiring the limit of $f$ to be zero at $x=0$, then it fails for all $a \leq -1$ and $a = -1$ is not a unique exception. If the choice is via $f(1) = 1/(a + 1)$ this is no better justified than, say, $f(7) = \cot(a+1)$ which breaks down at many other values, or $f(1)=0$ that works for all $a$ (i.e. uses $\int_1^{x}$ to select antiderivative). – zyx Sep 14 '11 at 21:16
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    Also, the scaling between integrals on $(s,t)$ and $(cs,ct)$, described in the second argument, works perfectly (and most simply of all) in the logarithmic case. It is only the requirement for a notationally convenient, but otherwise arbitrary, form of the anti-derivative that is in conflict with this scaling. – zyx Sep 14 '11 at 21:32
  • @zyx, sorry but I fail to see your point, nor how what you write would apply to what I wrote (where you will not find the expression "the antiderivative".) – Did Sep 14 '11 at 21:32
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    My point was that your answer, like every answer that boils down to "$1/(\alpha + 1)$ being undefined at $\alpha = -1$", is a non-answer. It takes as a premise the entire source of the difficulty, which is the assumption that an answer of type $Ax^B$ is required. It is very easy to show that the logarithm is a special case under such an assumption, but justifying the assumption is another matter entirely (if it is in fact possible to justify it at all). – zyx Sep 14 '11 at 21:37
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    By the way, in English at least, it is common to use quotation marks in phrases like "the" X to gently mock the idea that X is unique (and also to suggest that only "an X" or "a X" is the correct description). It does not necessarily imply or suppose that the expression "the X" was actually utilized, and when discussing a written text it does not (necessarily) imply that such an expression is a direct quotation from the text. – zyx Sep 14 '11 at 22:34
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    Didier: Although your answer is well formulated and gives several good explanations that $\alpha=-1$ is special, it somehow does not answer why it is special. E.g.: Why is your geometric argument failing (despite the fact that it is simply not working)? Is there a geometric way to see that for $\alpha=-1$ the antiderivative should scale like $\log$? – Dirk Sep 20 '11 at 06:58
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    Dirk: The geometric argument does not fail, rather it shows rigorously that no constant-plus-a-power-of-$x$ can be an antiderivative of $x\mapsto x^{-1}$. // Re a geometric way to see that log solves the case $\alpha=-1$, here is a partly geometric one. By the geometric proof in my post, by scaling, $F(ct)-F(cs)=F(t)-F(s)$ for every positive $c$, $t$ and $s$. The rest is analysis: $F(xy)=F(x)+F(y)-F(1)$, hence $G=F(\exp(\ ))-F(1)$ is such that $G(t+s)=G(t)+G(s)$. Adding the continuity of $G$, $G(t)=tG(1)$ hence $F$ is an affine function of the inverse of exp. – Did Sep 20 '11 at 10:47
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The algebra of all polynomials is closed under differentiation and integration, however as soon as one wants to include negative powers of $x$, integration is no longer closed. As this paper discusses,

Roman, Steven. The Logarithmic Binomial Formula. Amer. Math. Monthly. Vol. 99, No. 7, Aug.-Sept. 1992.

the smallest algebra of functions including both $x$ and $x^{-1}$ that is closed under both diff. and anti-diff. is generated by functions of the form $x^i (\log x)^j$, for $i, j \in \mathbb{Z}$.

As a (very loose) analogy, $\mathbb{R}$ is not closed under taking square roots, but by adjoining $i$, we get closure under arbitrary roots.

Hope this helps!

Shaun Ault
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    This seems to be more or less just a broader restatement of the fact that $x^{-1}$ is a special case. – anon Sep 14 '11 at 20:17
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    +1 for a nice reference, even if it does not answer the original question... – lhf Sep 20 '11 at 10:43
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For $\alpha \neq -1$, we have $$\int x^{\alpha} dx = \dfrac{x^{\alpha+1}}{\alpha+1} + \text{ constant} = \dfrac{x^{\alpha+1}-1}{\alpha+1} + \text{ another constant}$$ Now letting $\alpha \to -1$, and assuming some nice swapping of limit and integral, we get that $$\lim_{\alpha \to -1}\int x^{\alpha} dx = \int \lim_{\alpha \to -1}x^{\alpha} dx = \int \dfrac{dx}x= \lim_{\alpha \to -1}\dfrac{x^{\alpha+1}-1}{\alpha+1} + \text{ constant} = \log(x) + \text{constant}$$

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OK, suppose you know some calculus but you fall asleep as your professor talks about the exponential function. Your true love is algebra and you've noticed the following:

$x \mapsto x^{-1}$ is a homomorphism of multiplication. You look at different forms of the function on algebraic objects,
$\Bbb R^* \to \Bbb R^*$
$(0, +\infty) \to (0, +\infty)$
$(0, 1] \to [1, +\infty)$
taking the inverse of a product to the product of the inverses.

As you study the mapping $(0, +\infty) \to (0, +\infty)$ you find it interesting to look at how the set $\{\dots,6^{-1},5^{-1},4^{-1},3^{-1},2^{-1},1^0,2^1,3^1,4^1,5^1,6^1,\dots\}$
is mapped under inversion. It looks like multiplying by $-1$ on $\Bbb Z$, and as you explore it further, you wonder if you can create a homomorphism that takes multiplication to addition,
$(0, +\infty) \to \Bbb R$
that somehow uses the inversion function $1/x$. You are looking for a function that dilates and is negative when operating on arugments less than $1$, is equal to $0$ at $1$, and is a contraction and positive when operating on arguments greater than $1$.

But $1/x$ is greater than $1$ and explodes near 0 when $x \lt 1$, and is less than $1$ and shrinks to zero near $+\infty$ when $x \gt 1$. With a flash of insight you think, just maybe, the function

$F(x) =\int_1^x u^{-1}du$

will do the trick. It feels like it will stretch $(0, +\infty) \to \Bbb R$ in the right way and it takes $1$ to $0$ with $F^{'}(1) = 1$.

OK, time to see if integrating a multiplicative homomorphism with the lower limit the identity $1$ will give you a new homomorphism that takes multiplication to addition.

CopyPasteIt
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Let $b>a>0$. Neither $$lim_{b\to\infty}\int_a^b x^{\alpha}dx$$ nor $$lim_{a\to 0}\int_a^b x^{\alpha}dx$$ converge for $\alpha=-1$. For any other $\alpha$, at least one of the limits exists.

$\log(x)$ also goes to infinity for $x$ going to infinity and for $x$ going to zero. So it's a candidate to consider for the antiderivative. For a power function of the form $cx^\beta$, however, at least one way goes to zero. So a $cx^\beta$ cannot be "the" antiderivative on the entire real line.

Elsa
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imagine the following

$$ \int \frac{dx}{x} = \frac{x^{0}}{0}$$

substract the quantity $ \frac{1}{0} $ so we get $ \frac{x^{0}-1}{0} =log(x)$ by using the taylor expansion of x^{s} on 's' near $ s=0 $

Jose Garcia
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In the expression $\dfrac{x^{\alpha+1}}{\alpha+1}$, when $\alpha=-1$, then you're dividing by $0$. If you understand why you can divide by any other number but not by $0$, then that immediately gives you a reason to expect the answer to be quite different in that case.

(I'm surprised to see this comment absent from the other answers.)

Michael Hardy
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