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Normally when handling limits we consider a property of an expression to be valid at a limit of a parameter if that property is continuously true as we approach a limit of that parameter. For example: we can't divide by zero but we can divide by $a$ for arbitrarily small values of $a$ and show that this is valid $\lim_\limits{a\rightarrow 0}$.

Consider this:

We know that $\int x^{a}\mathrm{dx}$ is polynomial for $a\neq -1$

So surely $\lim_\limits{a\rightarrow -1} \int x^{a}$ is polynomial?

But we know this limit is actually $\ln(x)+C$ which is not a polynomial.

So the polynomial-ness of the result of integrating is violated $at$ the limit, even though it holds as we approach the limit.

So in what sense is this limit well-behaved?

spraff
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  • Things staying true in the limit is practically the definition of "being closed". Apparently, the set of polynomials isn't closed. – JonathanZ Apr 18 '23 at 16:14
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    $x^a$ and its integral are only polynomials when $a$ is a non-negative integer. If $a$ is restricted to integers, then you cannot take $\lim_{a\to -1}$, as the integers are discrete. – Paul Sinclair Apr 18 '23 at 16:15
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    Also bear in mind that $\lim_{a\to-1}\int x^a$ technically doesn't make any sense as the integral is indefinite. If you wished to define a function, say: $$f(x)=\lim_{a\to-1}\int_1^x t^a,\mathrm{d}t$$Then that makes sense – FShrike Apr 18 '23 at 16:18
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    You might want to check out the Weierstrass theorem: that essentially says on any finite closed interval, you can make any continuous function be a limit of polynomial functions. – Daniel Schepler Apr 18 '23 at 16:23
  • Related: https://math.stackexchange.com/questions/498339/demystify-integration-of-int-frac1x-mathrm-dx – Hans Lundmark Apr 18 '23 at 18:32

1 Answers1

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One approach: Because of the $+C$ in an indefinite integral, I may say if we want to that $$ \int x^a\;dx = \frac{x^{a+1}-1}{a+1}+C, \quad\text{for all } a \ne -1 , $$ and then we have the correct limit $$ \lim_{a\to -1}\frac{x^{a+1}-1}{a+1} = \ln x . $$

GEdgar
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