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If we take the right limit

$$\lim_{a\to-1}\int x^a dx=\lim_{a\to-1}\frac{x^{a+1}}{a+1}=+\infty$$

but on the other hand

$$\int\lim_{a\to-1} x^a dx=\ln x$$

I'm aware you can't just commute the limit and the integral, but I'd still like an explanation here. To me this is analogous to someone saying "The right limit of $1/x$ is $+\infty$ and the left one is $-\infty$ but $1/0$ is $7$ (or something)"

Is there an intuitive explanation to this break in continuity?

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  • $\begingroup$ $x^a$ has complex issues when $x \lt 0$ and $a$ is not an integer $\endgroup$
    – Henry
    Jun 29, 2015 at 11:56
  • $\begingroup$ $$\lim_{a\to-1}\int x^a dx=\lim_{a\to-1}\frac{x^{a+1}-1}{a+1}+C=\ln x+C$$ $\endgroup$
    – Did
    Jun 29, 2015 at 12:03
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    $\begingroup$ It's an indefinite integral, you forgot the constant. It means a lot here $\endgroup$
    – Venus
    Jun 29, 2015 at 12:04

3 Answers 3

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If $z>y>0$, then $$\lim \int_y^z x^a dx = \lim \frac {z^{a+1} - y^{a+1}}{a+1} \\ = \lim \left(\frac {\exp ((a+1)\log z) - \exp 0}{a+1} - \frac{\exp((a+1)\log y) - \exp 0}{a+1} \right) = \log z - \log y$$ by definition of what $\exp'(0)=1$ means.

So $\lim \int_y^z x^a dx = \int_y^z x^{-1} dx$ and so the function $a \mapsto \int_y^z x^a dx$ is continuous at $a= -1$ .

In your "choices" of indefinite integrals, you tried to choose alternatively $y=0$ and have $z$ vary (for $a > -1$) ; and then choose $z= \infty$ and have $y$ vary (for $a < -1$). So this is why you got nonsense when you looked at $a=-1$

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BTW, and assuming $\;x>0\;$:

$$\lim_{a\to-1}\frac{x^{a+1}-1}{a+1}\stackrel{l'H}=\lim_{a\to-1}x^{a+1}\log x=\log x\neq 0\;\;,\;\;\text{provided}\;\;x\neq 1$$

so that you in fact get the same in both.

Can you see where did the integration constant kick in?

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The second one is saying compute the integral of $x^a$ as $a \to -1$. Well we know this limit is quite clearly just $\frac{1}{x}$. So what's the integral of $\frac{1}{x}$. It's just $ln(x)$.

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