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$$\lim_{\epsilon \to 0} \int \frac1{x^{1+\epsilon}} \mathrm dx$$

How should I go about evaluating this integral? Does this integral converge to $\log_e x $ or to something else?

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Is the $1+\epsilon$ supposed to be an exponent? I see its left parenthesis raised, which could mean you meant to put it as a subscript but forgot to include braces. If not, you can just factor out the $1/(1+\epsilon)$ to get $\log x$ as your limit. If so, I believe the limit just doesn't converge -- integrate $x^{-(1+\epsilon)}$ like any other power of $x$. (edit: nm, it is.) –  Paul VanKoughnett Oct 12 '10 at 7:11
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It's strange to talk about a sequence of indefinite integrals, since they all only exist up to constants. You should fix upper and lower bounds. –  Qiaochu Yuan Oct 12 '10 at 11:53

4 Answers 4

up vote 6 down vote accepted

As it stands, the limit doesn't exist (see Tobias's answer). But if you consider the definite integral from 1 to $x$, then the limit is $\ln x$: $$ \int_1^x \frac{dt}{t^{1+\epsilon}} = \left[ \frac{t^{-\epsilon}}{-\epsilon} \right]_1^x = \frac{1-x^{-\epsilon}}{\epsilon} = \frac{1-e^{-\epsilon \ln x}}{\epsilon} = \frac{1-(1-\epsilon \ln x+O(\epsilon^2))}{\epsilon} \to \ln x$$ as $\epsilon\to 0$.

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+1 Maybe that's what my subconsciousness considered when I falsely applied l'Hospital... Please feel free to include my part in your answer to have one complete and correct answer for the OP to accept (and for me to delete that edit-mess) –  Tobias Kienzler Oct 12 '10 at 11:18
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It is not that it does not exist: it simply does not make sense. –  Mariano Suárez-Alvarez Oct 12 '10 at 13:20
    
@Mariano Suárez-Alvarez: I agree, that's a better way of putting it. –  Hans Lundmark Oct 12 '10 at 13:43
    
Congratulations! You answer was accepted on July 5th, 2014. This is $1362$ days after the answer was posted, new record for the site. –  900 sit-ups a day Jul 19 at 4:02
    
@Thisismuchhealthier.: Thank you! :-) –  Hans Lundmark Jul 19 at 10:35

edit The integral asked for was $\displaystyle\lim_{\epsilon\to0}\int\frac1{x^{1+\epsilon}}dx$, so it's $=\lim\int x^{-(1+\epsilon)}dx = \lim \frac1{-\epsilon}x^{-\epsilon} = -\mathrm{sign}(\epsilon)\cdot\infty$ (just use $\int x^a dx = \frac1{a+1} x^{a+1}$ for $a\neq -1$).

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Perhaps the formatting isn't so clear but x is to the power of $(1 + \epsilon)$ in the denominator. –  shuttle87 Oct 12 '10 at 7:05
    
@shuttle87: then you need to put a ^{ and } around the $(1+epsilon)$ edit looked at your source, you forgot the curly brackets, only the ( is elevated... –  Tobias Kienzler Oct 12 '10 at 7:06
    
@Tobias, thanks, that fixed it. –  shuttle87 Oct 12 '10 at 7:08
    
@shuttle87: ok, edited the answer –  Tobias Kienzler Oct 12 '10 at 7:09
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Also, l'Hopital doesn't apply here. As $\epsilon\rightarrow 0$, $x^{-\epsilon}\rightarrow 1$, so the limit just diverges. It approaches $-\infty$ from the right and $\infty$ from the left. (no worries, though... up way too late myself) –  Paul VanKoughnett Oct 12 '10 at 8:14

Yes, the limit quantity whether its outside the integral or inside the integral doesn't make any sense. So the answer is $$\lim_{\epsilon \to 0} \frac{1}{1+ \epsilon} \cdot \int \frac{1}{x} \ dx = \log{x}$$

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shuttle87 updated the question, the $1+\epsilon$ was intended as a power –  Tobias Kienzler Oct 12 '10 at 7:44
    
@Tobias: then your answer is correct. –  anonymous Oct 12 '10 at 7:45

Evaluate the integral first using u-substitution and you'll end up with (1/(1+epsilon))ln(u) and the the limit of this as epsilon approaches 0 is indeed ln(u).

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shuttle87 updated the question, the $1+\epsilon$ was intended as a power –  Tobias Kienzler Oct 12 '10 at 7:41

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