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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.)2010-10-12
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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.2010-10-12

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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)2010-10-12
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    It is not that it does not exist: it simply does not make sense.2010-10-12
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    @Mariano Suárez-Alvarez: I agree, that's a better way of putting it.2010-10-12
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    @Thisismuchhealthier.: Thank you! :-)2014-07-19
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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.2010-10-12
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    @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...2010-10-12
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    @Tobias, thanks, that fixed it.2010-10-12
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    @shuttle87: ok, edited the answer2010-10-12
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    Um, I think you're confusing integration and differentiation. $\int x^a\,dx=\frac{x^{a+1}}{a+1}$.2010-10-12
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    @Paul VanKoughnett: what the...? It's too early in the morning, thanks a lot! Editing right away **edit**ed even worse, it was a terrible mixture2010-10-12
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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)2010-10-12
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    @Paul that's right again, what am I up to today? Thank you2010-10-12
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    @Paul Hans' answer delivers a correct reasoning for using l'Hospital by taking the definite integral $\int_1^x$2010-10-12
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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 power2010-10-12
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    @Tobias: then your answer is correct.2010-10-12
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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 power2010-10-12