How do I evaluate this integral? $$\lim_{\epsilon \to 0} \int \frac1{x^{1+\epsilon}} \mathrm dx$$
How should I go about evaluating this integral?
Does this integral converge to $\log_e x $ or to something else?
 A: 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 $\epsilon\to 0$.
A: 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 $a\neq -1$).
A: Yes, the limit quantity whether its outside the integral or inside the integral doesn't make any sense. So the answer is $$\lim_{\epsilon \to 0} \frac{1}{1+ \epsilon} \cdot \int \frac{1}{x} \ dx = \log{x}$$
A: 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).
