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