Question about convergence of integrals Is the following correct, or am I missing something?
$\int_0^1 \frac{dx}{x}$ does not converge if we include the end point $0$. On the other hand, $\int_\epsilon^1 \frac{dx}{x}\ ,\ \epsilon>0$ does converge to $-log(\epsilon)$. As long as you integrate on the semi-open interval $(0,1]$, there is convergence. If you integrate on the closed interval $[0,1]$, there is divergence. 
In other words, we have:
$$\int_0^1 \frac{dx}{x}\neq\int_\epsilon^1 \frac{dx}{x}\ ,\ \epsilon>0$$
but we can say:
$$\int_0^1 \frac{dx}{x}= \lim_{\epsilon\to 0^+}\int_\epsilon^1\frac{dx}{x}$$
What is interesting here is that the integrals on the semi-open $(0,1]$ all converge (have a definite value $<\infty$), yet the integral on the closed $[0,1]$ diverges to $+\infty$. Is that a correct way of seeing things?
 A: You are (mostly) correct, though you need to be a bit careful.  Integrals do not distinguish between open and closed integrals, so saying that you obtain a different result when integrating on $(0,1]$ vs. $[0,1]$ doesn't make sense.  It is true, however, that $$\int_0^1 \frac{dx}{x}\neq\int_\varepsilon^1 \frac{dx}{x}\ ,\ \varepsilon>0$$ because the integral on the left diverges and the integral on the right converges.  What that means is that the integral diverges on $[0,1]$ (or equivalently $(0,1],[0,1)$, or $(0,1)$) and converges on the interval $[\varepsilon,1]$ for all $\varepsilon > 0$.  However, it is also worth noting that since $$\int_{\varepsilon}^{1} \frac{dx}{x} = -\log(\varepsilon),\quad\text{and}\quad\lim_{\varepsilon \to 0}-\log(\varepsilon) = \infty,$$ we recover the divergent case as we let $\varepsilon \to 0$ as expected.
A: $\int_\epsilon^1 \frac{dx}{x}$ doesn't converge because we're not taking any kind of limit here. You are correct it's value is $-\ln (\epsilon)$ though.
However, if you take the limit as $\epsilon \to 0$, then $$\lim_{\epsilon \to 0} \int_\epsilon^1\frac{dx}{x} = \lim_{\epsilon \to 0} - \ln (\epsilon) = - \lim_{\epsilon \to 0} \ln \epsilon = \infty$$
With lower bound $\epsilon > 0$, the integral is proper, so there's no issue.
