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?

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 $$\int_0^1 \frac{dx}{x}\
eq\int_\varepsilon^1 \frac{dx}{x}\ ,\ \varepsilon>0$$ 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.