This integral is defined ? $\displaystyle\int_0^0\frac 1x\:dx$

It is defined, and has value $0$: write $I=\int_{0}^{0}\frac{dx}{x}=\int_{-1}^{1}\frac{\chi_{\{0\}}dx}{x}$, where $\chi_{A}$ is the indicator function on set A, $\chi_{A}(x)=1$ if $x$ belongs to $A$, and is $0$ otherwise. Now $\{0\}$ has Lebesgue measure $0$, and outside of $\{0\},\frac{\chi_{\{0\}}}{x}=0$. Hence $I=\int_{-1}^{1}0=0$. Thus $I$ is defined, at least in the sense of Lebesgue.