Quick question about improper integral

Question

What do I do if in the point of lower bound of some first-odered improper intagral
integrand doesn't exists? For instance, $$\int _1^{\infty }\frac{dx}{x\log ^2x} $$

Answer 1

Assuming you are in the framework of integration theory according to Riemann, by definition $$ \int_1^\infty \frac{dx}{x \log^2 x} = \lim_{a \to 1+} \int_a^2 \frac{dx}{x \log^2 x} + \lim_{b \to +\infty} \int_2^b \frac{dx}{x \log^2 x}, $$ provided that both limits exist as real numbers. The choice of splitting the integral at $2$ is totally arbitrary, and you could choose any number at which the integrand function is continuous.