What do I do if in the point of lower bound of some first-odered improper intagral
integrand doesn't exists? For instance, $$\int _1^{\infty }\frac{dx}{x\log ^2x} $$
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PersonaNonGrata
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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.
Siminore
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I'm curious what if a product of $$\int_a^\infty f(x) dx = \lim_{\epsilon \to 0} \int_{a+\epsilon}^c f(x)dx + \lim_{b \to +\infty} \int_c^b f(x)dx$$ will be $\infty - \infty$? – PersonaNonGrata Jun 09 '16 at 14:56
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By definition the improper integral diverges. – Siminore Jun 09 '16 at 17:25
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that's not what I asked, but I figured it out already – PersonaNonGrata Jun 09 '16 at 18:16