I am reading Problems in Calculus of One Variable by I.A.Maron and the author seems to make the following statements: $$\int_{0}^{\pi/2}\frac{\sin t}{\sqrt t}\text{dt}\text{ is proper integral since }\lim_{t\to+0}\frac{\sin t}{\sqrt t}=0$$ $$ \int_{0}^{\pi/2}\frac{\sin t}{t}\text{dt}\text{ is proper integral since }\lim_{t\to+0}\frac{\sin t}{t}=1$$
Does this mean that $\int^{a}_{b}f(x)\text{dx}$ is a proper integral iff $\lim_{t\to+b}f(x)$ is a finite value and $b\neq\infty$?
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$\begingroup$There are two kinds of improper integrals on the real line:
- Those where the domain of integration is unbounded, on one or both sides (e.g. $\int_1^\infty x^{-2}\ dx=1$)
- Those where the integrand itself is unbounded at one or more points (e.g. $\int_{-1}^1x^{-2/3}\ dx=6$; unbounded at 0)
There are some improper integrals satisfying both these conditions (e.g. $\int_0^\infty\frac1{(x+1)\sqrt x}\ dx=\pi$). An integral is proper if and only if it meets neither of these conditions: the domain of integration and the integrand are bounded.
Hence the given integrals are proper in spite of their undefined points, because finite limits (approaching on a path wholly within the domain of integration) exist at those points.
$\endgroup$ $\begingroup$Not only $\lim_{x\to b^+}f(x)$ should exist but also $\lim_{x\to a^-}f(x)$ should sometimes. Then it is proper.
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