Suppose one has a function $ \ f(x) = (x-b)^ {-p} \ $ and ,of course, an asymptote at x = b and p is positive.
Is it correct that the integral of f(x) from a to c
(where a less than b less than c) will always converge of the following conditions are met:
(-p+1) is greater than 0. For example p could be 2/3 but not 3/2.
(p = $\frac{d}{q})$ where q is an odd number.
Note that if (p = $\frac{d}{q})$ where q is odd then (1 - p) must also have an odd denominator. The resulting function $ \ f(x) = (x-b)^ {-p} \ $ is therefore defined for all potential upper or lower bounds.
So any integral with an integrand of
$ \ f(x) = (x-b)^ {-p} \ $ ( where p = $\frac{d}{q}$ is less than 1 and q is odd) must converge for all possible finite boundaries.
I don't think this is profound in terms of concepts but might help students if pressed for time
(a student can at least rule out convergence just by looking at q).
Examples:
ex. $\int _a^b\ f(x) = (x-5)^ { {-2}/{3}} \ $ will converge for any finite a and b about 5.
ex. $\int _a^b\ f(x) = (x-5)^ {{-1}/{2}} \ $ will not converge for any finite a and b about 5.