Discussion
The improper integral \(\displaystyle\int_{a}^{b} \frac{dx}{(x - a)^{n}}\) converges :
Answer: D
The keyed answer is only if \(0 \lt n \lt 1\).
The integrand \(1/(x - a)^{n}\) blows up at the lower limit only when \(n \gt 0\), so that is when the integral is improper. For \(n \ne 1\):
\[ \begin{aligned} &\int_{a+\epsilon}^{b} \frac{dx}{(x - a)^{n}} \\ &= \frac{(b - a)^{1-n} - \epsilon^{1-n}}{1 - n} \end{aligned} \]
As \(\epsilon \to 0\), \(\epsilon^{1-n} \to 0\) if \(n \lt 1\) and \(\to \infty\) if \(n \gt 1\); for \(n = 1\) the integral is \(\log\) and diverges. So the improper integral converges exactly for \(0 \lt n \lt 1\). For \(n \le 0\) the integrand is continuous and the integral is an ordinary one, which is why the key states the range for the improper case.
It does not converge for every \(n\), it diverges for \(n \gt 1\), and integers \(n \ge 1\) all diverge.
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