Improper Integrals — Question 1

Question 1 – Solution

Step 1: Replace infinity with a limit. $$I=\lim_{b\to\infty}\int_1^b x^{-p}\,dx.$$ Step 2: First suppose $p\ne1$. $$\begin{align*} I&=\lim_{b\to\infty}\left[\frac{x^{1-p}}{1-p}\right]_1^b\\ &=\lim_{b\to\infty}\frac{b^{1-p}-1}{1-p}. \end{align*}$$ If $p>1$, then $1-p<0$ and $b^{1-p}\to0$, so $$I=\frac{-1}{1-p}=\frac1{p-1}.$$ If $p<1$, then $b^{1-p}\to\infty$, so the integral diverges.

Step 3: Check $p=1$ separately. $$\int_1^b\frac{dx}{x}=\ln b\to\infty.$$ $$\boxed{p>1:\ I=\frac1{p-1};\qquad p\le1:\text{diverges}}$$