Partial Fractions — Question 10

Question 10 – Solution

Part 1: Cauchy principal value.

Step 1: Approach zero by the same distance $\varepsilon$ from both sides. $$\begin{align*} \operatorname{PV}\int_{-1}^{1}\frac{dx}{x} &=\lim_{\varepsilon\to0^+} \left(\int_{-1}^{-\varepsilon}\frac{dx}{x} +\int_{\varepsilon}^{1}\frac{dx}{x}\right)\\ &=\lim_{\varepsilon\to0^+} \left([\ln|x|]_{-1}^{-\varepsilon} +[\ln|x|]_{\varepsilon}^{1}\right)\\ &=\lim_{\varepsilon\to0^+} \left((\ln\varepsilon-\ln1)+(\ln1-\ln\varepsilon)\right)\\ &=\lim_{\varepsilon\to0^+}0=0. \end{align*}$$ $$\boxed{\displaystyle \operatorname{PV}\int_{-1}^{1}\frac{dx}{x}=0}.$$

Part 2: Ordinary improper integral.

The two one-sided integrals must converge separately. On the right, $$\begin{align*} \int_0^1\frac{dx}{x} &=\lim_{\varepsilon\to0^+}[\ln x]_{\varepsilon}^{1}\\ &=\lim_{\varepsilon\to0^+}(0-\ln\varepsilon)=+\infty. \end{align*}$$ On the left, $$\begin{align*} \int_{-1}^0\frac{dx}{x} &=\lim_{\varepsilon\to0^+}[\ln|x|]_{-1}^{-\varepsilon}\\ &=\lim_{\varepsilon\to0^+}(\ln\varepsilon-0)=-\infty. \end{align*}$$ Therefore, the ordinary improper integral diverges. The principal value is finite only because these opposite divergences cancel in a symmetric limit.