Improper Integrals — Question 7

Question 7

Determine whether the ordinary improper integral converges: $$\int_{-1}^{1}\frac{dx}{x}.$$

Question 7 – Solution

Step 1: Split at the singularity. Ordinary convergence requires both one-sided integrals to converge.

Step 2: Evaluate the left side. $$\begin{align*} \int_{-1}^0\frac{dx}{x} &=\lim_{\varepsilon\to0^+}\int_{-1}^{-\varepsilon}\frac{dx}{x}\\ &=\lim_{\varepsilon\to0^+}(\ln\varepsilon-\ln1)=-\infty. \end{align*}$$ Step 3: Evaluate the right side. $$\begin{align*} \int_0^1\frac{dx}{x} &=\lim_{\varepsilon\to0^+}\int_\varepsilon^1\frac{dx}{x}\\ &=\lim_{\varepsilon\to0^+}(\ln1-\ln\varepsilon)=+\infty. \end{align*}$$ Since the one-sided integrals diverge, the ordinary integral diverges. $$\boxed{\text{diverges}}$$ The symmetric principal value is $0$, but it is not an ordinary integral value.