Improper Integrals — Question 8

Question 8 – Solution

Step 1: Replace the improper endpoint with a limit. $$I=\lim_{\varepsilon\to0^+}\int_\varepsilon^1\ln x\,dx.$$ Step 2: Integrate by parts. Let $u=\ln x$ and $dv=dx$. Then $du=dx/x$ and $v=x$: $$\begin{align*} \int\ln x\,dx &=x\ln x-\int x\left(\frac1x\right)dx\\ &=x\ln x-x. \end{align*}$$ Step 3: Evaluate the limit. $$\begin{align*} I&=\lim_{\varepsilon\to0^+}[x\ln x-x]_\varepsilon^1\\ &=\lim_{\varepsilon\to0^+} \left(-1-\varepsilon\ln\varepsilon+\varepsilon\right). \end{align*}$$ Since $\varepsilon\ln\varepsilon\to0$ and $\varepsilon\to0$, $$I=-1.$$ $$\boxed{-1}$$