Question

Use limits of definite integrals to calculate each of the improper integrals in Exercises 21–56.

${\int }_0^1\frac{\ln x}{x}dx$

Short answer

The answer is$\infty$.

Step-by-step solution

Step 1. Given Information.

The given function is${\int }_0^1\frac{\ln x}{x}dx$.

Step 2. Value of the integral.

On simplifying,

$$\begin{gathered} {\int }_0^1\frac{\ln x}{x}dx=\underset{t\to 1}{\lim }{\int }_0^t\ln x·\frac{1}{x}dx \\ \text{Substitute}\ln x=u\Rightarrow \frac{1}{x}dx=du \\ \text{Then}\int \ln x·\frac{1}{x}dx=\int u·du \\ =\frac{u^2}{2}=\frac{1}{2}(\ln x{)}^2 \\ \text{Therefore equation becomes} \\ {\int }_0^1\frac{\ln x}{x}dx=\underset{t\to 1}{\lim }{\int }_0^t\ln x\frac{1}{x}dx \\ =\underset{t\to 1}{\lim }{\left[\frac{1}{2}(\ln x{)}^2\right]}_0^t \\ =\underset{t\to 1}{\lim }\frac{1}{2}\left[(\ln t{)}^2-(\ln 0{)}^2\right] \\ =\frac{1}{2}\times \infty \\ =\infty \end{gathered}$$

That is divergent.