Question

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

${\int }_1^{\infty }\frac{1}{x\ln x}dx$

Short answer

The value is$\infty$.

Step-by-step solution

Step 1. Given information.

The given integral is${\int }_1^{\infty }\frac{1}{x\ln x}dx$.

Step 2. Value of the integral.

On simplifying,

$$\begin{gathered} \text{Let}u=\ln \left(x\right)\text{i.e}du=\frac{1}{x}dx \\ \int \frac{1}{x\ln \left(x\right)}dx=\int \frac{1}{u}du=\ln \left|u\right| \\ Therefore, \\ {\int }_1^{\infty }\frac{1}{x\ln x}dx=\underset{z\to 0}{\lim }{\int }_1^B\frac{1}{x\ln x}dx \\ =\underset{B\to \infty }{\lim }\left[\ln \right|\ln \left(x\right)|{]}_1^B \\ =\frac{1}{2}\underset{B\to \infty }{\lim }[\ln \left|\ln \right(B\left)\right|-\ln \left|\ln \right(1\left)\right|] \\ =\frac{1}{2}\underset{B\to \infty }{\lim }[\ln \left|\ln \right(B\left)\right|-\ln \left|0\right|] \\ =\infty (Diverges) \end{gathered}$$