Question

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

${\int }_e^{\infty }\frac{1}{x(\ln x{)}^2}dx$

Short answer

The value is$1\left(Convergent\right)$.

Step-by-step solution

Step 1. Given Information.

The given integral is${\int }_e^{\infty }\frac{1}{x(\ln x{)}^2}dx$.

Step 2. Value of the limit.

On simplifying,

${\int }_e^{\infty }\frac{1}{x(\ln x{)}^2}dx$

role="math" localid="1648888264087" $$\begin{gathered} Let, \\ u=\log x \\ Ondifferentiating, \\ dx=xdu \\ Now, \\ ={\int }_{u=1}^{u=\infty }\frac{1}{x{u}^2}xdu \\ =\frac{1}{-1}{\int }_1^{\infty }{u}^{-2}du \\ {{\left.=\frac{-1}{u}\right]}^{\infty }}_1 \\ =\frac{-1}{\infty }-\frac{-1}{1} \\ =1\left(converges\right) \end{gathered}$$