Question

Express each improper integral in Exercises15–20 as a sum of limits of proper definite integrals. Do not calculate any integrals or limits; just write them down.

${\int }_2^{\infty }\frac{{\displaystyle 1}}{{\displaystyle x\ln (x-2)}}dx$

Short answer

The integral is,

${\int }_2^3\frac{1}{x\ln (x-2)}dx+\underset{A\to \infty }{\lim }{\int }_3^A\frac{1}{x\ln (x-2)}dx\text{.}$

Step-by-step solution

Step 1. Given information.   

We are given an integrals,

${\int }_2^{\infty }\frac{1}{x\ln (x-2)}dx$

Step 2. Graph of function 

Let ,

The graph is as follows,

Step 3. Expressing the Integral. 

The integrand $\frac{1}{x\ln (x-2)}$is not defined when $x\ln (x-2)=0$

That is $x=0$or $\ln (x-2)=0$, which implies $x=0$or $3$

So $y=\frac{1}{x\ln (x-2)}$is continuous everywhere except at $x=0,3$.

Thus, split the interval $(2,\infty )$at $x=3$.

Rewriting the improper integral by splitting the interval $(2,\infty )$at $x=3$.

${\int }_2^{\infty }\frac{1}{x\ln (x-2)}dx={\int }_2^3\frac{1}{x\ln (x-2)}dx+{\int }_3^{\infty }\frac{1}{x\ln (x-2)}dx$

Write,

${\int }_3^{\infty }\frac{1}{x\ln (x-2)}dx=\underset{x\to \infty }{\lim }{\int }_3^A\frac{1}{x\ln (x-2)}dx$

Hence, the integral is,

${\int }_2^{\infty }\frac{1}{x\ln (x-2)}dx={\int }_2^3\frac{1}{x\ln (x-2)}dx+\underset{A\to \infty }{\lim }{\int }_3^A\frac{1}{x\ln (x-2)}dx\text{.}$