91Ó°ÊÓ

The improper integral

The goal is to use the limit comparison test to determine if the improper integral discussed above is convergent or divergent.

According to the limit comparison test, if two functions,$f\left(x\right)$and $g\left(x\right)$, exhibit comparable behaviour on $[a,∞)$, their incorrect integrals on $[a,∞)$should either converge or diverge in unison. The limit comparison test essentially states that if $\underset{x→∞}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=L$if some actual integer L is positive, then the incorrect integrals ${∫}_a^{∞}f\left(x\right)dx$and ${∫}_a^{∞}g\left(x\right)dx$either both converge or diverge.

Consider the following two functions, localid="1661324402295" $f\left(x\right)$and $g\left(x\right)$:

$f\left(x\right)=\frac{1}{x}\text{and}g\left(x\right)=\frac{1}{x+0.001}.$

Calculate the value of the following functions: $\underset{x→∞}{\lim }\frac{f\left(x\right)}{g\left(x\right)}$

$\underset{x→∞}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x→∞}{\lim }\frac{\left(\frac{1}{x}\right)}{\left(\frac{1}{x+0.001}\right)}=\underset{x→∞}{\lim }\frac{x+0.001}{x}=\underset{x→∞}{\lim }1+\frac{0.001}{x}\left(\right.Since\left.\frac{1}{∞}=0\right)=1\left(1isapositiverealnumber\right)$

As a result, both limit comparison testslocalid="1661324595683" role="math" ${∫}_1^{∞}\frac{1}{x}dx,{∫}_1^{∞}\frac{1}{x+0.001}dx$either converge or diverge$........\left(1\right)$

Use incorrect power function integrals ${∫}_1^{∞}\frac{1}{x}dx$diverges.

Consequently, the improper integral ${∫}_1^{∞}\frac{1}{x+0.001}dx$also diverges.