Question

Resolver a integral imprópria $\int\limits^a_1 {lnx/x^2} \, dx$    sendo que a é infinito

Answer 1

Olá, Antonnio.

Utilizaremos o método de integração por partes.

$\begin{cases} u(x)=\ln x\Rightarrow du=\frac1 x\,dx\\\\v(x)=\frac1 x\Rightarrow dv=-\frac{1}{x^2}\,dx\end{cases}$

Definidas as funções $u$ e $v$, aplicamos agora o método:

$\int{u\,dv}=uv-\int{v\,du}\Rightarrow \int{-\frac{\ln x}{x^2}}=\frac{\ln x}x-\int{\frac1 x\cdot\frac1 x\,dx}\Rightarrow$

$\int\limits_1^\infty{\frac{\ln x}{x^2}}=-\left\frac{\ln x}x\right|\limits_1^\infty+\int\limits_1^\infty{\frac1 x^2\,dx}=\\\\\\=\underbrace{-\lim_{x\to\infty}\frac{\ln x}{x}}_{\text{Regra de L'H\^opital}}+\underbrace{\frac{\ln 1}{1}}_{=0}+\left\left(-\frac1 x\right)\right|\limits_1^\infty=$

$=-\underbrace{\lim_{x\to\infty}\frac{(\frac1 x)}{1}}_{=0}-\left[\underbrace{\left(\lim_{x\to\infty}\frac{1}{x}\right)}_{=0}-1\right]=1\\\\\\ \therefore\boxed{\int\limits_1^\infty{\frac{\ln x}{x^2}}=1}$