Determine whether the integral is convergent or divergent. $\int^{\infty}_1 81\frac{\ln(x)}{x}dx$

Question

Determine whether the integral is convergent or divergent.$\int^{\infty}_1 81\frac{\ln(x)}{x}dx$

My answer is $\infty-0=\infty$

But I am unsure whether it is convergent or divergent.

Answer 1

The constant $81$ is irrelevant, so we can just consider $$ \int_{1}^{\infty}\frac{\ln x}{x}\,dx $$ The integral converges if and only if $$ \int_{e}^{\infty}\frac{\ln x}{x}\,dx $$ converges. Since $\ln x\ge1$ for $x\ge e$, we have $$ \frac{\ln x}{x}\ge\frac{1}{x} $$ However $$ \int_e^\infty\frac{1}{x}\,dx $$ does not converge, because $\lim_{x\to \infty}\ln x=\infty$, therefore also the given integral is divergent.

Answer 2

Hint. Observe that, for $x>0$, $$ \frac{d}{dx}\left(\ln x \right)^2=2\frac{\ln x}{x} $$ and that, as $x \to \infty$, $$ \left(\ln x \right)^2 \to \infty. $$

Answer 3

An antiderivative of the integrand is: \begin{align} \int 81 \frac{\ln(x)}{x} dx &= 81 \frac{\ln(x)^2}{2} + C \end{align}

We have $\ln(1) = 0$ and $\lim_{x\to\infty} \ln(x) = \infty$. This gives \begin{align} \int\limits_1^\infty 81 \frac{\ln(x)}{x} dx &= \frac{81}{2}\left[\ln(x)^2\right]_{x=1}^{x \to \infty} \\ &= \lim_{x\to\infty} \frac{81}{2} \ln(x)^2 = \infty \end{align}

Answer 4

You can prove that the integral is divergent by Bertrand criterion where here $\alpha=1$ and $\beta=-1.$