Checking convergence/ divergence of $\int_1^\infty \frac{\log{x}}{x^2}\,dx$

Question

I tried 2 tests, but both failed.

Test-1:

checking the integrand, $\log{x}\le x$ in the domain, $\frac{\log{x}}{x^2} \le \frac{x}{x^2} = \frac{1}{x^2}$ and $\int_1^\infty\frac{1}{x}\,dx$ is divergent by p-series test, but we cant say anything about the smaller integral.

Test-2:

Splitting the given integral as $\int_1^\infty$=$\int_1^3$+$\int_3^\infty$. The first term is a proper integral and hence finite, the second can be solved by considering fact that $\log{x}>1, \forall x \ge 3$, and hence, $\frac{\log{x}}{x^2} \ge \frac{1}{x^2}$ ($x\ge 3$) and we have $\int_3^\infty\frac{1}{x^2}\,dx$ convergent by p-series test, but we cant say anything about the larger integral.

So please help on how to approach this problem.

Answer 1

Since you already know that $\log x \leqslant x$, use $\log x = 2 \log x^{1/2}\leqslant 2 x^{1/2}$ to get the estimate

$$\frac{\log x}{x^2}\leqslant \frac{2x^{1/2}}{x^2}= \frac{2}{x^{3/2}}$$

The presence of a constant is irrelevant in applying the comparison test.

Answer 2

By parts, you get the integral $$\int_{1}^{\infty}\frac{\log{x}}{x^{2}} \ dx = -\frac{\log{x}}{x} \mid_{1}^{\infty} + \int_{1}^{\infty}\frac{1}{x^{2}} \ dx = \int_{1}^{\infty}\frac{1}{x^{2}} \ dx$$ And then the convergence is clear.

Answer 3

Let me provide an answer building upon your "Test-1" and discussions we had in the comments.

Similarly as $\log x \leq x$, there also holds $\log x \leq \sqrt{x}$ (see this question for a proof). Hence, the integrand is bounded above by $x^{-\frac 32}$, for which you can apply successfully your convergence test.