Prove that $\int_0^{\infty} \frac{\log (1+x)}{x^2}\text{d}x$ is divergent.

Question

Could you please tell me how to prove that $$\int_0^{\infty} \frac{\log (1+x)}{x^2}\text{d}x$$ is divergent? I calculated an indefinite integral but I don't know how to prove that it diverges.

Answer 1

One may observe that, as $x \to 0$, by using the Taylor expansion, we have $$ \frac{\log (1+x)}{x^2}=\frac{x+O(x^2)}{x^2}=\frac1x+O(1) $$ then, since $\displaystyle \int_0^1 \frac1x \:dx $ is divergent, the integral $\displaystyle \int_0^1 \frac{\log (1+x)}{x^2} \:dx $ is divergent too, so is the initial integral $\displaystyle \int_0^\infty \frac{\log (1+x)}{x^2} \:dx $.

Answer 2

Use the fact that $x-\frac{x^2}{2} < \log(1+x)$ and try the comparison test.

Answer 3

We don't need Taylor or any fancy inequalities. We just need to know
$\displaystyle \frac{\ln (1+x)}{x} \to 1$ as $x \to 0.$
That follows from the definition $\ln'(1),$ which equals $1.$

From this we know that for some $a > 0$, on the interval $(0,a):$
$\displaystyle \frac{\ln (1+x)}{x}> 1/2.$
Our integrand is thus at least $1/(2x)$ on this interval.
Since $\displaystyle \int_0^a dx/(2x) = \infty,$ the integral is divergent.