Why does $\int\limits_0^\infty {\ln(x) \over x}\ dx$ diverge?

Question

Why does $\int\limits_0^\infty {\ln(x) \over x}\ dx$ diverge?

I know it diverges from $1 \to \infty$, however how can I prove that the $-\infty$ from $0 \to 1$ is smaller than the area from $1 \to \infty$.

Answer 1

This integral diverges in the regular improper Riemann sense, since it is defined as $$ \lim_{\epsilon\to 0}\int_\epsilon^1\frac{\ln x}{x}\mathrm dx+\lim_{M\to \infty}\int_1^M\frac{\ln x}{x}\mathrm dx $$ the latter of which diverges.

Perhaps you are talking about the principal value integral? In which case we can indeed make sense of this integral using Cauchy principal value, in which case it is defined as $$ \lim_{\epsilon\to 0^+}\int_\epsilon^{1/\epsilon}\frac{\ln x}{x}\mathrm dx $$ and we can substitue $u=\ln x$ to find equality with $$ \int_{\ln(1/\epsilon)}^{\ln \epsilon} u\mathrm du=\frac12\lim_{\epsilon\to 0^+}([\ln(\epsilon)]^2-[\ln(1/\epsilon)]^2)\\ =\frac12\lim_{\epsilon\to 0^+}(\ln(\epsilon)-\ln(1/\epsilon))(\ln(\epsilon)+\ln(1/\epsilon)\\ =0\cdot\lim_{\epsilon \to 0^+}\ln(\epsilon)=0 $$

Answer 2

Between $0$ and $1$, set $x=\dfrac1t$, $\;\mathrm d x=-\dfrac {\mathrm dt}{t^2}$. You obtain $$\int_0^1\frac{\ln x}{x}\,\mathrm d x=\int_\infty^1-t\ln t\cdot-\frac {\mathrm dt}{t^2}=-\int_1^\infty\frac{\ln t}{t}\,\mathrm d t,$$ hence it also diverges.

Answer 3

Notice $$\int^1_0\frac{\ln x}{x}\,dx=-\int^{\infty}_1\frac{\ln y}{y}\,dy$$ after change of variable $x:=1/y$. Therefore $$\int^{\infty}_0\frac{\ln x}{x}\,dx=\int^{\infty}_1\frac{\ln x}{x}\,dx+\int^1_0\frac{\ln x}{x}\,dx=\int^{\infty}_1\frac{\ln x}{x}\,dx-\int^{\infty}_1\frac{\ln x}{x}\,dx=+\infty-\infty$$ This is an undetermined form. Hence if you want to make sense somehow of your integral you can consider the following integral for a given $a>1$: $$\int^{a}_{1/a}\frac{\ln x}{x}\,dx$$ and then let $a\uparrow+\infty$. In this case you would get: $$\lim_{a\to+\infty}\int^{a}_{1/a}\frac{\ln x}{x}\,dx=\lim_{a\to+\infty}\Big(\int^{a}_1\frac{\ln x}{x}\,dx-\int^{a}_1\frac{\ln x}{x}\,dx\Big)=\lim_{a\to+\infty}0=0$$

Answer 4

The improper integral

$$-\int\limits_0^1 {\ln(x) \over x}\ dx$$

diverges by limit comparison with

$$\int\limits_0^1 {1 \over x}\ dx$$