Is the improper integral $\int_0^\infty \frac{\ln x}{1+x^2}\,\mathrm dx$ convergent?

Question

I would like to know if

$$\int_{0}^{\infty}\frac{\ln x}{1+x^2}\,\mathrm dx$$

is convergent or divergent.

Answer 1

HINT:

Consider $$\int_{0}^{1}\frac{lnx}{1+x^2}dx$$ and $$\int_{1}^{\infty}\frac{lnx}{1+x^2}dx$$ separately.

Answer 2

It is. To see why there is not problem around $\infty$: Bertrand Integrals (in French); essentially, around $\infty$ it behaves like $x\mapsto\frac{\ln x}{x^2}$, which is $L^1$ (cf. link) because the power of $x$ is $ < -1$; and around $0$ it is equivalent to $x\mapsto\ln x$, which is integrable as it is for instance $o(\sqrt{x})$.

Answer 3

Near $x=0$: $\ln{x}/(1+x^2) \sim \ln{x}$ and

$$\int dx \, \ln{x} = x \, \ln{x} - x$$

Thus the integrand is integrable near $x=0$ because $\lim_{x \to 0} x \ln{x} = 0$.

Near $\infty$: $\ln{x}/(1+x^2) \sim \ln{x}/x^2$

and

$$\int dx \frac{\ln{x}}{x^2} = - \frac{\ln{x}-1}{x} - \frac{2}{x^2}$$

which goes to zero as $x \to \infty$. Thus the integral converges.

Answer 4

Hint: Note that under the change of variables $y=1/x$, $$ \int_a^1\frac{\log(x)}{1+x^2}\mathrm{d}x=-\int_1^{1/a}\frac{\log(y)}{1+y^2}\mathrm{d}y\tag{1} $$ Therefore, if the integral on the left side of $(1)$ converges as $a\to0$, so does the one on the right.

Since $\log(x)\le0$ on $[a,1]$ for $a\in(0,1)$, we have $$ \int_a^1\frac{\log(x)}{1}\mathrm{d}x \le\int_a^1\frac{\log(x)}{1+x^2}\mathrm{d}x \le\int_a^1\frac{\log(x)}{2}\mathrm{d}x\tag{2} $$ This reduces the problem to the convergence/divergence of $$ \int_0^1\log(x)\,\mathrm{d}x\tag{3} $$

Answer 5

Expanding eccstartup's answer: for the first integral, compare $\log x$ to $0 $, so it is obviously $0$. The second integral is upper-bounded by $$ I_2 < \int_{1}^{\infty}\frac{ \log x dx}{x^2} $$ Integrating by parts, $$ - \frac{\log x}{x}\Bigg|_{1}^{\infty}+\int_{1}^{\infty}\frac{dx}{x^3}=0 $$ The limit of the first fraction as $x \to \infty$ can be checked in many ways, incl L'Hospital's rule.

Answer 6

(i) One has $$\int_\epsilon^1\left|{\log x\over1+x^2}\right|\ dx\leq \int_\epsilon^1 \log{1\over x}\ dx=1-\epsilon+\epsilon\log\epsilon\leq C\qquad(0<\epsilon\leq1)\ .$$ Therefore the given improper integral converges at the left end.

(ii) There is an $a\geq1$ such that $\log x\leq\sqrt{x}$ for $x\geq a$. It follows that for all $b\geq a$ we have $$\int_a^b\left|{\log x\over1+x^2}\right|\ dx\leq\int_a^b{\sqrt{x}\over1+x^2}\ dx\leq \int_a^b x^{-3/2}\ dx\leq {2\over\sqrt{a}}\qquad(b\geq a)\ .$$ Therefore the given improper integral converges at the right end as well.