Convergence of $\int\frac{\arctan x}{x} dx$

Question

I can't find function to bound this integral in the intervals from $1$ to $+\infty$, to prove if it converges. $$ \int _1^{\infty }\frac{\arctan x}{x}dx $$

Any idea?

How can I refute this if it is not converges? Thanks (:

Answer 1

Timbuc already provided a method to show that the integral diverges. We can also give an asymptotic equivalent. Integrating by parts, we have $$\int_1^R\frac{\arctan(x)} x\mathrm dx=\left[\ln x\cdot\arctan x \right]_{x=1}^R-\int_1^R\frac{\ln x} {1+x^2}\mathrm dx=\ln(R)\arctan(R)-\int_1^R\frac{\ln x} {1+x^2}\mathrm dx.$$ Since $0\lt \ln (x) \lt x^{1/2}$ for $x$ large enough, we get $$\int_1^R\frac{\arctan(x)} x\mathrm dx\overset{R\to \infty} {\sim} \ln(R)\frac{\pi}4.$$

Answer 2

Since the numerator is ascending, we get

$$\frac{\arctan x}x\ge\frac{\arctan 1}{x}=\frac{\pi}{4x}$$

and the integral of the right side diverges...