does $\int_0^\infty \frac{x-\arctan(x)}{x(1+x^2)\arctan(x)} \,dx$ converge

Question

does the following integral converges?

$\int_0^\infty \frac{x-\arctan(x)}{x(1+x^2)\arctan(x)} \,dx$

I calculated $$\int \frac{x-\arctan(x)}{x(1+x^2)\arctan(x)} \,dx = \ln(\arctan(x)) - \ln(x) + 0.5\ln(1+x^2)$$

But wasnt able to detemine the convergence.

Answer 1

Converges. The limit at plus infinity can be computed by combining the last two logarithms and evaluating arctan(+infinity) and at 0 combine the first two logarithms and apply Hopital on arctanx/x.

Answer 2

Using the Taylor series $\arctan x= x-\frac{x^3}{3}+o(x^3)$ we see that the integrand can be extended at $0$. Moreover, at $\infty$ we have

$$\frac{x-\arctan x}{x(1+x^2)\arctan x}\sim_\infty\frac{2}{\pi}\frac1{1+x^2}\in L^1(1,\infty)$$ so we conclude the convergence of the given integral.

Answer 3

Notice, we have $$\int_{0}^{\infty}\frac{x-\tan^{-1}x}{x(1+x^2)\tan^{-1}x}dx$$ Let $\tan^{-1}x=\theta \implies \frac{dx}{1+x^2}=d\theta$ $$\int_{0}^{\pi/2}\frac{\tan\theta-\theta}{\theta\tan\theta}d\theta$$ $$\int_{0}^{\pi/2}\left(\frac{1}{\theta}-\cot\theta\right)d\theta$$

I hope you can take it from here.

Answer 4

\begin{align*} I&=\int_0^\infty \frac{x-\arctan(x)}{x(1+x^2)\arctan(x)}\,dx\\ &=\left[ \ln(\arctan(x)) - \ln(x) + 0.5\ln(1+x^2)\right]_0^\infty\\ &=\lim_{x\rightarrow \infty}\left[ \ln(\arctan(x)) - \ln(x) + 0.5\ln(1+x^2)\right]-\lim_{x\rightarrow \infty}\left[ \ln(\arctan(1/x)) - \ln(1/x) + 0.5\ln(1+1/x^2)\right]\\ &=\lim_{x\rightarrow \infty} \ln \left(\frac{\arctan(x) \sqrt{1+x^2}}{\arctan(1/x)x^2\sqrt{1+1/x^2}}\right)\\ &=\lim_{x\rightarrow \infty}\ln\left( \frac{\arctan(x)}{\arctan(1/x)\times x} \right) \\ \text{We know that:}\\ \\ \lim_{x\rightarrow \infty} x\arctan(1/x)=1\\ \text{Hence:} \\ \\ I&=\ln(\frac \pi 2) \end{align*}