Improper integral convergence problem

Question

I need to show that this integral is convergent $$\int_1^{\infty} \frac{\arctan 2x -\arctan x}{x} dx$$ Maybe I should express $\arctan$ with one variable using $\tan (\alpha - \beta) = \ldots$

Anyway, I could use some help with this problem. It would be great to see some different solutions. Thanks

Answer 1

We have $$\arctan x=\frac{\pi}{2}-\arctan\left(\frac 1 x\right),\quad x>0$$ and $$\arctan\left(\frac 1 x\right)\sim_\infty \frac 1 x$$ so we find $$\frac{\arctan 2x -\arctan x}{x}\sim_\infty\frac{ 1}{2 x^2}$$ so we deduce the desired result

Answer 2

\begin{eqnarray} 0\le \int_1^\infty\frac{\arctan2x-\arctan x}{x}\,dx&=&\int_1^\infty\frac{1}{x}\left(\int_x^{2x}\frac{1}{1+t^2}\,dt\right)\,dx\\ &\le&\int_1^\infty\frac1x\left(\int_x^{2x}\frac{1}{1+x^2}\,dt\right)\,dx\\ &=&\int_1^\infty\frac{1}{1+x^2}\,dx=\arctan x\Big|_1^\infty=\frac\pi2-\frac\pi4=\frac\pi4. \end{eqnarray}

Answer 3

Note that beyond $x=1$, we have $\arctan(2x) - \arctan(x)$ to be a decreasing function. We hence have \begin{align} \int_{2^0}^{\infty} \dfrac{\arctan(2x)-\arctan(x)}{x} dx & < \sum_{k=0}^{\infty} \dfrac{\arctan(2^{k+1})-\arctan(2^k)}{2^k} (2^{k+1} - 2^k)\\ & = \sum_{k=0}^{\infty} \dfrac{\arctan(2^{k+1})-\arctan(2^k)}{2^k} 2^k\\ & = \sum_{k=0}^{\infty} \left(\arctan(2^{k+1})-\arctan(2^k) \right)\\ & = \dfrac{\pi}2 - \dfrac{\pi}4 = \dfrac{\pi}4 \end{align} From $0$ to $1$, the integrand is bounded with a removable singularity at origin and hence we are done.