Check integral for convergence For the given integral check whether it converges/diverges:
$$\int_{0}^{\infty}{\frac{\operatorname{arctg}(\alpha x)-\operatorname{arctg}(\beta x)}{x}dx}$$
Literally have no ideas how to cope with it, have tried to estimate $\operatorname{tg}^{-1}(x)$ as $x$, however with no result.
 A: We initially assume that both $\alpha>0$ and $\beta>0$.  Then, we can write
$$\begin{align}
\int_\varepsilon^L \frac{\arctan(\alpha x)-\arctan(\beta x)}{x}\,dx&=\int_\varepsilon^L \frac{\arctan(\alpha x)}{x}\,dx-\int_\varepsilon^L \frac{\arctan(\beta x)}{x}\,dx\\\\
&=\int_{\alpha \varepsilon}^{\alpha L}\frac{\arctan(x)}{x}\,dx-\int_{\beta \varepsilon}^{\beta L}\frac{\arctan(x)}{x}\,dx\\\\
&=\int_{\alpha \varepsilon}^{\beta \varepsilon}\frac{\arctan(x)}{x}\,dx-\int_{\alpha L}^{\beta L}\frac{\arctan(x)}{x}\,dx\\\\
&=\int_\alpha^\beta \frac{\arctan(\varepsilon x)-\arctan(Lx)}{x}\,dx
\end{align}$$
Letting $\varepsilon\to 0$ and $L\to \infty$ we find that for $\alpha>0$ and $\beta>0$
$$\int_0^\infty \frac{\arctan(\alpha x)-\arctan(\beta x)}{x}\,dx=\frac\pi2 \log\left(\frac{\alpha}{\beta}\right)$$
If both $\alpha<0$ and $\beta<0$, then we have 
$$\int_0^\infty \frac{\arctan(\alpha x)-\arctan(\beta x)}{x}\,dx=-\frac\pi2 \log\left(\frac{\alpha}{\beta}\right)$$
The integral diverges if $\alpha \beta<0$.
A: Hint:
For convergence or divergence on the $\infty$ side, you may think of using the  relations
$$\arctan x+\arctan\frac 1x=\begin{cases}\phantom{-}\frac\pi 2&\text{if }x>0,\\[1ex]-\frac\pi 2&\text{if }x<0. \end{cases}$$
This will lead you to obtain an asymptotic equivalent of the integrand, which will depend on the signs of $\alpha$ and $\beta$.
Some details:
If $\alpha$ and $\beta$ have the same sign, say positive, we have
$$\arctan \alpha x =\frac \pi 2 -\arctan\frac1{\alpha x}=\frac\pi 2-\frac1{\alpha x}+o\Bigl(\frac1x\Bigr)$$
by Taylor's formula at order $1$. Similarly
$$\arctan \beta x =\frac\pi 2-\frac1{\beta x}+o\Bigl(\frac1x\Bigr),$$
so that we obtain
$$\arctan \alpha x-\arctan \beta x =\frac1{\beta x}-\frac1{\alpha x}+o\Bigl(\frac1x\Bigr),$$
which means that
$$\arctan \alpha x-\arctan \beta x\sim_\infty\frac{\alpha-\beta}{\alpha\beta }\frac1{x}\quad\text{if }\alpha\ne\beta,$$
and consequently
$$\frac{\arctan \alpha x-\arctan \beta x}{x}\sim_\infty\frac{\alpha-\beta}{\alpha\beta }\frac1{x^2},$$
which has a convergent integral on the $\infty$ side.
Can you examine the other cases?
