I have to analyse the convergence of $\displaystyle \int_{0}^{+\infty}\frac{\arctan \alpha x - \arctan \beta x}{x} dx$; $\alpha,\beta \in R$
I have written:
$\displaystyle \int_{0}^{+\infty}\frac{\arctan \alpha x - \arctan \beta x}{x} dx = \displaystyle \int_{0}^{+\infty}\frac{\arctan \frac {\alpha x - \beta x}{1 + \alpha \beta x^2}}{x}dx = \displaystyle \int_{0}^{1} + \displaystyle \int_{1}^{+\infty}$
But I don't know how I can conclude something about this.
By turning into a double integral and reversing the order of integration, we get that
$$\int_0^\infty \frac{\arctan(\alpha x) - \arctan(\beta x)}{x}\:dx = \int_\beta^\alpha \frac{\arctan(+\infty)-\arctan(0)}{x}\:dx $$
$$= \frac{\pi}{2} \int_\beta^\alpha \frac{dx}{x} = \frac{\pi}{2}\log\left(\frac{\alpha}{\beta}\right)$$
which converges if $\alpha,\beta > 0$
The only things to consider are behavior near zero and infinity. Near zero $\arctan(\alpha x) - \arctan(\beta x) = \alpha x - \beta x +O(x^3) $ and so the integrand tends to $\alpha-\beta$ as $x \to 0^{+} $ and the integral converges, while as $x \to \infty, \arctan(\alpha x) - \arctan(\beta x) = \frac{1}{\beta x} - \frac{1}{\alpha x} + O(\frac{1}{x^3}) $ (as long as $\alpha$ and $ \beta $ have the same sign) and so the integrand is $O(\frac{1}{x^2})$ so the integral also converges. If $\alpha$ and $\beta$ have opposite signs the numerator near infinity is $ \pm \pi + O(\frac{1}{x}) $ so the integral diverges.
Another way to obtain the value is by differentiating under the integral sign. Suppose $\alpha, \beta >0 $. Set $$ f(\alpha) = \int_0^{\infty} \frac{\arctan(\alpha x)-\arctan(\beta x)}{x} \, dx $$ Then $$ f'(\alpha) = \int_0^{\infty} \frac{1}{1+\alpha^2 x^2} \, dx = \frac{1}{\alpha} \arctan(\alpha x) \left|_0^{\infty} \right. = \frac{\pi}{2\alpha} $$ So $$ f(\alpha) = \frac{\pi}{2} \log (\alpha) + c $$ To find $c$ note that $ f(\beta) = 0 $. This is a special case of a more general result. See https://en.wikipedia.org/wiki/Frullani_integral
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 $\alpha \beta<0$.
Writing the inverse trigonometric functions in terms of logarithms, it is "easy" to show that $$\int \frac{\arctan( x) }x\,dx=-\dfrac{\mathrm{i}\left(\operatorname{Li}_2\left(\mathrm{i}x\right)-\operatorname{Li}_2\left(-\mathrm{i}x\right)\right)}{2}$$ which makes $$I_c=\int_0^t \frac{\arctan(c x) }x\,dx=-\dfrac{\mathrm{i}\left(\operatorname{Li}_2\left(\mathrm{i}ct\right)-\operatorname{Li}_2\left(-\mathrm{i}ct\right)\right)}{2}$$
Now, expanding as series for large values of $t$
$$c > 0 \implies I_c=\frac{1}{2} \pi \log (c t)+\frac{1}{c t}+O\left(\frac{1}{t^3}\right)$$ $$c < 0 \implies I_c=-\frac{1}{2} \pi \log (-ct )+\frac{1}{c t}+O\left(\frac{1}{t^3}\right)$$ and then the results @Mark Viola gave in his answer.
