Definite integral of trigonometric functions In evaluating this integration
$$
I = \int_{0}^{\infty}\frac{\arctan\left(\beta x\right) - \arctan\left(x\right)}{x}\,\mathrm{d}x\,,
$$
I separated the $1^{\mathrm{st}}$ term, multiply numerator and denominator with $\beta$ . 
$$
\mbox{That gives me}\
\int_{0}^{\infty}\frac{\arctan\left(\beta x\right)}{\beta x}\, \beta\,\mathrm{d}x
=\int_{0}^{\infty}\frac{\arctan\left(y\right)}{y}\,\mathrm{d}y\,,
$$
where, $y = \beta x$. It is the negative of $2^{\mathrm{nd}}$ term . So, $I = 0$
But my teacher said that I can't do this for this type of trigonometric functions. I can't understand his logic . Have anyone any better explaination ?. 
 A: It can be solved with a direct application of the Frullani's theorem (assuming $\beta>0$
 ) $$\int_{0}^{\infty}\frac{\tan^{-1}\left(\beta x\right)-\tan^{-1}\left(x\right)}{x}dx=\left(\tan^{-1}\left(0\right)-\tan^{-1}\left(\infty\right)\right)\log\left(\frac{1}{\beta}\right)=\color{red}{\frac{\pi}{2}\log\left(\beta\right)}.$$
A: $$
\begin{align}
&\int_0^\infty\frac{\tan^{-1}(\beta x)-\tan^{-1}(x)}{x}\,\mathrm{d}x\\
&=\lim_{a\to\infty}\int_0^a\frac{\tan^{-1}(\beta x)-\tan^{-1}(x)}{x}\,\mathrm{d}x\tag{1}\\
&=\lim_{a\to\infty}\left[\int_0^a\frac{\tan^{-1}(\beta x)}{x}\,\mathrm{d}x-\int_0^a\frac{\tan^{-1}(x)}{x}\,\mathrm{d}x\right]\tag{2}\\
&=\lim_{a\to\infty}\left[\int_0^{a\beta}\frac{\tan^{-1}(x)}{x}\,\mathrm{d}x-\int_0^a\frac{\tan^{-1}(x)}{x}\,\mathrm{d}x\right]\tag{3}\\
&=\lim_{a\to\infty}\int_a^{a\beta}\frac{\tan^{-1}(x)}{x}\,\mathrm{d}x\tag{4}\\
&=\lim_{a\to\infty}\int_1^\beta\frac{\tan^{-1}(ax)}{x}\,\mathrm{d}x\tag{5}\\
&=\int_1^\beta\frac{\pi/2}{x}\,\mathrm{d}x\tag{6}\\[6pt]
&=\log(\beta)\frac\pi2\tag{7}
\end{align}
$$
Explanation:
$(1)$: write the improper integral as a limit
$(2)$: separate the integrals
$(3)$: substitute $x\mapsto x/\beta$ in the left-hand integral
$(4)$: combine the integrals
$(5)$: substitute $x\mapsto ax$
$(6)$: dominated convergence, $\tan^{-1}(ax)\le\frac\pi2$, and $\lim\limits_{a\to\infty}\tan^{-1}(ax)=\frac\pi2$
$(7)$: integrate
