Question Number 184925 by cortano1 last updated on 14/Jan/23
Commented by Frix last updated on 14/Jan/23
$$\mathrm{I}\:\mathrm{think}\:\mathrm{that} \\ $$$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\mathrm{tan}^{−\mathrm{1}} \:{ax}\:−\mathrm{tan}^{−\mathrm{1}} \:{bx}}{{x}}{dx}=\frac{\pi}{\mathrm{2}}\mathrm{ln}\:\frac{{a}}{{b}} \\ $$
Answered by ARUNG_Brandon_MBU last updated on 14/Jan/23
$${I}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{tan}^{−\mathrm{1}} \left({ax}\right)−\mathrm{tan}^{−\mathrm{1}} \left({bx}\right)}{{x}}{dx} \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{tan}^{−\mathrm{1}} \left({ax}\right)}{{x}}{dx}−\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{tan}^{−\mathrm{1}} \left({bx}\right)}{{x}}{dx}={I}\left({a}\right)−{I}\left({b}\right) \\ $$$$\Rightarrow{I}\:'\left({a}\right)−{I}\:'\left({b}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\mathrm{1}+\left({ax}\right)^{\mathrm{2}} }−\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\mathrm{1}+\left({bx}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{{a}}\left[\mathrm{tan}^{−\mathrm{1}} \left({ax}\right)\right]_{\mathrm{0}} ^{\infty} −\frac{\mathrm{1}}{{b}}\left[\mathrm{tan}^{−\mathrm{1}} \left({bx}\right)\right]_{\mathrm{0}} ^{\infty} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\pi}{\mathrm{2}{a}}−\frac{\pi}{\mathrm{2}{b}}\:\Rightarrow{I}\left({a}\right)−{I}\left({b}\right)=\frac{\pi}{\mathrm{2}}\mathrm{ln}{a}−\frac{\pi}{\mathrm{2}}\mathrm{ln}{b}+{C} \\ $$$${I}\left(\mathrm{1}\right)−{I}\left(\mathrm{1}\right)=\mathrm{0}={C}\:\Rightarrow{I}={I}\left({a}\right)−{I}\left({b}\right)=\frac{\pi}{\mathrm{2}}\mathrm{ln}\left(\frac{{a}}{{b}}\right) \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{tan}^{−\mathrm{1}} \left(\pi{x}\right)−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}{x}\right)}{{x}}{dx}={I}\left(\pi\right)−{I}\left(\mathrm{2}\right)=\frac{\pi}{\mathrm{2}}\mathrm{ln}\left(\frac{\pi}{\mathrm{2}}\right)\bigstar \\ $$