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Question Number 31097 by abdo imad last updated on 02/Mar/18

$$calculateintermsofaandbtheintegral$$ $${\int }_0^{\mathrm{\infty }}\frac{arctan\left(bt\right)-arctan\left(at\right)}{t}dtwithaandb>0.$$

Commented byabdo imad last updated on 05/Mar/18

$$letputI={\int }_0^{\mathrm{\infty }}\frac{arctan\left(bt\right)-arctan\left(at\right)}{t}dtandfor\xi >0$$ $$I\left(\xi \right)={\int }_0^{\xi }\frac{arctan\left(bt\right)-arctan\left(at\right)}{t}dtwehave$$ $$I={lim}_{\xi \to +\mathrm{\infty }}I\left(\xi \right)butI\left(\xi \right)={\int }_0^{\xi }\frac{arctan\left(bt\right)}{t}dt-{\int }_0^{\xi }\frac{arctan\left(at\right)}{t}dt$$ $$but{\int }_0^{\xi }\frac{arctan\left(bt\right)}{t}dt{=}_{bt=x}{\int }_0^{b\xi }\frac{arctanx}{x}dx$$ $${\int }_0^{\xi }\frac{artan\left(at\right)}{t}dt={}_{at=x}{\int }_0^{a\xi }\frac{arctanx}{x}dx\Rightarrow$$ $$I\left(\xi \right)={\int }_0^{b\xi }\frac{arctanx}{x}dx-{\int }_0^{a\xi }\frac{arctanx}{x}dx={\int }_{a\xi }^{b\xi }\frac{arctanx}{x}dx$$ $$\mathrm{\exists }c\in ]a\xi ,b\xi [/I\left(\xi \right)=arctan\xi {\int }_{a\xi }^{b\xi }\frac{dx}{x}=ln\left(\frac{b}{a}\right)arctan\xi \Rightarrow$$ $$lim{}_{\xi \to +\mathrm{\infty }}I\left(\xi \right)=\frac{\pi }{2}ln\left(\frac{b}{a}\right)\Rightarrow I=\frac{\pi }{2}(ln\left(b\right)-ln\left(a\right)).$$

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