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Question Number 29551 by abdo imad last updated on 09/Feb/18

$$findthevalueof{\int }_0^{\mathrm{\infty }}\frac{arctan\left(2x\right)-arctanx}{x}dx.$$

Commented by prof Abdo imad last updated on 11/Feb/18

$$letputI={\int }_0^{\mathrm{\infty }}\frac{arctan\left(2x\right)-arctanx}{x}dxwehave$$$$I=lim{}_{\xi \to +\mathrm{\infty }}I\left(\xi \right)/I\left(\xi \right)={\int }_0^{\xi }\frac{arctan\left(2x\right)-arctanx}{x}dx$$$$={\int }_0^{\xi }\frac{arctan\left(2x\right)}{x}dx-{\int }_0^{\xi }\frac{srctanx}{x}dx$$$$={\int }_0^{2\xi }\frac{arctant}{\frac{t}{2}}\frac{dt}{2}-{\int }_0^{\xi }\frac{arctanx}{x}dx$$$$={\int }_{\xi }^{2\xi }\frac{arctanx}{x}dxbut\mathrm{\exists }c\in ]\xi ,2\xi [/$$$$I\left(\xi \right)=arctan\xi {\int }_{\xi }^{2\xi }\frac{dx}{x}=ln\left(2\right)arctan\left(\xi \right)and$$$${lim}_{\xi \to +\mathrm{\infty }}I\left(\xi \right)=\frac{\pi }{2}ln\left(2\right).\Rightarrow I=\frac{\pi }{2}ln\left(2\right).$$

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