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Question Number 88424 by abdomathmax last updated on 10/Apr/20

$$calculate{U}_n={\int }_0^{\mathrm{\infty }}\frac{arctan\left(n^{2}x\right)-arctan\left(nx\right)}{x}dx$$$$andxeterminenatureoftheserie\mathrm{\Sigma }{U}_n$$

Commented by mathmax by abdo last updated on 11/Apr/20

$$letI\left(\xi \right)={\int }_0^{\xi }\frac{arctan\left(ax\right)-arctan\left(bx\right)}{x}dx\Rightarrow$$$$I\left(\xi \right)={\int }_0^{\xi }\frac{arctan\left(ax\right)}{x}dx-{\int }_0^{\xi }\frac{arctan\left(bx\right)}{x}dxbut$$$${\int }_0^{\xi }\frac{arctan\left(ax\right)}{x}dx{=}_{ax=t}{\int }_0^{a\xi }\frac{arctan\left(t\right)}{\frac{t}{a}}\times \frac{dt}{a}={\int }_0^{a\xi }\frac{arctan\left(t\right)}{t}dt$$$$also{\int }_0^{\xi }\frac{arctan\left(bx\right)}{x}dx={\int }_0^{b\xi }\frac{arctan\left(t\right)}{t}dt\Rightarrow$$$$I\left(\xi \right)={\int }_0^{a\xi }\frac{arctant}{t}dt+{\int }_{b\xi }^0\frac{arctan\left(t\right)}{t}dt={\int }_{b\xi }^{a\xi }\frac{arctan\left(t\right)}{t}dt$$$$\mathrm{\exists }c\in ]b\xi ,a\xi [/I\left(\xi \right)=arctan\left(\xi \right){\int }_{b\xi }^{a\xi }\frac{dt}{t}=arctan\left(\xi \right)ln\mid \frac{a}{b}\mid \Rightarrow$$$${lim}_{\xi \to +\mathrm{\infty }}I\left(\xi \right)={\int }_0^{\mathrm{\infty }}\frac{arctan\left(ax\right)-arctan\left(bx\right)}{x}dx=\frac{\pi }{2}ln\mid \frac{a}{b}\mid$$$$\Rightarrow U_n=\frac{\pi }{2}arctan\left(\frac{n^2}{n}\right)=\frac{\pi }{2}arctan\left(n\right)$$$$wehave{lim}_{n\to +\mathrm{\infty }}{U}_n=\frac{{\pi }^2}{4}\ne 0\Rightarrow \mathrm{\Sigma }{U}_{n}diverges$$

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