Question Number 67011 by mathmax by abdo last updated on 21/Aug/19
![calculate U_n =∫_1 ^(+∞) ((arctan(n[x]))/x^2 )dx](https://www.tinkutara.com/question/Q67011.png)
$${calculate}\:{U}_{{n}} =\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{arctan}\left({n}\left[{x}\right]\right)}{{x}^{\mathrm{2}} }{dx} \\ $$
Commented by mathmax by abdo last updated on 26/Aug/19
![U_n =Σ_(k=1) ^∞ ∫_k ^(k+1) ((arctan(nk))/x^2 )dx =Σ_(k=1) ^(+∞) arctan(nk)[−(1/x)]_k ^(k+1) =Σ_(k=1) ^∞ arctan(nk){(1/k)−(1/(k+1))} =Σ_(k=1) ^∞ ((arctan(nk))/(k^2 +k)) ...be continued....](https://www.tinkutara.com/question/Q67365.png)
$${U}_{{n}} =\sum_{{k}=\mathrm{1}} ^{\infty} \:\int_{{k}} ^{{k}+\mathrm{1}} \:\frac{{arctan}\left({nk}\right)}{{x}^{\mathrm{2}} }{dx}\:=\sum_{{k}=\mathrm{1}} ^{+\infty} \:{arctan}\left({nk}\right)\left[−\frac{\mathrm{1}}{{x}}\right]_{{k}} ^{{k}+\mathrm{1}} \\ $$$$=\sum_{{k}=\mathrm{1}} ^{\infty} \:{arctan}\left({nk}\right)\left\{\frac{\mathrm{1}}{{k}}−\frac{\mathrm{1}}{{k}+\mathrm{1}}\right\}\:=\sum_{{k}=\mathrm{1}} ^{\infty} \:\frac{{arctan}\left({nk}\right)}{{k}^{\mathrm{2}} \:+{k}} \\ $$$$…{be}\:{continued}…. \\ $$