Question Number 67012 by mathmax by abdo last updated on 21/Aug/19
![find ∫_1 ^(+∞) ((arctan([x]))/x^3 )dx](https://www.tinkutara.com/question/Q67012.png)
$${find}\:\int_{\mathrm{1}} ^{+\infty} \:\frac{{arctan}\left(\left[{x}\right]\right)}{{x}^{\mathrm{3}} }{dx} \\ $$
Commented by mathmax by abdo last updated on 24/Aug/19
![let A =∫_1 ^(+∞) ((arctan([x]))/x^3 )dx ⇒A =Σ_(n=1) ^∞ ∫_n ^(n+1) ((actan(n))/x^3 )dx =Σ_(n=1) ^∞ arctan(n)∫_n ^(n+1) (dx/x^3 ) but ∫_n ^(n+1) (dx/x^3 ) =∫_n ^(n+1) x^(−3) dx =[(1/(−2))x^(−2) ]_n ^(n+1) =−(1/2){(1/((n+1)^2 ))−(1/n^2 )} ⇒ A =Σ_(n=1) ^∞ arctan(n){(1/n^2 )−(1/((n+1)^2 ))} =Σ_(n=1) ^∞ ((arctan(n))/n^2 )−Σ_(n=1) ^∞ ((arctan(n))/((n+1)^2 )) =Σ_(n=1) ^∞ ((arctan(n))/n^2 )−Σ_(n=2) ^∞ ((arctan(n+1))/n^2 ) A=(π/4) +Σ_(n=2) ^∞ ((arctan(n)−arctan(n+1))/n^2 ) ....be continued....](https://www.tinkutara.com/question/Q67221.png)
$${let}\:{A}\:=\int_{\mathrm{1}} ^{+\infty} \:\frac{{arctan}\left(\left[{x}\right]\right)}{{x}^{\mathrm{3}} }{dx}\:\Rightarrow{A}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\int_{{n}} ^{{n}+\mathrm{1}} \:\frac{{actan}\left({n}\right)}{{x}^{\mathrm{3}} }{dx} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:{arctan}\left({n}\right)\int_{{n}} ^{{n}+\mathrm{1}} \:\frac{{dx}}{{x}^{\mathrm{3}} }\:\:\:{but}\: \\ $$$$\int_{{n}} ^{{n}+\mathrm{1}} \:\frac{{dx}}{{x}^{\mathrm{3}} }\:=\int_{{n}} ^{{n}+\mathrm{1}} {x}^{−\mathrm{3}} {dx}\:=\left[\frac{\mathrm{1}}{−\mathrm{2}}{x}^{−\mathrm{2}} \right]_{{n}} ^{{n}+\mathrm{1}} \:=−\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right\}\:\Rightarrow \\ $$$${A}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:{arctan}\left({n}\right)\left\{\frac{\mathrm{1}}{{n}^{\mathrm{2}} }−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\right\} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{arctan}\left({n}\right)}{{n}^{\mathrm{2}} }−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{arctan}\left({n}\right)}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{arctan}\left({n}\right)}{{n}^{\mathrm{2}} }−\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{{arctan}\left({n}+\mathrm{1}\right)}{{n}^{\mathrm{2}} } \\ $$$${A}=\frac{\pi}{\mathrm{4}}\:+\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\frac{{arctan}\left({n}\right)−{arctan}\left({n}+\mathrm{1}\right)}{{n}^{\mathrm{2}} }\:….{be}\:{continued}…. \\ $$