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Question Number 72390 by mathmax by abdo last updated on 28/Oct/19

calculate U_n =∫_0 ^∞ ((arctan(1+x^4 ))/((x^2 +n^2 )^3 ))dx and determine nature of the serie Σ U_n

$${calculate}\:{U}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left(\mathrm{1}+{x}^{\mathrm{4}} \right)}{\left({x}^{\mathrm{2}} \:+{n}^{\mathrm{2}} \right)^{\mathrm{3}} }{dx} \\ $$$${and}\:{determine}\:{nature}\:{of}\:{the}\:{serie}\:\Sigma\:{U}_{{n}} \\ $$

Commented by mathmax by abdo last updated on 31/Oct/19

changement x=nt give U_n =∫_0 ^∞ ((arctan(1+n^4 t^4 ))/(n^6 (t^2 +1)^3 )) (n)dt =(1/n^5 ) ∫_0 ^∞ ((arctan(1+n^4 t^4 ))/((t^2 +1)^3 ))dt ⇒2n^5 U_n =∫_(−∞) ^(+∞) ((arctan(1+n^4 t^4 ))/((t^2 +1)^3 ))dt let ϕ(z)=((arctan(1+n^4 z^4 ))/((z^2 +1)^3 )) ⇒ϕ(z)=((arctan(1+n^4 z^4 ))/((z−i)^3 (z+i)^3 )) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) Res(ϕ,i) =lim_(z→i) (1/((3−1)!)){(z−i)^3 ϕ(z)}^((2)) =lim_(z→i) (1/2){ ((arctan(n^4 z^4 +1))/((z+i)^3 ))}^((2)) 2Res(ϕ,i) =lim_(z→i) {((((4n^4 z^3 )/(1+(n^4 z^4 +1)^2 ))×(z+i)^3 −3(z+i)^2 arctan(n^4 z^4 +1))/((z+i)^6 ))}^((1)) =lim_(z→i) { ((4n^4 z^3 (z+i)−3arctan(n^4 z^4 +1))/((z+i)^4 {1+(n^4 z^4 +1)^2 }))}^((1)) ....be continued....

$${changement}\:{x}={nt}\:{give}\:{U}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left(\mathrm{1}+{n}^{\mathrm{4}} {t}^{\mathrm{4}} \right)}{{n}^{\mathrm{6}} \left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} }\:\left({n}\right){dt} \\ $$$$=\frac{\mathrm{1}}{{n}^{\mathrm{5}} }\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left(\mathrm{1}+{n}^{\mathrm{4}} {t}^{\mathrm{4}} \right)}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} }{dt}\:\Rightarrow\mathrm{2}{n}^{\mathrm{5}} \:{U}_{{n}} =\int_{−\infty} ^{+\infty} \:\frac{{arctan}\left(\mathrm{1}+{n}^{\mathrm{4}} {t}^{\mathrm{4}} \right)}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} }{dt} \\ $$$${let}\:\varphi\left({z}\right)=\frac{{arctan}\left(\mathrm{1}+{n}^{\mathrm{4}} {z}^{\mathrm{4}} \right)}{\left({z}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }\:\Rightarrow\varphi\left({z}\right)=\frac{{arctan}\left(\mathrm{1}+{n}^{\mathrm{4}} {z}^{\mathrm{4}} \right)}{\left({z}−{i}\right)^{\mathrm{3}} \left({z}+{i}\right)^{\mathrm{3}} } \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\right) \\ $$$${Res}\left(\varphi,{i}\right)\:={lim}_{{z}\rightarrow{i}} \:\:\frac{\mathrm{1}}{\left(\mathrm{3}−\mathrm{1}\right)!}\left\{\left({z}−{i}\right)^{\mathrm{3}} \varphi\left({z}\right)\right\}^{\left(\mathrm{2}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\:\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\:\frac{{arctan}\left({n}^{\mathrm{4}} {z}^{\mathrm{4}} +\mathrm{1}\right)}{\left({z}+{i}\right)^{\mathrm{3}} }\right\}^{\left(\mathrm{2}\right)} \\ $$$$\mathrm{2}{Res}\left(\varphi,{i}\right)\:={lim}_{{z}\rightarrow{i}} \:\:\:\left\{\frac{\frac{\mathrm{4}{n}^{\mathrm{4}} {z}^{\mathrm{3}} }{\mathrm{1}+\left({n}^{\mathrm{4}} {z}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{2}} }×\left({z}+{i}\right)^{\mathrm{3}} −\mathrm{3}\left({z}+{i}\right)^{\mathrm{2}} \:{arctan}\left({n}^{\mathrm{4}} {z}^{\mathrm{4}} +\mathrm{1}\right)}{\left({z}+{i}\right)^{\mathrm{6}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\:\left\{\:\frac{\mathrm{4}{n}^{\mathrm{4}} {z}^{\mathrm{3}} \left({z}+{i}\right)−\mathrm{3}{arctan}\left({n}^{\mathrm{4}} {z}^{\mathrm{4}} +\mathrm{1}\right)}{\left({z}+{i}\right)^{\mathrm{4}} \left\{\mathrm{1}+\left({n}^{\mathrm{4}} {z}^{\mathrm{4}} \:+\mathrm{1}\right)^{\mathrm{2}} \right\}}\right\}^{\left(\mathrm{1}\right)} \\ $$$$....{be}\:{continued}.... \\ $$

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