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


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

$c a l c u l a t e U_{n} = \int_{0}^{\infty} \frac{a r c t a n (1 + x^{4})}{{(x^{2} + n^{2})}^{3}} d x$ $a n d d e t e r m i n e n a t u r e o f t h e s e r i e Σ U_{n}$
Commented by mathmax by abdo last updated on 31/Oct/19
$changement x=nt give U_n =∫_0 ^∞ ((arctan(1+n^4 t^4 ))/(n^6 (t^2 +1)^3 )) (n)dt =(1/n^5 ) ∫_0 ^∞ ((arctan(1+n^4 t^4 ))/((t^2 +1)^3 ))dt ⇒2n^5 U_n =∫_(−∞) ^(+∞) ((arctan(1+n^4 t^4 ))/((t^2 +1)^3 ))dt let ϕ(z)=((arctan(1+n^4 z^4 ))/((z^2 +1)^3 )) ⇒ϕ(z)=((arctan(1+n^4 z^4 ))/((z−i)^3 (z+i)^3 )) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) Res(ϕ,i) =lim_(z→i) (1/((3−1)!)){(z−i)^3 ϕ(z)}^((2)) =lim_(z→i) (1/2){ ((arctan(n^4 z^4 +1))/((z+i)^3 ))}^((2)) 2Res(ϕ,i) =lim_(z→i) {((((4n^4 z^3 )/(1+(n^4 z^4 +1)^2 ))×(z+i)^3 −3(z+i)^2 arctan(n^4 z^4 +1))/((z+i)^6 ))}^((1)) =lim_(z→i) { ((4n^4 z^3 (z+i)−3arctan(n^4 z^4 +1))/((z+i)^4 {1+(n^4 z^4 +1)^2 }))}^((1)) ....be continued....$
$c h a n g e m e n t x = n t g i v e U_{n} = \int_{0}^{\infty} \frac{a r c t a n (1 + n^{4} t^{4})}{n^{6} {(t^{2} + 1)}^{3}} (n) d t$ $= \frac{1}{n^{5}} \int_{0}^{\infty} \frac{a r c t a n (1 + n^{4} t^{4})}{{(t^{2} + 1)}^{3}} d t \Rightarrow 2 n^{5} U_{n} = \int_{- \infty}^{+ \infty} \frac{a r c t a n (1 + n^{4} t^{4})}{{(t^{2} + 1)}^{3}} d t$ $l e t φ (z) = \frac{a r c t a n (1 + n^{4} z^{4})}{{(z^{2} + 1)}^{3}} \Rightarrow φ (z) = \frac{a r c t a n (1 + n^{4} z^{4})}{{(z - i)}^{3} {(z + i)}^{3}}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i)$ $R e s (φ, i) = {l i m}_{z \to i} \frac{1}{(3 - 1)!} {{(z - i)}^{3} φ (z)}^{(2)}$ $= {l i m}_{z \to i} \frac{1}{2} {\frac{a r c t a n (n^{4} z^{4} + 1)}{{(z + i)}^{3}}}^{(2)}$ $2 R e s (φ, i) = {l i m}_{z \to i} {\frac{\frac{4 n^{4} z^{3}}{1 + {(n^{4} z^{4} + 1)}^{2}} \times {(z + i)}^{3} - 3 {(z + i)}^{2} a r c t a n (n^{4} z^{4} + 1)}{{(z + i)}^{6}}}^{(1)}$ $= {l i m}_{z \to i} {\frac{4 n^{4} z^{3} (z + i) - 3 a r c t a n (n^{4} z^{4} + 1)}{{(z + i)}^{4} {1 + {(n^{4} z^{4} + 1)}^{2}}}}^{(1)}$ $. . . . b e c o n t i n u e d . . . .$
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