let w(x)=∫_0 ^∞ ((lnt)/((x^2 +t^2 )^2 ))dt 1) explicit w(x) 2) calculate U_n =∫_0 ^∞ ((lnt)/((n^2 +t^2 )^2 ))dt find lim_(n→+∞) n^4 U_n and determine nature of tbe serie Σ U


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Question Number 73327 by mathmax by abdo last updated on 10/Nov/19

$l e t w (x) = \int_{0}^{\infty} \frac{l n t}{{(x^{2} + t^{2})}^{2}} d t$ $1) e x p l i c i t w (x)$ $2) c a l c u l a t e U_{n} = \int_{0}^{\infty} \frac{l n t}{{(n^{2} + t^{2})}^{2}} d t$ $f i n d {l i m}_{n \to + \infty} n^{4} U_{n} a n d d e t e r m i n e n a t u r e o f t b e s e r i e Σ U_{n}$
Commented by mathmax by abdo last updated on 11/Nov/19
$1) we have w(x)=∫_0 ^∞ ((ln(t))/((x^2 +t^2 )^2 ))dt let f(x)=∫_0 ^∞ ((ln(t))/(x^2 +t^2 ))dt we have f^′ (x)=−∫_0 ^∞ ((2xln(t))/((x^2 +t^2 )))dt =−2x w(x) ⇒w(x)=−(1/(2x))f^′ (x) we have f(x)=_(t=xu) ∫_0 ^∞ ((ln(xu))/(x^2 (1+u^(2)) )) xdu (we suppose x>0 because f is even) ⇒f(x)=(1/x) ∫_0 ^∞ ((ln(x)+ln(u))/(1+u^2 ))du =((lnx)/x) ∫_0 ^∞ (du/(1+u^2 )) +(1/x)∫_0 ^∞ ((lnu)/(1+u^2 ))du ( ∫_0 ^∞ ((lnu)/(1+u^2 ))du =0) =((πlnx)/(2x)) ⇒f^′ (x)=(π/2){ ((1−lnx)/x^2 )} =((π(1−lnx))/(2x^2 )) ⇒ w(x)=−(1/(2x))×((π(1−lnx))/(2x^2 )) =((π(lnx−1))/(4x^3 )) 2) U_n =∫_0 ^∞ ((lnt)/((n^2 +t^2 )^2 ))dt ⇒ U_n =((π(ln(n)−1))/(4n^3 )) ⇒ lim_(n→+∞) n^4 U_n =lim_(n→+∞) ((nπ(ln(n)−1))/4) =+∞ U_n =(π/4)((ln(n))/n^3 ) −(π/4) (1/n^3 ) Σ((ln(n))/n^3 ) conv. and Σ (1/n^3 ) ⇒ Σ U_n converges.$
$1) w e h a v e w (x) = \int_{0}^{\infty} \frac{l n (t)}{{(x^{2} + t^{2})}^{2}} d t l e t f (x) = \int_{0}^{\infty} \frac{l n (t)}{x^{2} + t^{2}} d t$ $w e h a v e f^{^{'}} (x) = - \int_{0}^{\infty} \frac{2 x l n (t)}{(x^{2} + t^{2})} d t = - 2 x w (x) \Rightarrow w (x) = - \frac{1}{2 x} f^{^{'}} (x)$ $w e h a v e f (x) =_{t = x u} \int_{0}^{\infty} \frac{l n (x u)}{x^{2} (1 + u^{2)}} x d u (w e s u p p o s e x > 0 b e c a u s e f$ $i s e v e n) \Rightarrow f (x) = \frac{1}{x} \int_{0}^{\infty} \frac{l n (x) + l n (u)}{1 + u^{2}} d u$ $= \frac{l n x}{x} \int_{0}^{\infty} \frac{d u}{1 + u^{2}} + \frac{1}{x} \int_{0}^{\infty} \frac{l n u}{1 + u^{2}} d u (\int_{0}^{\infty} \frac{l n u}{1 + u^{2}} d u = 0)$ $= \frac{π l n x}{2 x} \Rightarrow f^{^{'}} (x) = \frac{π}{2} {\frac{1 - l n x}{x^{2}}} = \frac{π (1 - l n x)}{2 x^{2}} \Rightarrow$ $w (x) = - \frac{1}{2 x} \times \frac{π (1 - l n x)}{2 x^{2}} = \frac{π (l n x - 1)}{4 x^{3}}$ $2) U_{n} = \int_{0}^{\infty} \frac{l n t}{{(n^{2} + t^{2})}^{2}} d t \Rightarrow U_{n} = \frac{π (l n (n) - 1)}{4 n^{3}} \Rightarrow$ ${l i m}_{n \to + \infty} n^{4} U_{n} = {l i m}_{n \to + \infty} \frac{n π (l n (n) - 1)}{4} = + \infty$ $U_{n} = \frac{π}{4} \frac{l n (n)}{n^{3}} - \frac{π}{4} \frac{1}{n^{3}}$ $Σ \frac{l n (n)}{n^{3}} c o n v . a n d Σ \frac{1}{n^{3}} \Rightarrow Σ U_{n} c o n v e r g e s .$
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