find f(λ) =∫_0 ^∞ ((arctan(λx))/(1+λx^2 ))dx with λ>0


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Question Number 50683 by maxmathsup by imad last updated on 18/Dec/18

$f i n d f (λ) = \int_{0}^{\infty} \frac{a r c t a n (λ x)}{1 + λ x^{2}} d x w i t h λ > 0$
Commented byAbdo msup. last updated on 21/Dec/18
$changement (√λ)x=t give f(λ)=(1/(√λ))∫_0 ^∞ ((arctan(λ(t/(√λ))))/(1+t^2 ))dt=(1/(√λ))∫_0 ^∞ ((arctan((√λ)t))/(1+t^2 ))dt and ∫_0 ^∞ ((arctan((√λ)t))/(1+t^2 ))dt =W((√λ)) with W(x)=∫_0 ^∞ ((arctan(xt))/(1+t^2 ))dt (x>0) let determine W(x) W^′ (x)=∫_0 ^∞ (t/((1+x^2 t^2 )(1+t^2 )))dt =_(xt =u) ∫_0 ^∞ (u/(x(1+u^2 )(1+(u^2 /x^2 )))) (du/x) =∫_0 ^∞ ((udu)/((u^2 +x^2 )(u^2 +1))) let devompose F(u)=(u/((u^2 +x^2 )(u^2 +1))) F(u)=((au+b)/(u^2 +x^2 )) +((cu +d)/(u^2 +1)) F(−u)=−F(u) ⇒((−au+b)/(u^2 +x^2 )) +((−cu +d)/(u^2 +1)) =((−au−b)/(u^2 +x^2 )) +((−cu−d)/(u^(2 ) +1)) ⇒b=d=0 ⇒ F(u)=((au)/(u^2 +x^2 )) +((cu)/(u^2 +1)) lim_(u→+∞) uF(u)=0=a+c ⇒c=−a ⇒ F(u)=((au)/(u^2 +x^2 )) −((au)/(u^2 +1)) F(1)=(1/(2(x^2 +1))) =(a/(x^2 +1)) −(a/2) ⇒ (1/2) =a−((a(x^2 +1))/2) ⇒1=2a −(x^2 +1)a =(1−x^2 )a ⇒ a=(1/(1−x^2 )) (we suppose x^2 ≠1) ⇒ F(u) =(1/(1−x^2 )){ (u/(u^2 +x^2 )) −(u/(u^2 +1))} ⇒ ∫_0 ^∞ F(u)du =(1/(1−x^2 ))( ∫_0 ^∞ ((udu)/(u^2 +x^2 ))−∫_0 ^∞ (u/(1+u^2 ))du) but ∫_0 ^∞ ((udu)/(1+u^2 ))du = (1/2)ln(1+u^2 ) also ∫_0 ^∞ ((udu)/(u^2 +x^2 )) du =_(u=xα) ∫_0 ^∞ ((xα )/(x^2 α^2 +x^2 )) xdα =∫_0 ^∞ ((αdα)/(α^2 +1)) ⇒∫_0 ^∞ F(u)du =0 ⇒W^′ (x)=0 ⇒ W(x)=c =W(1) = ∫_0 ^∞ ((arctant)/(1+t^2 )) dt changement t=(1/u) give W(1)=−∫_0 ^∞ (((π/2)−arctanu)/(1+(1/u^2 ))) ((−du)/u^2 ) =∫_0 ^∞ (((π/2)−arctanu)/(u^2 +1)) du =(π/2)∫_0 ^∞ (du/(1+u^2 )) −∫_0 ^∞ ((arctanu)/(1+u^2 )) du =(π^2 /4) −W(1) ⇒ 2W(1)=(π^2 /4) ⇒W(1)=(π^2 /8) ⇒f(λ)=(π^2 /(8(√λ))) .$
$c h a n g e m e n t \sqrt{λ} x = t g i v e$ $f (λ) = \frac{1}{\sqrt{λ}} \int_{0}^{\infty} \frac{a r c t a n (λ \frac{t}{\sqrt{λ}})}{1 + t^{2}} d t = \frac{1}{\sqrt{λ}} \int_{0}^{\infty} \frac{a r c t a n (\sqrt{λ} t)}{1 + t^{2}} d t$ $a n d \int_{0}^{\infty} \frac{a r c t a n (\sqrt{λ} t)}{1 + t^{2}} d t = W (\sqrt{λ}) w i t h$ $W (x) = \int_{0}^{\infty} \frac{a r c t a n (x t)}{1 + t^{2}} d t (x > 0) l e t d e t e r m i n e W (x)$ $W^{^{'}} (x) = \int_{0}^{\infty} \frac{t}{(1 + x^{2} t^{2}) (1 + t^{2})} d t$ $=_{x t = u} \int_{0}^{\infty} \frac{u}{x (1 + u^{2}) (1 + \frac{u^{2}}{x^{2}})} \frac{d u}{x}$ $= \int_{0}^{\infty} \frac{u d u}{(u^{2} + x^{2}) (u^{2} + 1)} l e t d e v o m p o s e$ $F (u) = \frac{u}{(u^{2} + x^{2}) (u^{2} + 1)}$ $F (u) = \frac{a u + b}{u^{2} + x^{2}} + \frac{c u + d}{u^{2} + 1}$ $F (- u) = - F (u) \Rightarrow \frac{- a u + b}{u^{2} + x^{2}} + \frac{- c u + d}{u^{2} + 1}$ $= \frac{- a u - b}{u^{2} + x^{2}} + \frac{- c u - d}{u^{2} + 1} \Rightarrow b = d = 0 \Rightarrow$ $F (u) = \frac{a u}{u^{2} + x^{2}} + \frac{c u}{u^{2} + 1}$ ${l i m}_{u \to + \infty} u F (u) = 0 = a + c \Rightarrow c = - a \Rightarrow$ $F (u) = \frac{a u}{u^{2} + x^{2}} - \frac{a u}{u^{2} + 1}$ $F (1) = \frac{1}{2 (x^{2} + 1)} = \frac{a}{x^{2} + 1} - \frac{a}{2} \Rightarrow$ $\frac{1}{2} = a - \frac{a (x^{2} + 1)}{2} \Rightarrow 1 = 2 a - (x^{2} + 1) a = (1 - x^{2}) a \Rightarrow$ $a = \frac{1}{1 - x^{2}} (w e s u p p o s e x^{2} \neq 1) \Rightarrow$ $F (u) = \frac{1}{1 - x^{2}} {\frac{u}{u^{2} + x^{2}} - \frac{u}{u^{2} + 1}} \Rightarrow$ $\int_{0}^{\infty} F (u) d u = \frac{1}{1 - x^{2}} (\int_{0}^{\infty} \frac{u d u}{u^{2} + x^{2}} - \int_{0}^{\infty} \frac{u}{1 + u^{2}} d u) b u t$ $\int_{0}^{\infty} \frac{u d u}{1 + u^{2}} d u = \frac{1}{2} l n (1 + u^{2}) a l s o$ $\int_{0}^{\infty} \frac{u d u}{u^{2} + x^{2}} d u =_{u = x α} \int_{0}^{\infty} \frac{x α}{x^{2} α^{2} + x^{2}} x d α$ $= \int_{0}^{\infty} \frac{α d α}{α^{2} + 1} \Rightarrow \int_{0}^{\infty} F (u) d u = 0 \Rightarrow W^{^{'}} (x) = 0 \Rightarrow$ $W (x) = c = W (1) = \int_{0}^{\infty} \frac{a r c t a n t}{1 + t^{2}} d t c h a n g e m e n t$ $t = \frac{1}{u} g i v e W (1) = - \int_{0}^{\infty} \frac{\frac{π}{2} - a r c t a n u}{1 + \frac{1}{u^{2}}} \frac{- d u}{u^{2}}$ $= \int_{0}^{\infty} \frac{\frac{π}{2} - a r c t a n u}{u^{2} + 1} d u$ $= \frac{π}{2} \int_{0}^{\infty} \frac{d u}{1 + u^{2}} - \int_{0}^{\infty} \frac{a r c t a n u}{1 + u^{2}} d u = \frac{π^{2}}{4} - W (1) \Rightarrow$ $2 W (1) = \frac{π^{2}}{4} \Rightarrow W (1) = \frac{π^{2}}{8} \Rightarrow f (λ) = \frac{π^{2}}{8 \sqrt{λ}} .$
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