find f(x)=∫_0 ^∞ ((arctan(xt))/(1+t^2 ))dt with x real .


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Question Number 53284 by maxmathsup by imad last updated on 20/Jan/19

$f i n d f (x) = \int_{0}^{\infty} \frac{a r c t a n (x t)}{1 + t^{2}} d t w i t h x r e a l .$
Commented by prof Abdo imad last updated on 21/Jan/19
$we have f^′ (x)=∫_0 ^∞ (t/((1+x^2 t^2 )(1+t^2 )))dt =_(xt =u) ∫_0 ^∞ (u/(x(1+u^2 )(1+(u^2 /x^2 )))) (du/x) =∫_0 ^∞ (u/((u^2 +1))(u^2 +x^2 )))du let decompose F(u)=(u/((u^2 +1)(u^2 +x^2 ))) ⇒ F(u)=((au+b)/(u^2 +1)) +((cu +d)/(u^2 +x^2 )) F(−u)=−F(u) ⇒((−au +b)/(u^2 +1)) +((−cu +d)/(u^2 +x^2 )) =((−au−b)/(u^2 +1)) +((−cu−d)/(u^2 +x^2 )) ⇒b=d=0 ⇒ F(u) =((au)/(u^2 +1)) +((cu)/(u^2 +x^2 )) lim_(u→+∞) u F(u)=0 =a+c ⇒c=−a ⇒ F(u) =((au)/(u^2 +1)) −((au)/(u^2 +x^2 )) F(1)= (1/(2(1+x^2 ))) =(a/2)−(a/(1+x^2 ))=((a +ax^2 −2a)/(2(1+x^2 ))) ⇒ (x^2 −1)a =1 ⇒a =(1/(x^2 −1)) (we suppose x≠+^− 1) ⇒ F(u) =(1/(x^2 −1)){ (u/(u^2 +1)) −(u/(u^2 +x^2 ))} ⇒ f^′ (x) =(1/(x^2 −1))∫_0 ^∞ ((u/(u^2 +1)) −(u/(u^2 +x^2 )))du =(1/(2(x^2 −1)))[ln∣((u^(2 ) +1)/(u^2 +x^2 ))∣]_0 ^(+∞) =(1/(2(x^2 −1)))(2ln∣x∣) =((ln∣x∣)/(x^2 −1)) let suppose x>1 ⇒ f^′ (x) =((ln(x))/(x^2 −1)) ⇒ f(x)=∫_1 ^x ((ln(t))/(t^2 −1)) dt +λ λ =f(1) =∫_0 ^∞ ((arctan(t))/(1+t^2 ))dt by parts λ =[arctan^2 (t)]_0 ^∞ −∫_0 ^∞ ((arctan(t))/(1+t^2 ))dt =(π^2 /4) −λ ⇒2λ =(π^2 /4) ⇒λ=(π^2 /8) ⇒ f(x)=∫_1 ^x ((ln(t))/(t^2 −1))dt +(π^2 /8)$
$w e h a v e f^{^{'}} (x) = \int_{0}^{\infty} \frac{t}{(1 + x^{2} t^{2}) (1 + t^{2})} d t$ $=_{x t = u} \int_{0}^{\infty} \frac{u}{x (1 + u^{2}) (1 + \frac{u^{2}}{x^{2}})} \frac{d u}{x}$ $= \int_{0}^{\infty} \frac{u}{(u^{2} + 1)) (u^{2} + x^{2})} d u l e t d e c o m p o s e$ $F (u) = \frac{u}{(u^{2} + 1) (u^{2} + x^{2})} \Rightarrow$ $F (u) = \frac{a u + b}{u^{2} + 1} + \frac{c u + d}{u^{2} + x^{2}}$ $F (- u) = - F (u) \Rightarrow \frac{- a u + b}{u^{2} + 1} + \frac{- c u + d}{u^{2} + x^{2}}$ $= \frac{- a u - b}{u^{2} + 1} + \frac{- c u - d}{u^{2} + x^{2}} \Rightarrow b = d = 0 \Rightarrow$ $F (u) = \frac{a u}{u^{2} + 1} + \frac{c u}{u^{2} + x^{2}}$ ${l i m}_{u \to + \infty} u F (u) = 0 = a + c \Rightarrow c = - a \Rightarrow$ $F (u) = \frac{a u}{u^{2} + 1} - \frac{a u}{u^{2} + x^{2}}$ $F (1) = \frac{1}{2 (1 + x^{2})} = \frac{a}{2} - \frac{a}{1 + x^{2}} = \frac{a + {a x}^{2} - 2 a}{2 (1 + x^{2})} \Rightarrow$ $(x^{2} - 1) a = 1 \Rightarrow a = \frac{1}{x^{2} - 1} (w e s u p p o s e x \neq \bar{+} 1) \Rightarrow$ $F (u) = \frac{1}{x^{2} - 1} {\frac{u}{u^{2} + 1} - \frac{u}{u^{2} + x^{2}}} \Rightarrow$ $f^{^{'}} (x) = \frac{1}{x^{2} - 1} \int_{0}^{\infty} (\frac{u}{u^{2} + 1} - \frac{u}{u^{2} + x^{2}}) d u$ $= \frac{1}{2 (x^{2} - 1)} {[l n ∣ \frac{u^{2} + 1}{u^{2} + x^{2}} ∣]}_{0}^{+ \infty} = \frac{1}{2 (x^{2} - 1)} (2 l n ∣ x ∣)$ $= \frac{l n ∣ x ∣}{x^{2} - 1} l e t s u p p o s e x > 1 \Rightarrow$ $f^{^{'}} (x) = \frac{l n (x)}{x^{2} - 1} \Rightarrow f (x) = \int_{1}^{x} \frac{l n (t)}{t^{2} - 1} d t + λ$ $λ = f (1) = \int_{0}^{\infty} \frac{a r c t a n (t)}{1 + t^{2}} d t b y p a r t s$ $λ = {[{a r c t a n}^{2} (t)]}_{0}^{\infty} - \int_{0}^{\infty} \frac{a r c t a n (t)}{1 + t^{2}} d t$ $= \frac{π^{2}}{4} - λ \Rightarrow 2 λ = \frac{π^{2}}{4} \Rightarrow λ = \frac{π^{2}}{8} \Rightarrow$ $f (x) = \int_{1}^{x} \frac{l n (t)}{t^{2} - 1} d t + \frac{π^{2}}{8}$
Commented by prof Abdo imad last updated on 21/Jan/19
$let detetmine ∫_1 ^x ((ln(t))/(t^2 −1))dt chsng.t =(1/u) give ∫_1 ^x ((ln(t))/(t^2 −1))dt =∫_1 ^(1/x) ((−ln(u))/(((1/u^2 )−1))) (−(du/u^2 )) =∫_1 ^(1/x) ((ln(u))/(u^2 −1)) du =∫_(1/x) ^1 ((ln(u))/(1−u^2 ))du =(1/2)∫_(1/x) ^1 ln(u){ (1/(1−u)) +(1/(1+u))}du =(1/2) ∫_(1/x) ^1 ((ln(u))/(1−u))du +(1/2) ∫_(1/x) ^1 ((ln(u))/(1+u))du ∫_(1/x) ^1 ((ln(u))/(1−u))du =∫_(1/x) ^1 ln(u){Σ_(n=0) ^∞ u^n )du =Σ_(n=0) ^∞ ∫_(1/x) ^1 u^n ln(u)du =Σ_(n=0) ^∞ A_n by parts A_n =[(1/(n+1))u^(n+1) ln(u)]_(1/x) ^1 −∫_(1/x) ^1 (u^n /(n+1))du =((ln(x))/((n+1)x^(n+1) )) −(1/((n+1)))[(1/(n+1)) u^(n+1) ]_(1/x) ^1 =((ln(x))/((n+1)x^(n+1) )) −(1/((n+1)^2 )){1−(1/x^(n+1) )} ⇒ ∫_(1/x) ^1 ((ln(u))/(1−u)) du =ln(x)Σ_(n=0) ^∞ (1/((n+1)x^(n+1) )) −Σ_(n=0) ^∞ (1/((n+1)^2 )) +Σ_(n=0) ^∞ (1/((n+1)^2 x^(n+1) )) Σ_(n=0) ^∞ (1/((n+1)^2 )) =ξ(2) =(π^2 /6) let find Σ_(n=0) ^∞ (1/((n+1)x^(n+1) )) be continued...$
$l e t d e t e t m i n e \int_{1}^{x} \frac{l n (t)}{t^{2} - 1} d t c h s n g . t = \frac{1}{u} g i v e$ $\int_{1}^{x} \frac{l n (t)}{t^{2} - 1} d t = \int_{1}^{\frac{1}{x}} \frac{- l n (u)}{(\frac{1}{u^{2}} - 1)} (- \frac{d u}{u^{2}})$ $= \int_{1}^{\frac{1}{x}} \frac{l n (u)}{u^{2} - 1} d u = \int_{\frac{1}{x}}^{1} \frac{l n (u)}{1 - u^{2}} d u$ $= \frac{1}{2} \int_{\frac{1}{x}}^{1} l n (u) {\frac{1}{1 - u} + \frac{1}{1 + u}} d u$ $= \frac{1}{2} \int_{\frac{1}{x}}^{1} \frac{l n (u)}{1 - u} d u + \frac{1}{2} \int_{\frac{1}{x}}^{1} \frac{l n (u)}{1 + u} d u$ $\int_{\frac{1}{x}}^{1} \frac{l n (u)}{1 - u} d u = \int_{\frac{1}{x}}^{1} l n (u) {\sum_{n = 0}^{\infty} u^{n}) d u$ $= \sum_{n = 0}^{\infty} \int_{\frac{1}{x}}^{1} u^{n} l n (u) d u = \sum_{n = 0}^{\infty} A_{n}$ $b y p a r t s A_{n} = {[\frac{1}{n + 1} u^{n + 1} l n (u)]}_{\frac{1}{x}}^{1} - \int_{\frac{1}{x}}^{1} \frac{u^{n}}{n + 1} d u$ $= \frac{l n (x)}{(n + 1) x^{n + 1}} - \frac{1}{(n + 1)} {[\frac{1}{n + 1} u^{n + 1}]}_{\frac{1}{x}}^{1}$ $= \frac{l n (x)}{(n + 1) x^{n + 1}} - \frac{1}{{(n + 1)}^{2}} {1 - \frac{1}{x^{n + 1}}} \Rightarrow$ $\int_{\frac{1}{x}}^{1} \frac{l n (u)}{1 - u} d u = l n (x) \sum_{n = 0}^{\infty} \frac{1}{(n + 1) x^{n + 1}}$ $- \sum_{n = 0}^{\infty} \frac{1}{{(n + 1)}^{2}} + \sum_{n = 0}^{\infty} \frac{1}{{(n + 1)}^{2} x^{n + 1}}$ $\sum_{n = 0}^{\infty} \frac{1}{{(n + 1)}^{2}} = ξ (2) = \frac{π^{2}}{6}$ $l e t f i n d \sum_{n = 0}^{\infty} \frac{1}{(n + 1) x^{n + 1}} b e c o n t i n u e d . . .$
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