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Question Number 38373 by tanmay.chaudhury50@gmail.com last updated on 24/Jun/18

Commented by tanmay.chaudhury50@gmail.com last updated on 25/Jun/18
$I(a)=∫_0 ^∞ ((tan^(−1) (ax))/(x(1+x^2 )))dx ((dI(a))/da)=∫_0 ^∞ (∂/∂a)((tan^(−1) ax)/(x(1+x^2 )))dx =∫_0 ^∞ ((xdx)/(x(1+x^2 )(1+a^2 x^2 ))) =∫_0 ^∞ (a^2 /((a^2 +a^2 x^2 )(1+a^2 x^2 )))dx =(a^2 /(a^2 −1))∫_0 ^∞ (((a^2 +a^2 x^2 )−(1+a^2 x^2 ))/((a^2 +a^2 x^2 )(1+a^2 x^2 )))dx =(a^2 /(a^2 −1)){∫_0 ^∞ (dx/(1+a^2 x^2 ))−∫_0 ^∞ (dx/(a^2 +a^2 x^2 ))} =(a^2 /(a^2 −1)){(1/a^2 )∫_0 ^∞ (dx/(((1/a^2 )+x^2 )))−(1/a^2 )∫_0 ^∞ (dx/(1+x^2 ))} =(1/(a^2 −1))∣{atan^(−1) (ax)−tan^(−1) x}∣_0 ^∞ =(1/(a^2 −1)){atan^(−1) (∞)−tan^(−1) (∞)} =(1/(a^2 −1)){a(Π/2)−(Π/2)} (dI/da)=(Π/2)×((a−1)/((a+1)(a−1))) dI=(Π/2)(da/((a+1))) I=(Π/2)ln(a+1)+c when a=0 I=0 so c=0 I=(Π/2)ln(a+1)$
$I (a) = \int_{0}^{\infty} \frac{{t a n}^{- 1} (a x)}{x (1 + x^{2})} d x$ $\frac{d I (a)}{d a} = \int_{0}^{\infty} \frac{\partial}{\partial a} \frac{{t a n}^{- 1} a x}{x (1 + x^{2})} d x$ $= \int_{0}^{\infty} \frac{x d x}{x (1 + x^{2}) (1 + a^{2} x^{2})}$ $= \int_{0}^{\infty} \frac{a^{2}}{(a^{2} + a^{2} x^{2}) (1 + a^{2} x^{2})} d x$ $= \frac{a^{2}}{a^{2} - 1} \int_{0}^{\infty} \frac{(a^{2} + a^{2} x^{2}) - (1 + a^{2} x^{2})}{(a^{2} + a^{2} x^{2}) (1 + a^{2} x^{2})} d x$ $= \frac{a^{2}}{a^{2} - 1} {\int_{0}^{\infty} \frac{d x}{1 + a^{2} x^{2}} - \int_{0}^{\infty} \frac{d x}{a^{2} + a^{2} x^{2}}}$ $= \frac{a^{2}}{a^{2} - 1} {\frac{1}{a^{2}} \int_{0}^{\infty} \frac{d x}{(\frac{1}{a^{2}} + x^{2})} - \frac{1}{a^{2}} \int_{0}^{\infty} \frac{d x}{1 + x^{2}}}$ $= \frac{1}{a^{2} - 1} ∣ {{a t a n}^{- 1} (a x) - {t a n}^{- 1} x} ∣_{0}^{\infty}$ $= \frac{1}{a^{2} - 1} {{a t a n}^{- 1} (\infty) - {t a n}^{- 1} (\infty)}$ $= \frac{1}{a^{2} - 1} {a \frac{Π}{2} - \frac{Π}{2}}$ $\frac{d I}{d a} = \frac{Π}{2} \times \frac{a - 1}{(a + 1) (a - 1)}$ $d I = \frac{Π}{2} \frac{d a}{(a + 1)}$ $I = \frac{Π}{2} l n (a + 1) + c$ $w h e n a = 0 I = 0 s o c = 0$ $I = \frac{Π}{2} l n (a + 1)$
Commented by math khazana by abdo last updated on 25/Jun/18

$y o u a n s w e r i s n o t c o r r e c t s i r T a n m a y . . . t h r e i s$ $s o m e t h i n g w r o n g . . .$
Commented by tanmay.chaudhury50@gmail.com last updated on 25/Jun/18

$n o i h a v e c h e c k e d i t . . . a n d i t i s r i g h t . . .$
Commented by tanmay.chaudhury50@gmail.com last updated on 25/Jun/18

Commented by tanmay.chaudhury50@gmail.com last updated on 25/Jun/18

Commented by tanmay.chaudhury50@gmail.com last updated on 25/Jun/18

Commented by prof Abdo imad last updated on 25/Jun/18
$5) let f(a)= ∫_0 ^∞ ((arctan(ax))/(x(1+x^2 ))) dx we have f^′ (a) = ∫_0 ^∞ (x/(x(1+x^2 )(1+a^2 x^2 )))dx =∫_0 ^∞ (dx/((1+x^2 )(1+a^2 x^2 ))) let decompose F(x)= (1/((1+x^2 )(1+a^2 x^2 ))) F(x)= ((αx+b)/(1+x^2 )) +((cx+d)/(1+a^2 x^2 )) F(−x)=F(x) ⇒((−αx +b)/(1+x^2 )) +((−cx +d)/(1+a^2 x^2 )) =F(x)⇒ α=c=0 ⇒F(x)= (b/(1+x^2 )) +(d/(1+a^2 x^2 )) lim_(x→+∞) x^2 F(x)=b +(d/a^2 ) =0 ⇒d +a^2 b=0 ⇒ d=−a^2 b ⇒F(x)=(b/(1+x^2 )) −a^2 (b/(1+a^2 x^2 )) F(0)=1 =b −a^2 b =(1−a^2 )b ⇒b= (1/(1−a^2 )) (we suppose that a^2 ≠1) ⇒ F(x)= (1/(1−a^2 )){ (1/(1+x^2 )) −(a^2 /(1+a^2 x^2 ))} ⇒ f^′ (a) = (1/(1−a^2 )) ∫_0 ^∞ (dx/(1+x^2 )) −(a^2 /(1−a^2 )) ∫_0 ^∞ (dx/(1+a^2 x^2 )) changement ax=t give( a>0) ∫_0 ^∞ (dx/(1+a^2 x^2 )) = ∫_0 ^∞ (1/(1+t^2 )) (dt/a) = (π/(2a)) ⇒ f^′ (a) = (π/(2(1−a^2 ))) −(a^2 /(1−a^2 )) (π/(2a)) = (π/(2(1−a^2 ))){ 1 −a} = (π/(2(1+a))) ⇒ f(a) =(π/2) ∫ (da/(1+a)) +c= (π/2)ln(1+a) +c c=f(0)=0 ⇒ f(a)=(π/2)ln(1+a) if a<0 and a≠−1 changegent ax=−t give ∫_0 ^∞ (1/(1+t^2 )) ((−dt)/a) =−(π/(2a)) ⇒ f^′ (a)= (π/(2(1−a^2 ))) −(a^2 /(1−a^2 ))(−(π/(2a))) = (π/(2(1−a^2 ))){ 1+a} = (π/(2(1−a))) ⇒ f(a) = (π/2) ∫ (da/(1−a)) =−(π/2)ln∣1−a∣ +c but c=f(0)=0 ⇒f(a) =−(π/2)ln(1−a).$
$5) l e t f (a) = \int_{0}^{\infty} \frac{a r c t a n (a x)}{x (1 + x^{2})} d x w e h a v e$ $f^{^{'}} (a) = \int_{0}^{\infty} \frac{x}{x (1 + x^{2}) (1 + a^{2} x^{2})} d x$ $= \int_{0}^{\infty} \frac{d x}{(1 + x^{2}) (1 + a^{2} x^{2})} l e t d e c o m p o s e$ $F (x) = \frac{1}{(1 + x^{2}) (1 + a^{2} x^{2})}$ $F (x) = \frac{α x + b}{1 + x^{2}} + \frac{c x + d}{1 + a^{2} x^{2}}$ $F (- x) = F (x) \Rightarrow \frac{- α x + b}{1 + x^{2}} + \frac{- c x + d}{1 + a^{2} x^{2}} = F (x) \Rightarrow$ $α = c = 0 \Rightarrow F (x) = \frac{b}{1 + x^{2}} + \frac{d}{1 + a^{2} x^{2}}$ ${l i m}_{x \to + \infty} x^{2} F (x) = b + \frac{d}{a^{2}} = 0 \Rightarrow d + a^{2} b = 0 \Rightarrow$ $d = - a^{2} b \Rightarrow F (x) = \frac{b}{1 + x^{2}} - a^{2} \frac{b}{1 + a^{2} x^{2}}$ $F (0) = 1 = b - a^{2} b = (1 - a^{2}) b \Rightarrow b = \frac{1}{1 - a^{2}}$ $(w e s u p p o s e t h a t a^{2} \neq 1) \Rightarrow$ $F (x) = \frac{1}{1 - a^{2}} {\frac{1}{1 + x^{2}} - \frac{a^{2}}{1 + a^{2} x^{2}}} \Rightarrow$ $f^{^{'}} (a) = \frac{1}{1 - a^{2}} \int_{0}^{\infty} \frac{d x}{1 + x^{2}} - \frac{a^{2}}{1 - a^{2}} \int_{0}^{\infty} \frac{d x}{1 + a^{2} x^{2}}$ $c h a n g e m e n t a x = t g i v e (a > 0)$ $\int_{0}^{\infty} \frac{d x}{1 + a^{2} x^{2}} = \int_{0}^{\infty} \frac{1}{1 + t^{2}} \frac{d t}{a} = \frac{π}{2 a} \Rightarrow$ $f^{^{'}} (a) = \frac{π}{2 (1 - a^{2})} - \frac{a^{2}}{1 - a^{2}} \frac{π}{2 a}$ $= \frac{π}{2 (1 - a^{2})} {1 - a} = \frac{π}{2 (1 + a)} \Rightarrow$ $f (a) = \frac{π}{2} \int \frac{d a}{1 + a} + c = \frac{π}{2} l n (1 + a) + c$ $c = f (0) = 0 \Rightarrow f (a) = \frac{π}{2} l n (1 + a)$ $i f a < 0 a n d a \neq - 1 c h a n g e g e n t a x = - t g i v e$ $\int_{0}^{\infty} \frac{1}{1 + t^{2}} \frac{- d t}{a} = - \frac{π}{2 a} \Rightarrow$ $f^{^{'}} (a) = \frac{π}{2 (1 - a^{2})} - \frac{a^{2}}{1 - a^{2}} (- \frac{π}{2 a})$ $= \frac{π}{2 (1 - a^{2})} {1 + a} = \frac{π}{2 (1 - a)} \Rightarrow$ $f (a) = \frac{π}{2} \int \frac{d a}{1 - a} = - \frac{π}{2} l n ∣ 1 - a ∣ + c b u t c = f (0) = 0$ $\Rightarrow f (a) = - \frac{π}{2} l n (1 - a) .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 25/Jun/18

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