calculate F(x) = ∫_0 ^∞ (dt/(1+(1+x(1+t^2 ))^2 ))


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Question Number 39389 by maxmathsup by imad last updated on 05/Jul/18

$c a l c u l a t e F (x) = \int_{0}^{\infty} \frac{d t}{1 + {(1 + x (1 + t^{2}))}^{2}}$
Commented by prof Abdo imad last updated on 24/Jul/18
$changement t=tanθ give F(x) = ∫_0 ^(π/2) (((1+tan^2 θ))/(1+{1+x(1+tan^2 θ)}^2 ))dθ = ∫_0 ^(π/2) (1/(cos^2 θ{1+(x/(cos^2 θ))}^2 ))dθ = ∫_0 ^(π/2) ((cos^2 θ)/({x +cos^2 θ}^2 ))dθ =∫_0 ^(π/2) ((x+cos^2 θ−x)/((x+cos^2 θ)^2 ))dθ = ∫_0 ^(π/2) (dθ/(x +cos^2 θ)) −x ∫_0 ^(π/2) (dθ/((x+cos^2 θ)^2 )) but ∫_0 ^(π/2) (dθ/(x+cos^2 θ)) = ∫_0 ^(π/2) (dθ/(x+((1+cos(2θ))/2))) = ∫_0 ^(π/2) ((2dθ)/(2x+1+cos(2θ))) =_(2θ=t) ∫_0 ^π (dt/(2x+1+cos(t))) =_(u=tan((t/2))) ∫_0 ^∞ (1/(2x+1 +((1−u^2 )/(1+u^2 )))) ((2du)/(1+u^2 )) = ∫_0 ^∞ ((2du)/((2x+1)u^2 +1−u^2 )) = ∫_0 ^∞ ((2du)/(2xu^2 +1)) if x>0 ∫_0 ^∞ ((2du)/(1+2xu^2 )) =_((√(2x))u=α) ∫_0 ^∞ (2/(1+α^2 )) (dα/(√(2x))) =((√2)/(√x)) .(π/2) if x<0 ∫_0 ^∞ ((2du)/(1+2xu^2 )) = ∫_0 ^∞ ((2du)/(1−((√(−2x))u)^2 )) =∫_0 ^∞ ((2du)/((1−(√(2x))u)(1+(√(2x))u))) = ∫_0 ^∞ { (1/(1−(√(2x))u)) +(1/(1+(√(2x))u))}du =[ln∣((1+(√(2x))u)/(1−(√(2x))u))∣]_0 ^(+∞) =0 so ∫_0 ^(π/2) (dθ/(x+cos^2 θ)) =((π(√2))/(√(2x))) if x>0 ∫_0 ^(π/2) (dθ/(x+cos^2 θ)) =0 if x<0$
$c h a n g e m e n t t = t a n θ g i v e$ $F (x) = \int_{0}^{\frac{π}{2}} \frac{(1 + {t a n}^{2} θ)}{1 + {1 + x (1 + {t a n}^{2} θ)}^{2}} d θ$ $= \int_{0}^{\frac{π}{2}} \frac{1}{{c o s}^{2} θ {1 + \frac{x}{{c o s}^{2} θ}}^{2}} d θ$ $= \int_{0}^{\frac{π}{2}} \frac{{c o s}^{2} θ}{{x + {c o s}^{2} θ}^{2}} d θ$ $= \int_{0}^{\frac{π}{2}} \frac{x + {c o s}^{2} θ - x}{{(x + {c o s}^{2} θ)}^{2}} d θ$ $= \int_{0}^{\frac{π}{2}} \frac{d θ}{x + {c o s}^{2} θ} - x \int_{0}^{\frac{π}{2}} \frac{d θ}{{(x + {c o s}^{2} θ)}^{2}} b u t$ $\int_{0}^{\frac{π}{2}} \frac{d θ}{x + {c o s}^{2} θ} = \int_{0}^{\frac{π}{2}} \frac{d θ}{x + \frac{1 + c o s (2 θ)}{2}}$ $= \int_{0}^{\frac{π}{2}} \frac{2 d θ}{2 x + 1 + c o s (2 θ)} =_{2 θ = t} \int_{0}^{π} \frac{d t}{2 x + 1 + c o s (t)}$ $=_{u = t a n (\frac{t}{2})} \int_{0}^{\infty} \frac{1}{2 x + 1 + \frac{1 - u^{2}}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}}$ $= \int_{0}^{\infty} \frac{2 d u}{(2 x + 1) u^{2} + 1 - u^{2}}$ $= \int_{0}^{\infty} \frac{2 d u}{2 {x u}^{2} + 1}$ $i f x > 0 \int_{0}^{\infty} \frac{2 d u}{1 + 2 {x u}^{2}} =_{\sqrt{2 x} u = α} \int_{0}^{\infty} \frac{2}{1 + α^{2}} \frac{d α}{\sqrt{2 x}}$ $= \frac{\sqrt{2}}{\sqrt{x}} . \frac{π}{2}$ $i f x < 0 \int_{0}^{\infty} \frac{2 d u}{1 + 2 {x u}^{2}} = \int_{0}^{\infty} \frac{2 d u}{1 - {(\sqrt{- 2 x} u)}^{2}}$ $= \int_{0}^{\infty} \frac{2 d u}{(1 - \sqrt{2 x} u) (1 + \sqrt{2 x} u)}$ $= \int_{0}^{\infty} {\frac{1}{1 - \sqrt{2 x} u} + \frac{1}{1 + \sqrt{2 x} u}} d u$ $= {[l n ∣ \frac{1 + \sqrt{2 x} u}{1 - \sqrt{2 x} u} ∣]}_{0}^{+ \infty} = 0 s o$ $\int_{0}^{\frac{π}{2}} \frac{d θ}{x + {c o s}^{2} θ} = \frac{π \sqrt{2}}{\sqrt{2 x}} i f x > 0$ $\int_{0}^{\frac{π}{2}} \frac{d θ}{x + {c o s}^{2} θ} = 0 i f x < 0$
Commented by prof Abdo imad last updated on 24/Jul/18
$let w(x)= ∫_0 ^(π/2) (dθ/(x+cos^2 θ)) ⇒ w^′ (x) =−∫_0 ^(π/2) (dθ/((x +cos^2 θ)^2 )) ⇒ ∫_0 ^(π/2) (dθ/((x+cos^2 θ)^2 )) =−w^′ (x) so if x>0 w(x)=((π(√2))/(2(√x))) ⇒w^′ (x)=(π/(√2)) (−(1/(2x(√x)))) =−(π/(2(√2)x(√x))) ⇒ ∫_0 ^(π/2) (dθ/((x+cos^2 θ)^2 )) = (π/(2(√2)x(√x))) and F(x)=((π(√2))/(2(√x))) −x (π/(2(√2)x(√x))) =((π(√2))/(2(√x))) −(π/(2(√2)(√x))) =(π/(√x)){((√2)/2) −(1/(2(√2)))}=(π/(√x)){(2/(4(√2)))} = (π/(2(√(2x)))) if x<0 w(x)=0 ⇒ ∫_0 ^(π/2) (dθ/((x+cos^2 θ)^2 )) =0 ⇒ F(x) = 0 because ∫_0 ^(π/2) (dθ/(x+cos^2 θ)) =0$
$l e t w (x) = \int_{0}^{\frac{π}{2}} \frac{d θ}{x + {c o s}^{2} θ} \Rightarrow$ $w^{^{'}} (x) = - \int_{0}^{\frac{π}{2}} \frac{d θ}{{(x + {c o s}^{2} θ)}^{2}} \Rightarrow$ $\int_{0}^{\frac{π}{2}} \frac{d θ}{{(x + {c o s}^{2} θ)}^{2}} = - w^{^{'}} (x) s o$ $i f x > 0 w (x) = \frac{π \sqrt{2}}{2 \sqrt{x}} \Rightarrow w^{^{'}} (x) = \frac{π}{\sqrt{2}} (- \frac{1}{2 x \sqrt{x}})$ $= - \frac{π}{2 \sqrt{2} x \sqrt{x}} \Rightarrow \int_{0}^{\frac{π}{2}} \frac{d θ}{{(x + {c o s}^{2} θ)}^{2}} = \frac{π}{2 \sqrt{2} x \sqrt{x}}$ $a n d F (x) = \frac{π \sqrt{2}}{2 \sqrt{x}} - x \frac{π}{2 \sqrt{2} x \sqrt{x}}$ $= \frac{π \sqrt{2}}{2 \sqrt{x}} - \frac{π}{2 \sqrt{2} \sqrt{x}} = \frac{π}{\sqrt{x}} {\frac{\sqrt{2}}{2} - \frac{1}{2 \sqrt{2}}} = \frac{π}{\sqrt{x}} {\frac{2}{4 \sqrt{2}}}$ $= \frac{π}{2 \sqrt{2 x}}$ $i f x < 0 w (x) = 0 \Rightarrow \int_{0}^{\frac{π}{2}} \frac{d θ}{{(x + {c o s}^{2} θ)}^{2}} = 0 \Rightarrow$ $F (x) = 0 b e c a u s e \int_{0}^{\frac{π}{2}} \frac{d θ}{x + {c o s}^{2} θ} = 0$
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