let f(x)=∫_0 ^∞ arctan(xt^2 )dt . find a explicite form of f^′ (x)


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Question Number 41279 by math khazana by abdo last updated on 04/Aug/18

$l e t f (x) = \int_{0}^{\infty} a r c t a n ({x t}^{2}) d t .$ $f i n d a e x p l i c i t e f o r m o f f^{^{'}} (x)$
Commented by maxmathsup by imad last updated on 05/Aug/18
$we have f^′ (x)= ∫_0 ^∞ (t^2 /(1+x^2 t^4 )) dt =(1/2) ∫_(−∞) ^(+∞) (t^2 /(1+x^2 t^4 )) dt let consider the complex function ϕ(z) = (z^2 /(1+x^2 z^4 )) we have for x>0 ϕ(z) = (z^2 /((xz^2 2)^2 −i^2 )) =(z^2 /((xz^2 −i)(xz^2 +i))) = (z^2 /(x^2 (t^2 −(i/x))(z^2 +(i/x)))) =(z^2 /(x^2 (t−((√i)/(√x)))(t+((√i)/(√x)))(t−((√(−i))/((√x) )))(t+((√(−i))/(√x))))) = (z^2 /(x^2 ( t−(e^(i(π/4)) /(√x)))(t +(e^((iπ)/4) /(√x)))(t−(e^(−((iπ)/4)) /((√x) )))(t +(e^(−((iπ)/4)) /(√x))))) so the poles of ϕ are +^− (e^((iπ)/4) /(√x)) and +^− (e^(−((iπ)/4)) /(√x)) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,(e^((iπ)/4) /(√x))) +Res(ϕ,−(e^(−((iπ)/4)) /(√x)))} but Res(ϕ, (e^((iπ)/4) /(√x))) =(i/(x x^2 (2(e^((iπ)/4) /(√x)))(2(i/x)))) = ((√x)/(4x^2 e^((iπ)/4) )) =((√x)/(4x^2 )) e^(−((iπ)/4)) Res(ϕ,−(e^(−((iπ)/4)) /(√x))) =((−i)/(x(−2(e^(−((iπ)/4)) /(√x)))x^2 (−2(i/x)))) =−((√x)/(4x^2 )) e^((iπ)/4) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =−2iπ((√x)/(4x^2 )){e^((iπ)/4) −e^(−((iπ)/4)) } =−2iπ ((√x)/(4x^2 )) 2i sin((π/4)) =((√x)/x^2 ) ((√2)/2) and f^′ (x)=(1/2) ∫_(−∞) ^(+∞) ϕ(z)dz = ((√2)/(4x(√x))) with x>0 if x< we put x =−u ⇒f(x) = f(−u) =∫_0 ^∞ arctan(−ut^2 )dt =−f(u) ⇒f^′ (x)=−f^′ (u) =−((√2)/(4u(√u))) =−((√2)/(−4x(√(−x)))) =((√2)/(4x(√(−x)))) .$
$w e h a v e f^{^{'}} (x) = \int_{0}^{\infty} \frac{t^{2}}{1 + x^{2} t^{4}} d t = \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{t^{2}}{1 + x^{2} t^{4}} d t l e t c o n s i d e r t h e c o m p l e x$ $f u n c t i o n φ (z) = \frac{z^{2}}{1 + x^{2} z^{4}} w e h a v e f o r x > 0$ $φ (z) = \frac{z^{2}}{{({x z}^{2} 2)}^{2} - i^{2}} = \frac{z^{2}}{({x z}^{2} - i) ({x z}^{2} + i)} = \frac{z^{2}}{x^{2} (t^{2} - \frac{i}{x}) (z^{2} + \frac{i}{x})}$ $= \frac{z^{2}}{x^{2} (t - \frac{\sqrt{i}}{\sqrt{x}}) (t + \frac{\sqrt{i}}{\sqrt{x}}) (t - \frac{\sqrt{- i}}{\sqrt{x}}) (t + \frac{\sqrt{- i}}{\sqrt{x}})}$ $= \frac{z^{2}}{x^{2} (t - \frac{e^{i \frac{π}{4}}}{\sqrt{x}}) (t + \frac{e^{\frac{i π}{4}}}{\sqrt{x}}) (t - \frac{e^{- \frac{i π}{4}}}{\sqrt{x}}) (t + \frac{e^{- \frac{i π}{4}}}{\sqrt{x}})} s o t h e p o l e s o f φ a r e$ $\bar{+} \frac{e^{\frac{i π}{4}}}{\sqrt{x}} a n d \bar{+} \frac{e^{- \frac{i π}{4}}}{\sqrt{x}} r e s i d u s t h e o r e m g i v e$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, \frac{e^{\frac{i π}{4}}}{\sqrt{x}}) + R e s (φ, - \frac{e^{- \frac{i π}{4}}}{\sqrt{x}})} b u t$ $R e s (φ, \frac{e^{\frac{i π}{4}}}{\sqrt{x}}) = \frac{i}{x x^{2} (2 \frac{e^{\frac{i π}{4}}}{\sqrt{x}}) (2 \frac{i}{x})} = \frac{\sqrt{x}}{4 x^{2} e^{\frac{i π}{4}}} = \frac{\sqrt{x}}{4 x^{2}} e^{- \frac{i π}{4}}$ $R e s (φ, - \frac{e^{- \frac{i π}{4}}}{\sqrt{x}}) = \frac{- i}{x (- 2 \frac{e^{- \frac{i π}{4}}}{\sqrt{x}}) x^{2} (- 2 \frac{i}{x})} = - \frac{\sqrt{x}}{4 x^{2}} e^{\frac{i π}{4}} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) d z = - 2 i π \frac{\sqrt{x}}{4 x^{2}} {e^{\frac{i π}{4}} - e^{- \frac{i π}{4}}} = - 2 i π \frac{\sqrt{x}}{4 x^{2}} 2 i s i n (\frac{π}{4})$ $= \frac{\sqrt{x}}{x^{2}} \frac{\sqrt{2}}{2} a n d f^{^{'}} (x) = \frac{1}{2} \int_{- \infty}^{+ \infty} φ (z) d z = \frac{\sqrt{2}}{4 x \sqrt{x}} w i t h x > 0$ $i f x < w e p u t x = - u \Rightarrow f (x) = f (- u) = \int_{0}^{\infty} a r c t a n (- {u t}^{2}) d t$ $= - f (u) \Rightarrow f^{^{'}} (x) = - f^{^{'}} (u) = - \frac{\sqrt{2}}{4 u \sqrt{u}} = - \frac{\sqrt{2}}{- 4 x \sqrt{- x}} = \frac{\sqrt{2}}{4 x \sqrt{- x}} .$
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