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Question Number 41279 by math khazana by abdo last updated on 04/Aug/18
let f(x)=∫_0 ^∞ arctan(xt^2 )dt . find a explicite form of f^′ (x)
$${let}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\:{arctan}\left({xt}^{\mathrm{2}} \right){dt}\:. \\ $$$${find}\:\:{a}\:{explicite}\:{form}\:{of}\:{f}^{'} \left({x}\right) \\ $$
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)))) .
$${we}\:{have}\:{f}^{'} \left({x}\right)=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{t}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} {t}^{\mathrm{4}} }\:{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\infty} ^{+\infty} \:\:\frac{{t}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} {t}^{\mathrm{4}} }\:{dt}\:{let}\:{consider}\:{the}\:{complex} \\ $$$${function}\:\varphi\left({z}\right)\:=\:\frac{{z}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} {z}^{\mathrm{4}} }\:\:{we}\:{have}\:\:{for}\:{x}>\mathrm{0} \\ $$$$\varphi\left({z}\right)\:=\:\frac{{z}^{\mathrm{2}} }{\left({xz}^{\mathrm{2}} \mathrm{2}\right)^{\mathrm{2}} −{i}^{\mathrm{2}} }\:=\frac{{z}^{\mathrm{2}} }{\left({xz}^{\mathrm{2}} −{i}\right)\left({xz}^{\mathrm{2}} \:+{i}\right)}\:=\:\frac{{z}^{\mathrm{2}} }{{x}^{\mathrm{2}} \left({t}^{\mathrm{2}} \:−\frac{{i}}{{x}}\right)\left({z}^{\mathrm{2}} \:+\frac{{i}}{{x}}\right)} \\ $$$$=\frac{{z}^{\mathrm{2}} }{{x}^{\mathrm{2}} \left({t}−\frac{\sqrt{{i}}}{\:\sqrt{{x}}}\right)\left({t}+\frac{\sqrt{{i}}}{\:\sqrt{{x}}}\right)\left({t}−\frac{\sqrt{−{i}}}{\:\sqrt{{x}}\:}\right)\left({t}+\frac{\sqrt{−{i}}}{\:\sqrt{{x}}}\right)}\: \\ $$$$=\:\frac{{z}^{\mathrm{2}} }{{x}^{\mathrm{2}} \left(\:{t}−\frac{{e}^{{i}\frac{\pi}{\mathrm{4}}} }{\:\sqrt{{x}}}\right)\left({t}\:+\frac{{e}^{\frac{{i}\pi}{\mathrm{4}}} }{\:\sqrt{{x}}}\right)\left({t}−\frac{{e}^{−\frac{{i}\pi}{\mathrm{4}}} }{\:\sqrt{{x}}\:}\right)\left({t}\:+\frac{{e}^{−\frac{{i}\pi}{\mathrm{4}}} }{\:\sqrt{{x}}}\right)}\:{so}\:{the}\:{poles}\:{of}\:\varphi\:{are} \\ $$$$\overset{−} {+}\:\frac{{e}^{\frac{{i}\pi}{\mathrm{4}}} }{\:\sqrt{{x}}}\:{and}\:\:\overset{−} {+}\:\:\frac{{e}^{−\frac{{i}\pi}{\mathrm{4}}} }{\:\sqrt{{x}}}\:\:{residus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:{Res}\left(\varphi,\frac{{e}^{\frac{{i}\pi}{\mathrm{4}}} }{\:\sqrt{{x}}}\right)\:+{Res}\left(\varphi,−\frac{{e}^{−\frac{{i}\pi}{\mathrm{4}}} }{\:\sqrt{{x}}}\right)\right\}\:\:{but} \\ $$$${Res}\left(\varphi,\:\frac{{e}^{\frac{{i}\pi}{\mathrm{4}}} }{\:\sqrt{{x}}}\right)\:=\frac{{i}}{{x}\:{x}^{\mathrm{2}} \left(\mathrm{2}\frac{{e}^{\frac{{i}\pi}{\mathrm{4}}} }{\:\sqrt{{x}}}\right)\left(\mathrm{2}\frac{{i}}{{x}}\right)}\:=\:\frac{\sqrt{{x}}}{\mathrm{4}{x}^{\mathrm{2}} \:{e}^{\frac{{i}\pi}{\mathrm{4}}} }\:=\frac{\sqrt{{x}}}{\mathrm{4}{x}^{\mathrm{2}} }\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \\ $$$${Res}\left(\varphi,−\frac{{e}^{−\frac{{i}\pi}{\mathrm{4}}} }{\:\sqrt{{x}}}\right)\:=\frac{−{i}}{{x}\left(−\mathrm{2}\frac{{e}^{−\frac{{i}\pi}{\mathrm{4}}} }{\:\sqrt{{x}}}\right){x}^{\mathrm{2}} \left(−\mathrm{2}\frac{{i}}{{x}}\right)}\:=−\frac{\sqrt{{x}}}{\mathrm{4}{x}^{\mathrm{2}} }\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=−\mathrm{2}{i}\pi\frac{\sqrt{{x}}}{\mathrm{4}{x}^{\mathrm{2}} }\left\{{e}^{\frac{{i}\pi}{\mathrm{4}}} \:−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right\}\:=−\mathrm{2}{i}\pi\:\frac{\sqrt{{x}}}{\mathrm{4}{x}^{\mathrm{2}} }\:\mathrm{2}{i}\:{sin}\left(\frac{\pi}{\mathrm{4}}\right) \\ $$$$=\frac{\sqrt{{x}}}{{x}^{\mathrm{2}} }\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\:{and}\:{f}^{'} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\:\frac{\sqrt{\mathrm{2}}}{\mathrm{4}{x}\sqrt{{x}}}\:\:{with}\:{x}>\mathrm{0} \\ $$$${if}\:{x}<\:{we}\:{put}\:{x}\:=−{u}\:\Rightarrow{f}\left({x}\right)\:=\:{f}\left(−{u}\right)\:=\int_{\mathrm{0}} ^{\infty} {arctan}\left(−{ut}^{\mathrm{2}} \right){dt} \\ $$$$=−{f}\left({u}\right)\:\Rightarrow{f}^{'} \left({x}\right)=−{f}^{'} \left({u}\right)\:=−\frac{\sqrt{\mathrm{2}}}{\mathrm{4}{u}\sqrt{{u}}}\:=−\frac{\sqrt{\mathrm{2}}}{−\mathrm{4}{x}\sqrt{−{x}}}\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}{x}\sqrt{−{x}}}\:. \\ $$

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