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Question Number 36336 by prof Abdo imad last updated on 31/May/18
find f(x)= ∫_0 ^∞ arctan(xt^2 )dt with x>0
$${find}\:{f}\left({x}\right)=\:\int_{\mathrm{0}} ^{\infty} \:\:{arctan}\left({xt}^{\mathrm{2}} \right){dt}\:{with}\:{x}>\mathrm{0} \\ $$
Commented by abdo.msup.com last updated on 01/Jun/18
we have f^′ (x)=∫_0 ^∞ (t^2 /(1+x^2 t^4 ))dt 2f^′ (x)=∫_(−∞) ^(+∞) (t^2 /(x^2 t^4 +1))dt let consider the complex function ϕ(z) =(z^2 /(x^2 z^4 +1)) we have ϕ(z) = (z^2 /(x^2 { z^4 +(1/x^2 )})) = (z^2 /(x^2 {z^2 −(i/x)}(z^2 +(i/x)))) =(z^2 /(x^2 (z −(1/( (√x)))e^(i(π/4)) )(z +(1/( (√x)))e^(i(π/4)) )(z−(1/( (√x)))e^(−i(π/4)) )(z +(1/( (√x)))e^(−((iπ)/4)) ))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,(1/( (√x)))e^((iπ)/4) )+Res(ϕ,−(1/( (√x)))e^(−((iπ)/4)) )} Res(ϕ, (1/( (√x)))e^(i(π/4)) ) = (i/(x^3 ((2/( (√x)))e^((iπ)/4) )(((2i)/x)))) = ((√x)/(4x^2 )) e^(−((iπ)/4)) Res(ϕ,−(1/( (√x)))e^(−((iπ)/4)) )= ((−i)/(x^3 (((−2)/( (√x)))e^(−((iπ)/4)) )(((−2i)/x)))) =−((√x)/(4x^2 )) e^((iπ)/4) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ .((√x)/(4x^2 )){ e^(−((iπ)/4)) −e^((iπ)/4) } =((iπ(√x))/(2x^2 )){−2isin((π/4))} =((π(√x))/x^2 ) ((√2)/2) = ((π(√x))/(x^2 (√2)))
$${we}\:{have}\:{f}^{'} \left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} {t}^{\mathrm{4}} }{dt} \\ $$$$\mathrm{2}{f}^{'} \left({x}\right)=\int_{−\infty} ^{+\infty} \:\:\:\frac{{t}^{\mathrm{2}} }{{x}^{\mathrm{2}} {t}^{\mathrm{4}} \:+\mathrm{1}}{dt}\:{let}\:{consider} \\ $$$${the}\:{complex}\:{function} \\ $$$$\varphi\left({z}\right)\:=\frac{{z}^{\mathrm{2}} }{{x}^{\mathrm{2}} {z}^{\mathrm{4}} \:+\mathrm{1}}\:{we}\:{have} \\ $$$$\varphi\left({z}\right)\:=\:\frac{{z}^{\mathrm{2}} }{{x}^{\mathrm{2}} \left\{\:{z}^{\mathrm{4}} \:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right\}} \\ $$$$=\:\frac{{z}^{\mathrm{2}} }{{x}^{\mathrm{2}} \left\{{z}^{\mathrm{2}} \:−\frac{{i}}{{x}}\right\}\left({z}^{\mathrm{2}} \:+\frac{{i}}{{x}}\right)} \\ $$$$=\frac{{z}^{\mathrm{2}} }{{x}^{\mathrm{2}} \left({z}\:−\frac{\mathrm{1}}{\:\sqrt{{x}}}{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)\left({z}\:+\frac{\mathrm{1}}{\:\sqrt{{x}}}{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)\left({z}−\frac{\mathrm{1}}{\:\sqrt{{x}}}{e}^{−{i}\frac{\pi}{\mathrm{4}}} \right)\left({z}\:+\frac{\mathrm{1}}{\:\sqrt{{x}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)} \\ $$$$\int_{−\infty} ^{+\infty} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:{Res}\left(\varphi,\frac{\mathrm{1}}{\:\sqrt{{x}}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)+{Res}\left(\varphi,−\frac{\mathrm{1}}{\:\sqrt{{x}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\right\} \\ $$$${Res}\left(\varphi,\:\frac{\mathrm{1}}{\:\sqrt{{x}}}{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)\:=\:\frac{{i}}{{x}^{\mathrm{3}} \left(\frac{\mathrm{2}}{\:\sqrt{{x}}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left(\frac{\mathrm{2}{i}}{{x}}\right)} \\ $$$$=\:\frac{\sqrt{{x}}}{\mathrm{4}{x}^{\mathrm{2}} }\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \\ $$$${Res}\left(\varphi,−\frac{\mathrm{1}}{\:\sqrt{{x}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)=\:\frac{−{i}}{{x}^{\mathrm{3}} \left(\frac{−\mathrm{2}}{\:\sqrt{{x}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left(\frac{−\mathrm{2}{i}}{{x}}\right)} \\ $$$$=−\frac{\sqrt{{x}}}{\mathrm{4}{x}^{\mathrm{2}} }\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:.\frac{\sqrt{{x}}}{\mathrm{4}{x}^{\mathrm{2}} }\left\{\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:−{e}^{\frac{{i}\pi}{\mathrm{4}}} \right\} \\ $$$$=\frac{{i}\pi\sqrt{{x}}}{\mathrm{2}{x}^{\mathrm{2}} }\left\{−\mathrm{2}{isin}\left(\frac{\pi}{\mathrm{4}}\right)\right\} \\ $$$$=\frac{\pi\sqrt{{x}}}{{x}^{\mathrm{2}} }\:\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:=\:\frac{\pi\sqrt{{x}}}{{x}^{\mathrm{2}} \sqrt{\mathrm{2}}} \\ $$
Commented by abdo.msup.com last updated on 01/Jun/18
⇒f^′ (x) = (π/(2(√2))) ((√x)/x^2 ) ⇒ f(x)= (π/(2(√2))) ∫ (((√x) dx)/x^2 ) +c but ∫ (((√x)dx)/x^2 ) =_((√x)=t) ∫ (t/t^4 ) 2t dt = 2 ∫ (dt/t^2 ) = −(2/t) +λ = ((−2)/( (√x))) +λ ⇒ f(x) = (π/(2(√2))) ((−2)/( (√x))) +c = ((−π)/( (√(2x)))) +c
$$\Rightarrow{f}^{'} \left({x}\right)\:=\:\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:\:\frac{\sqrt{{x}}}{{x}^{\mathrm{2}} }\:\Rightarrow \\ $$$${f}\left({x}\right)=\:\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:\int\:\:\:\:\:\:\frac{\sqrt{{x}}\:{dx}}{{x}^{\mathrm{2}} }\:+{c}\:\:{but} \\ $$$$\int\:\:\frac{\sqrt{{x}}{dx}}{{x}^{\mathrm{2}} }\:=_{\sqrt{{x}}={t}} \:\int\:\:\:\:\:\frac{{t}}{{t}^{\mathrm{4}} }\:\mathrm{2}{t}\:{dt} \\ $$$$=\:\mathrm{2}\:\int\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} }\:=\:−\frac{\mathrm{2}}{{t}}\:+\lambda\:=\:\frac{−\mathrm{2}}{\:\sqrt{{x}}}\:+\lambda\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\:\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:\frac{−\mathrm{2}}{\:\sqrt{{x}}}\:+{c}\:=\:\frac{−\pi}{\:\sqrt{\mathrm{2}{x}}}\:\:+{c} \\ $$
Commented by abdo.msup.com last updated on 01/Jun/18
we have f(1) = −(π/2) +c ⇒ c=(π/2) +f(1) =(π/2) +∫_0 ^∞ arctan(x^2 ) dx f(x)= −(π/(2(√x))) +(π/2) + ∫_0 ^∞ arctan(x^2 )dx . the value of this integral is easy to find.(by parts)
$${we}\:{have}\:{f}\left(\mathrm{1}\right)\:=\:−\frac{\pi}{\mathrm{2}}\:+{c}\:\Rightarrow \\ $$$${c}=\frac{\pi}{\mathrm{2}}\:+{f}\left(\mathrm{1}\right)\:=\frac{\pi}{\mathrm{2}}\:+\int_{\mathrm{0}} ^{\infty} \:{arctan}\left({x}^{\mathrm{2}} \right)\:{dx} \\ $$$${f}\left({x}\right)=\:−\frac{\pi}{\mathrm{2}\sqrt{{x}}}\:\:+\frac{\pi}{\mathrm{2}}\:+\:\int_{\mathrm{0}} ^{\infty} \:{arctan}\left({x}^{\mathrm{2}} \right){dx}\:. \\ $$$${the}\:{value}\:{of}\:{this}\:{integral}\:{is}\:{easy}\:{to} \\ $$$${find}.\left({by}\:{parts}\right) \\ $$
Commented by prof Abdo imad last updated on 01/Jun/18
let put L =lim_(x→+∞) f(x) ⇒c=Lso f(x)= L−(π/( (√(2x)))) .
$$\:{let}\:{put}\:\:{L}\:={lim}_{{x}\rightarrow+\infty} {f}\left({x}\right)\:\Rightarrow{c}={Lso} \\ $$$${f}\left({x}\right)=\:{L}−\frac{\pi}{\:\sqrt{\mathrm{2}{x}}}\:. \\ $$

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