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answer-to-25955-we-introduce-the-parametric-function-F-t-0-ln-1-1-x-2-t-1-x-2-1-dx-after-verifying-that-F-is-derivable-on-0-we-find-F-t-0-1-1-x-2-t-1-dx-F-

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Question Number 25960 by abdo imad last updated on 16/Dec/17
answer to 25955.we introduce the parametric function F(t) =∫_0 ^∞ ln(1+(1+x^2_ )t)(1+x^2 )^(−1) dx after verifying that F is derivable on[0.∝[ we find ∂F/∂t= ∫_0 ^∞ ( (1+(1+x^2 )t)^(−1) dx ∂F/∂t=1/2 ∫_R (tx^2 +t+1)^(−1) dx we put f(z) =(tz^2 +z+1)^(−1) let find the poles of f..tz^2 +z+1=0 <−> z=+−i((t+1)t^(−1) )^(1/2) and the poles are z_0 =i((t+1)t^(−1) )^(1/2) and z_1 =−i((t+1)t^(−1) )^(1/2) and f(t) =(t(t−z_0 )(t−z_1 ))^(−1) by residus theorem ∫_R f(z)dz =2iπ R(f.z_0 ) =2iπ (t(z_0 −z_1 ))^(−1) =π t^(−1/2) (t+1)^(−1/2 ) −>∂F/∂t =π 2^(−1) t^(−1/2) (1+t)^(−1/2) −>F(t) =π 2^(−1 ) ∫_0 ^t x^(−1/2) (1+x)^(−1/2) dx +α α=F(0)=0 and F(t) =π2^(−1) ∫_0 ^t x^(−1/2) (1+x)^(−1/2) dx and by the changement x^(1/2) =u we find F(t) = π ln( t^(1/2) +(1+t)^(1/2) ) so ∫_0 ^∞ ln(2+x^2 )(1+x^2 )^(−1) dx=F(1)=πln(1+2^(1/2) )
$${answer}\:{to}\:\mathrm{25955}.{we}\:{introduce}\:{the}\:{parametric}\:{function} \\ $$$${F}\left({t}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:{ln}\left(\mathrm{1}+\left(\mathrm{1}+{x}^{\mathrm{2}_{} } \right){t}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\mathrm{1}} {dx}\:{after}\:{verifying}\:{that}\: \\ $$$${F}\:{is}\:{derivable}\:{on}\left[\mathrm{0}.\propto\left[\:\:{we}\:{find}\:\:\:\partial{F}/\partial{t}=\:\:\int_{\mathrm{0}} ^{\infty} \left(\:\left(\mathrm{1}+\left(\mathrm{1}+{x}^{\mathrm{2}} \right){t}\right)^{−\mathrm{1}} {dx}\right.\right.\right. \\ $$$$\partial{F}/\partial{t}=\mathrm{1}/\mathrm{2}\:\int_{{R}} \left({tx}^{\mathrm{2}} +{t}+\mathrm{1}\right)^{−\mathrm{1}} {dx}\:\:\:{we}\:{put}\:\:{f}\left({z}\right)\:=\left({tz}^{\mathrm{2}} +{z}+\mathrm{1}\right)^{−\mathrm{1}} \\ $$$${let}\:{find}\:{the}\:{poles}\:{of}\:{f}..{tz}^{\mathrm{2}} +{z}+\mathrm{1}=\mathrm{0}\:<−>\:\:{z}=+−{i}\left(\left({t}+\mathrm{1}\right){t}^{−\mathrm{1}} \right)^{\mathrm{1}/\mathrm{2}} \\ $$$${and}\:{the}\:{poles}\:{are}\:\:{z}_{\mathrm{0}} ={i}\left(\left({t}+\mathrm{1}\right){t}^{−\mathrm{1}} \right)^{\mathrm{1}/\mathrm{2}} \:\:{and}\:\:\:{z}_{\mathrm{1}} =−{i}\left(\left({t}+\mathrm{1}\right){t}^{−\mathrm{1}} \right)^{\mathrm{1}/\mathrm{2}} \\ $$$${and}\:\:{f}\left({t}\right)\:\:=\left({t}\left({t}−{z}_{\mathrm{0}} \right)\left({t}−{z}_{\mathrm{1}} \right)\right)^{−\mathrm{1}} \:\:\:{by}\:{residus}\:{theorem} \\ $$$$ \\ $$$$\int_{{R}} {f}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{R}\left({f}.{z}_{\mathrm{0}} \right)\:\:=\mathrm{2}{i}\pi\:\left({t}\left({z}_{\mathrm{0}} −{z}_{\mathrm{1}} \right)\right)^{−\mathrm{1}} \\ $$$$=\pi\:{t}^{−\mathrm{1}/\mathrm{2}} \left({t}+\mathrm{1}\right)^{−\mathrm{1}/\mathrm{2}\:} \:\:\:\:−>\partial{F}/\partial{t}\:=\pi\:\mathrm{2}^{−\mathrm{1}} \:{t}^{−\mathrm{1}/\mathrm{2}} \left(\mathrm{1}+{t}\right)^{−\mathrm{1}/\mathrm{2}} \\ $$$$−>{F}\left({t}\right)\:\:=\pi\:\mathrm{2}^{−\mathrm{1}\:} \int_{\mathrm{0}} ^{{t}} \:\:\:{x}^{−\mathrm{1}/\mathrm{2}} \left(\mathrm{1}+{x}\right)^{−\mathrm{1}/\mathrm{2}} {dx}\:+\alpha \\ $$$$\alpha={F}\left(\mathrm{0}\right)=\mathrm{0}\:{and}\:\:{F}\left({t}\right)\:=\pi\mathrm{2}^{−\mathrm{1}} \int_{\mathrm{0}} ^{{t}} \:{x}^{−\mathrm{1}/\mathrm{2}} \left(\mathrm{1}+{x}\right)^{−\mathrm{1}/\mathrm{2}} {dx} \\ $$$${and}\:{by}\:{the}\:{changement}\:\:{x}^{\mathrm{1}/\mathrm{2}} ={u}\:\:\:{we}\:{find} \\ $$$$ \\ $$$${F}\left({t}\right)\:=\:\pi\:{ln}\left(\:{t}^{\mathrm{1}/\mathrm{2}} +\left(\mathrm{1}+{t}\right)^{\mathrm{1}/\mathrm{2}} \right)\:{so}\:\:\int_{\mathrm{0}} ^{\infty} {ln}\left(\mathrm{2}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\mathrm{1}} {dx}={F}\left(\mathrm{1}\right)=\pi{ln}\left(\mathrm{1}+\mathrm{2}^{\mathrm{1}/\mathrm{2}} \right) \\ $$

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