Question Number 25851 by abdo imad last updated on 15/Dec/17
$${find}\:{the}\:{value}\:{of}\:{integral}\:\:\:\int_{{R}} \left({z}−{a}\right)^{−\mathrm{1}} {dz}\:\:{with}\:{a}\:{from}\:{C}\:\:{aplly} \\ $$$${this}\:{result}\:{to}\:{find}\:{the}\:{value}\:{of}\:\:\:\int_{\mathrm{0}} ^{\infty} \left(\mathrm{2}\:+{x}^{\mathrm{4}_{} } \right)^{−\mathrm{1}} {dx}. \\ $$
Commented by abdo imad last updated on 21/Dec/17
![let put I(ξ)= ∫_(−ξ) ^ξ (dx/(x−a)) and a=α+iβ and β not 0 I(ξ) = ∫_(−ξ) ^ξ ((x−α+ iβ)/((x−α)^2 +β^2^ ))dx= A(ξ)+B(ξ) where A(ξ) = ∫_(−ξ) ^ξ ((x−α)/((x−α)^2 +β^2 ))dx and B(ξ) =iβ∫_(−ξ) ^ξ (dx/((x−α)^2 +β^2 )) but A(ξ)= (1/2)ln( (((ξ−α)^2 +β^2 )/((ξ+α)^2 +β^2 )) ) and lim _(ξ−>∝) A(ξ)=0 and by the changement x−α=βt B(ξ)=iβ ∫_((−ξ−α)/β) ^((ξ−α)/β) (β/(β^2 t^2 +β^2 ))dt =i( artan(((ξ−α)/β) ) +artan( ((ξ+α)/β))) −−>if β>0 lim B(ξ)_(ξ−>∝) =iπ if β<0 lim B(ξ)_(ξ−>∝) =−iπ...look that β =im(a) value of I=∫_0 ^∞ (dx/(2+x^4 )) I= (1/2) ∫_R (dx/(2+x^4 )) and by changement x= 2^(1/4) t ...I= (2^(1/4) /4) ∫_R (dt/(t^4 +1)) let find the poles of f(z)= (1/(z^4 +1)) the roots of z^4 +1=0 are the complex z_k = e^(i(2k+1)(π/4)) k from[[0.3]] z_0 = e^(i(π/4)) .. z_1 = e^(i((3π)/4)) .. z_2 = e^(i((5π)/4)) ..z_3 = e^(i((7π_ )/4)) f(z) = Σ_(k=0) ^(k=3) (λ_k /(z−z_k )) and λ_k = (1/(der( 1+z^4 )))=−(1/4)z_k (der means derivative ) ∫_R f(z) dz = ((−1)/4) ( ∫_R (z_0 /(z−z_0 ))dz + ∫_R (z_1 /(z−z_1 ))dz +∫_R (z_2 /(z−z_2 ))dz +∫_R (z_3 /(z−z_3 ))dz ) = ((−1)/4) (iπ z_0 +iπ z_1 −iπz_2 −iπ z_3 ) because im(z_0 )>0 im(z_1 )>0.. im(z_2 )<0 im(z_3 )<0 ∫_R f(z)dz = ((−iπ)/4) ( z_0 +z_1 +z_2 +z_3 ) =((−iπ)/4) ( z_0 − z_0 ^− +z_0 −z_0 ^− ) =((−iπ)/2)(2i) im(z_0 ) = ((π (√2))/2)−−−> ∫_0 ^∞ (dx/(2+x^4 )) = 2^(1/4) .4^(−1) .((π(√2))/2) −−−> ∫_R (dx/(2+x^4 )) = ((π(√2).2^(1/4) )/8) ...](https://www.tinkutara.com/question/Q26151.png)
$${let}\:{put}\:{I}\left(\xi\right)=\:\int_{−\xi} ^{\xi} \frac{{dx}}{{x}−{a}}\:\:{and}\:{a}=\alpha+{i}\beta\:{and}\:\beta\:{not}\:\mathrm{0} \\ $$$${I}\left(\xi\right)\:=\:\int_{−\xi} ^{\xi} \frac{{x}−\alpha+\:{i}\beta}{\left({x}−\alpha\right)^{\mathrm{2}} \:+\beta^{\mathrm{2}^{} } }{dx}=\:{A}\left(\xi\right)+{B}\left(\xi\right)\:{where} \\ $$$${A}\left(\xi\right)\:=\:\int_{−\xi} ^{\xi} \:\frac{{x}−\alpha}{\left({x}−\alpha\right)^{\mathrm{2}} +\beta^{\mathrm{2}} }{dx}\:{and}\:{B}\left(\xi\right)\:={i}\beta\int_{−\xi} ^{\xi} \:\frac{{dx}}{\left({x}−\alpha\right)^{\mathrm{2}} +\beta^{\mathrm{2}} } \\ $$$${but}\:{A}\left(\xi\right)=\:\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\:\frac{\left(\xi−\alpha\right)^{\mathrm{2}} +\beta^{\mathrm{2}} }{\left(\xi+\alpha\right)^{\mathrm{2}} +\beta^{\mathrm{2}} }\:\right)\:\:\:{and}\:{lim}\:_{\xi−>\propto} {A}\left(\xi\right)=\mathrm{0} \\ $$$${and}\:{by}\:{the}\:{changement}\:{x}−\alpha=\beta{t}\:\: \\ $$$${B}\left(\xi\right)={i}\beta\:\int_{\frac{−\xi−\alpha}{\beta}} ^{\frac{\xi−\alpha}{\beta}} \:\:\:\frac{\beta}{\beta^{\mathrm{2}} {t}^{\mathrm{2}} \:+\beta^{\mathrm{2}} }{dt}\:={i}\left(\:{artan}\left(\frac{\xi−\alpha}{\beta}\:\right)\:+{artan}\left(\:\frac{\xi+\alpha}{\beta}\right)\right) \\ $$$$−−>{if}\:\beta>\mathrm{0}\:\:\:{lim}\:\:{B}\left(\xi\right)_{\xi−>\propto} \:\:={i}\pi \\ $$$${if}\:\beta<\mathrm{0}\:\:\:{lim}\:{B}\left(\xi\right)_{\xi−>\propto} \:\:=−{i}\pi…{look}\:{that}\:\beta\:={im}\left({a}\right) \\ $$$${value}\:{of}\:\:{I}=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\mathrm{2}+{x}^{\mathrm{4}} }\:\:\:\:\:{I}=\:\frac{\mathrm{1}}{\mathrm{2}}\:\:\int_{{R}} \:\frac{{dx}}{\mathrm{2}+{x}^{\mathrm{4}} }\:\:{and}\:{by}\:{changement} \\ $$$${x}=\:\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{t}\:\:\:\:\:…{I}=\:\:\:\frac{\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} }{\mathrm{4}}\:\:\int_{{R}} \:\:\:\frac{{dt}}{{t}^{\mathrm{4}} +\mathrm{1}}\:\:\:{let}\:{find}\:{the}\:{poles}\:{of}\:{f}\left({z}\right)=\:\frac{\mathrm{1}}{{z}^{\mathrm{4}} +\mathrm{1}} \\ $$$${the}\:{roots}\:{of}\:\:{z}^{\mathrm{4}} \:\:+\mathrm{1}=\mathrm{0}\:\:{are}\:{the}\:{complex}\:\:{z}_{{k}} \:=\:\:{e}^{{i}\left(\mathrm{2}{k}+\mathrm{1}\right)\frac{\pi}{\mathrm{4}}} \:\:\:{k}\:{from}\left[\left[\mathrm{0}.\mathrm{3}\right]\right] \\ $$$${z}_{\mathrm{0}} =\:{e}^{{i}\frac{\pi}{\mathrm{4}}} \:\:\:..\:{z}_{\mathrm{1}} \:\:=\:{e}^{{i}\frac{\mathrm{3}\pi}{\mathrm{4}}} \:\:\:..\:{z}_{\mathrm{2}} \:\:=\:{e}^{{i}\frac{\mathrm{5}\pi}{\mathrm{4}}} \:\:\:\:..{z}_{\mathrm{3}} \:\:=\:{e}^{{i}\frac{\mathrm{7}\pi_{} }{\mathrm{4}}} \\ $$$${f}\left({z}\right)\:\:=\:\sum_{{k}=\mathrm{0}} ^{{k}=\mathrm{3}} \:\:\:\frac{\lambda_{{k}} }{{z}−{z}_{{k}} }\:\:\:{and}\:\:\:\:\:\lambda_{{k}} \:\:=\:\:\:\frac{\mathrm{1}}{{der}\left(\:\mathrm{1}+{z}^{\mathrm{4}} \right)}=−\frac{\mathrm{1}}{\mathrm{4}}{z}_{{k}} \:\:\:\left({der}\:\:{means}\:{derivative}\:\:\right) \\ $$$$\int_{{R}} \:\:{f}\left({z}\right)\:{dz}\:\:=\:\:\:\frac{−\mathrm{1}}{\mathrm{4}}\:\:\left(\:\:\int_{{R}} \:\frac{{z}_{\mathrm{0}} }{{z}−{z}_{\mathrm{0}} }{dz}\:\:+\:\int_{{R}} \:\:\:\frac{{z}_{\mathrm{1}} }{{z}−{z}_{\mathrm{1}} }{dz}\:\:+\int_{{R}} \:\:\frac{{z}_{\mathrm{2}} }{{z}−{z}_{\mathrm{2}} }{dz}\:\:+\int_{{R}} \:\:\frac{{z}_{\mathrm{3}} }{{z}−{z}_{\mathrm{3}} }{dz}\:\:\:\right) \\ $$$$=\:\frac{−\mathrm{1}}{\mathrm{4}}\:\left({i}\pi\:{z}_{\mathrm{0}} \:\:+{i}\pi\:{z}_{\mathrm{1}} \:−{i}\pi{z}_{\mathrm{2}} \:\:−{i}\pi\:{z}_{\mathrm{3}} \:\:\:\right)\:{because}\:\:{im}\left({z}_{\mathrm{0}} \:\:\right)>\mathrm{0} \\ $$$${im}\left({z}_{\mathrm{1}} \:\right)>\mathrm{0}..\:{im}\left({z}_{\mathrm{2}} \:\:\right)<\mathrm{0}\:\:\:\:{im}\left({z}_{\mathrm{3}} \:\right)<\mathrm{0} \\ $$$$\int_{{R}} \:\:{f}\left({z}\right){dz}\:\:=\:\frac{−{i}\pi}{\mathrm{4}}\:\:\left(\:{z}_{\mathrm{0}} \:\:+{z}_{\mathrm{1}} \:\:+{z}_{\mathrm{2}} \:\:+{z}_{\mathrm{3}} \:\:\:\right)\:\:=\frac{−{i}\pi}{\mathrm{4}}\:\:\left(\:{z}_{\mathrm{0}} \:\:−\:{z}_{\mathrm{0}} ^{−} \:+{z}_{\mathrm{0}} \:\:−{z}_{\mathrm{0}} ^{−} \:\:\:\right) \\ $$$$=\frac{−{i}\pi}{\mathrm{2}}\left(\mathrm{2}{i}\right)\:{im}\left({z}_{\mathrm{0}} \:\right)\:\:=\:\:\frac{\pi\:\sqrt{\mathrm{2}}}{\mathrm{2}}−−−>\:\:\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\mathrm{2}+{x}^{\mathrm{4}} }\:\:=\:\:\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} \:\:.\mathrm{4}^{−\mathrm{1}} \:.\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$−−−>\:\int_{{R}} \frac{{dx}}{\mathrm{2}+{x}^{\mathrm{4}} }\:\:=\:\:\frac{\pi\sqrt{\mathrm{2}}.\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} }{\mathrm{8}}\:\:\:… \\ $$