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Question Number 25851 by abdo imad last updated on 15/Dec/17
find the value of integral   ∫_R (z−a)^(−1) dz  with a from C  aplly  this result to find the value of   ∫_0 ^∞ (2 +x^4_  )^(−1) dx.
$${find}\:{the}\:{value}\:{of}\:{integral}\:\:\:\int_{{R}} \left({z}−{a}\right)^{−\mathrm{1}} {dz}\:\:{with}\:{a}\:{from}\:{C}\:\:{aplly} \\ $$$${this}\:{result}\:{to}\:{find}\:{the}\:{value}\:{of}\:\:\:\int_{\mathrm{0}} ^{\infty} \left(\mathrm{2}\:+{x}^{\mathrm{4}_{} } \right)^{−\mathrm{1}} {dx}. \\ $$
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)   ...
$${let}\:{put}\:{I}\left(\xi\right)=\:\int_{−\xi} ^{\xi} \frac{{dx}}{{x}−{a}}\:\:{and}\:{a}=\alpha+{i}\beta\:{and}\:\beta\:{not}\:\mathrm{0} \\ $$$${I}\left(\xi\right)\:=\:\int_{−\xi} ^{\xi} \frac{{x}−\alpha+\:{i}\beta}{\left({x}−\alpha\right)^{\mathrm{2}} \:+\beta^{\mathrm{2}^{} } }{dx}=\:{A}\left(\xi\right)+{B}\left(\xi\right)\:{where} \\ $$$${A}\left(\xi\right)\:=\:\int_{−\xi} ^{\xi} \:\frac{{x}−\alpha}{\left({x}−\alpha\right)^{\mathrm{2}} +\beta^{\mathrm{2}} }{dx}\:{and}\:{B}\left(\xi\right)\:={i}\beta\int_{−\xi} ^{\xi} \:\frac{{dx}}{\left({x}−\alpha\right)^{\mathrm{2}} +\beta^{\mathrm{2}} } \\ $$$${but}\:{A}\left(\xi\right)=\:\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\:\frac{\left(\xi−\alpha\right)^{\mathrm{2}} +\beta^{\mathrm{2}} }{\left(\xi+\alpha\right)^{\mathrm{2}} +\beta^{\mathrm{2}} }\:\right)\:\:\:{and}\:{lim}\:_{\xi−>\propto} {A}\left(\xi\right)=\mathrm{0} \\ $$$${and}\:{by}\:{the}\:{changement}\:{x}−\alpha=\beta{t}\:\: \\ $$$${B}\left(\xi\right)={i}\beta\:\int_{\frac{−\xi−\alpha}{\beta}} ^{\frac{\xi−\alpha}{\beta}} \:\:\:\frac{\beta}{\beta^{\mathrm{2}} {t}^{\mathrm{2}} \:+\beta^{\mathrm{2}} }{dt}\:={i}\left(\:{artan}\left(\frac{\xi−\alpha}{\beta}\:\right)\:+{artan}\left(\:\frac{\xi+\alpha}{\beta}\right)\right) \\ $$$$−−>{if}\:\beta>\mathrm{0}\:\:\:{lim}\:\:{B}\left(\xi\right)_{\xi−>\propto} \:\:={i}\pi \\ $$$${if}\:\beta<\mathrm{0}\:\:\:{lim}\:{B}\left(\xi\right)_{\xi−>\propto} \:\:=−{i}\pi…{look}\:{that}\:\beta\:={im}\left({a}\right) \\ $$$${value}\:{of}\:\:{I}=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\mathrm{2}+{x}^{\mathrm{4}} }\:\:\:\:\:{I}=\:\frac{\mathrm{1}}{\mathrm{2}}\:\:\int_{{R}} \:\frac{{dx}}{\mathrm{2}+{x}^{\mathrm{4}} }\:\:{and}\:{by}\:{changement} \\ $$$${x}=\:\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{t}\:\:\:\:\:…{I}=\:\:\:\frac{\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} }{\mathrm{4}}\:\:\int_{{R}} \:\:\:\frac{{dt}}{{t}^{\mathrm{4}} +\mathrm{1}}\:\:\:{let}\:{find}\:{the}\:{poles}\:{of}\:{f}\left({z}\right)=\:\frac{\mathrm{1}}{{z}^{\mathrm{4}} +\mathrm{1}} \\ $$$${the}\:{roots}\:{of}\:\:{z}^{\mathrm{4}} \:\:+\mathrm{1}=\mathrm{0}\:\:{are}\:{the}\:{complex}\:\:{z}_{{k}} \:=\:\:{e}^{{i}\left(\mathrm{2}{k}+\mathrm{1}\right)\frac{\pi}{\mathrm{4}}} \:\:\:{k}\:{from}\left[\left[\mathrm{0}.\mathrm{3}\right]\right] \\ $$$${z}_{\mathrm{0}} =\:{e}^{{i}\frac{\pi}{\mathrm{4}}} \:\:\:..\:{z}_{\mathrm{1}} \:\:=\:{e}^{{i}\frac{\mathrm{3}\pi}{\mathrm{4}}} \:\:\:..\:{z}_{\mathrm{2}} \:\:=\:{e}^{{i}\frac{\mathrm{5}\pi}{\mathrm{4}}} \:\:\:\:..{z}_{\mathrm{3}} \:\:=\:{e}^{{i}\frac{\mathrm{7}\pi_{} }{\mathrm{4}}} \\ $$$${f}\left({z}\right)\:\:=\:\sum_{{k}=\mathrm{0}} ^{{k}=\mathrm{3}} \:\:\:\frac{\lambda_{{k}} }{{z}−{z}_{{k}} }\:\:\:{and}\:\:\:\:\:\lambda_{{k}} \:\:=\:\:\:\frac{\mathrm{1}}{{der}\left(\:\mathrm{1}+{z}^{\mathrm{4}} \right)}=−\frac{\mathrm{1}}{\mathrm{4}}{z}_{{k}} \:\:\:\left({der}\:\:{means}\:{derivative}\:\:\right) \\ $$$$\int_{{R}} \:\:{f}\left({z}\right)\:{dz}\:\:=\:\:\:\frac{−\mathrm{1}}{\mathrm{4}}\:\:\left(\:\:\int_{{R}} \:\frac{{z}_{\mathrm{0}} }{{z}−{z}_{\mathrm{0}} }{dz}\:\:+\:\int_{{R}} \:\:\:\frac{{z}_{\mathrm{1}} }{{z}−{z}_{\mathrm{1}} }{dz}\:\:+\int_{{R}} \:\:\frac{{z}_{\mathrm{2}} }{{z}−{z}_{\mathrm{2}} }{dz}\:\:+\int_{{R}} \:\:\frac{{z}_{\mathrm{3}} }{{z}−{z}_{\mathrm{3}} }{dz}\:\:\:\right) \\ $$$$=\:\frac{−\mathrm{1}}{\mathrm{4}}\:\left({i}\pi\:{z}_{\mathrm{0}} \:\:+{i}\pi\:{z}_{\mathrm{1}} \:−{i}\pi{z}_{\mathrm{2}} \:\:−{i}\pi\:{z}_{\mathrm{3}} \:\:\:\right)\:{because}\:\:{im}\left({z}_{\mathrm{0}} \:\:\right)>\mathrm{0} \\ $$$${im}\left({z}_{\mathrm{1}} \:\right)>\mathrm{0}..\:{im}\left({z}_{\mathrm{2}} \:\:\right)<\mathrm{0}\:\:\:\:{im}\left({z}_{\mathrm{3}} \:\right)<\mathrm{0} \\ $$$$\int_{{R}} \:\:{f}\left({z}\right){dz}\:\:=\:\frac{−{i}\pi}{\mathrm{4}}\:\:\left(\:{z}_{\mathrm{0}} \:\:+{z}_{\mathrm{1}} \:\:+{z}_{\mathrm{2}} \:\:+{z}_{\mathrm{3}} \:\:\:\right)\:\:=\frac{−{i}\pi}{\mathrm{4}}\:\:\left(\:{z}_{\mathrm{0}} \:\:−\:{z}_{\mathrm{0}} ^{−} \:+{z}_{\mathrm{0}} \:\:−{z}_{\mathrm{0}} ^{−} \:\:\:\right) \\ $$$$=\frac{−{i}\pi}{\mathrm{2}}\left(\mathrm{2}{i}\right)\:{im}\left({z}_{\mathrm{0}} \:\right)\:\:=\:\:\frac{\pi\:\sqrt{\mathrm{2}}}{\mathrm{2}}−−−>\:\:\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\mathrm{2}+{x}^{\mathrm{4}} }\:\:=\:\:\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} \:\:.\mathrm{4}^{−\mathrm{1}} \:.\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$−−−>\:\int_{{R}} \frac{{dx}}{\mathrm{2}+{x}^{\mathrm{4}} }\:\:=\:\:\frac{\pi\sqrt{\mathrm{2}}.\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} }{\mathrm{8}}\:\:\:… \\ $$

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