Question Number 65293 by mathmax by abdo last updated on 27/Jul/19
$$\left.\mathrm{1}\right)\:{calculate}\:\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{{x}−{a}}\:\:{with}\:{a}\:\in{C} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{values}\:{of}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{{x}^{\mathrm{4}} \:+\mathrm{1}}\:\:{and}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{{x}^{\mathrm{6}} \:+\mathrm{1}} \\ $$$${by}\:{using}\:{the}\:{decomposition}\:{inside}\:{C}\left({x}\right). \\ $$
Commented by mathmax by abdo last updated on 01/Aug/19
![1) let I(ξ) =∫_(−ξ) ^ξ (dx/(x−a)) let a =α+iβ we have ∫_(−∞) ^(+∞) (dx/(x−a)) =lim_(ξ→+∞) I(ξ) I(ξ) =∫_(−ξ) ^ξ (dx/(x−α−iβ)) =∫_(−ξ) ^ξ ((x−α+iβ)/((x−α)^2 +β^2 ))dx =(1/2)[ln{(x−α)^2 +β^2 }]_(−ξ) ^ξ +iβ ∫_(−ξ) ^ξ (dx/((x−α)^2 +β^2 )) =(1/2)ln((((ξ−α)^2 +β^2 )/((ξ+α)^2 )+β^2 ))) +iβ ∫_(−ξ) ^ξ (dx/((x−α)^2 +β^2 )) cha7gement x−α =∣β∣u give ∫_(−ξ) ^ξ (dx/((x−α)^2 +β^2 )) = ∫_((−ξ−α)/(∣β∣)) ^((ξ−α)/(∣β∣)) ((∣β∣du)/(β^2 (1+u^2 ))) =(1/(∣β∣)){ arctan(((ξ−α)/(∣β∣)))+arctan(((ξ+α)/(∣β∣)))} case 1 β>0 ⇒lim_(ξ→+∞) I(ξ) =iβ(1/β)((π/2)+(π/2))=iπ case 2 β<0 ⇒lim_(ξ→+∞) I(ξ) =−iβ(1/β)((π/2)+(π/2))=−iπ .](https://www.tinkutara.com/question/Q65628.png)
$$\left.\mathrm{1}\right)\:{let}\:{I}\left(\xi\right)\:=\int_{−\xi} ^{\xi} \:\:\frac{{dx}}{{x}−{a}}\:\:{let}\:{a}\:=\alpha+{i}\beta\:\:\:{we}\:{have}\:\int_{−\infty} ^{+\infty} \:\frac{{dx}}{{x}−{a}}\:={lim}_{\xi\rightarrow+\infty} {I}\left(\xi\right) \\ $$$${I}\left(\xi\right)\:=\int_{−\xi} ^{\xi} \:\:\frac{{dx}}{{x}−\alpha−{i}\beta}\:=\int_{−\xi} ^{\xi} \:\frac{{x}−\alpha+{i}\beta}{\left({x}−\alpha\right)^{\mathrm{2}} \:+\beta^{\mathrm{2}} }{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\left\{\left({x}−\alpha\right)^{\mathrm{2}} \:+\beta^{\mathrm{2}} \right\}\right]_{−\xi} ^{\xi} \:+{i}\beta\:\int_{−\xi} ^{\xi} \:\:\frac{{dx}}{\left({x}−\alpha\right)^{\mathrm{2}} \:+\beta^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\left(\xi−\alpha\right)^{\mathrm{2}} \:+\beta^{\mathrm{2}} }{\left.\left(\xi+\alpha\right)^{\mathrm{2}} \:\right)+\beta^{\mathrm{2}} }\right)\:+{i}\beta\:\int_{−\xi} ^{\xi} \:\frac{{dx}}{\left({x}−\alpha\right)^{\mathrm{2}} \:+\beta^{\mathrm{2}} } \\ $$$${cha}\mathrm{7}{gement}\:{x}−\alpha\:=\mid\beta\mid{u}\:{give}\:\:\int_{−\xi} ^{\xi} \:\:\frac{{dx}}{\left({x}−\alpha\right)^{\mathrm{2}} \:+\beta^{\mathrm{2}} } \\ $$$$=\:\int_{\frac{−\xi−\alpha}{\mid\beta\mid}} ^{\frac{\xi−\alpha}{\mid\beta\mid}} \:\:\:\:\:\:\frac{\mid\beta\mid{du}}{\beta^{\mathrm{2}} \left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\:=\frac{\mathrm{1}}{\mid\beta\mid}\left\{\:{arctan}\left(\frac{\xi−\alpha}{\mid\beta\mid}\right)+{arctan}\left(\frac{\xi+\alpha}{\mid\beta\mid}\right)\right\} \\ $$$${case}\:\mathrm{1}\:\:\beta>\mathrm{0}\:\Rightarrow{lim}_{\xi\rightarrow+\infty} \:{I}\left(\xi\right)\:={i}\beta\frac{\mathrm{1}}{\beta}\left(\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}\right)={i}\pi \\ $$$${case}\:\mathrm{2}\:\:\beta<\mathrm{0}\:\:\Rightarrow{lim}_{\xi\rightarrow+\infty} \:{I}\left(\xi\right)\:=−{i}\beta\frac{\mathrm{1}}{\beta}\left(\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}\right)=−{i}\pi\:. \\ $$
Commented by mathmax by abdo last updated on 01/Aug/19
$${finally}\:\:\int_{−\infty} ^{+\infty} \:\frac{{dx}}{{x}−{a}}\:={i}\pi\:{if}\:{im}\left({a}\right)>\mathrm{0}\:\:{and}\:−{i}\pi\:{if}\:{im}\left({a}\right)<\mathrm{0} \\ $$
Commented by mathmax by abdo last updated on 01/Aug/19
![2) let decompose F(x)=(1/(x^4 +1)) x^4 +1 =0 ⇒x^4 =−1 ⇒(re^(iθ) )^4 =e^(i(2k+1)π) ⇒r =1 and θ =(2k+1)(π/4) so the roots are z_k =e^(i(2k+1)(π/4)) k∈[[0,3]] z_0 =e^((iπ)/4) , z_1 =e^(i((3π)/4)) ,z_2 =e^(i(((5π)/4))) ,z_3 =e^(i(((7π)/4))) F(x) =Σ_(i=0) ^3 (λ_i /(x−z_i )) and λ_i =(1/(4z_i ^3 )) =−(1/4)z_i ⇒ ∫_0 ^(+∞) (dx/(x^4 +1)) =(1/2)∫_(−∞) ^(+∞) F(x)dx =(1/2)∫_(−∞) ^(+∞) Σ_(i=0) ^3 (−(1/4))(z_i /(x−z_i ))dx =−(1/8) Σ_(i=0) ^3 ∫_(−∞) ^(+∞) (z_i /(x−z_i ))dx =−(1/8){ z_0 ∫_(−∞) ^(+∞) (dx/(x−z_0 )) +z_1 ∫_(−∞) ^(+∞) (dx/(x−z_1 )) +z_2 ∫_(−∞) ^(+∞) (dx/(x−z_2 )) +z_3 ∫_(−∞) ^(+∞) (dx/(x−z_3 ))} =−(1/8){ iπ(z_0 )+iπ z_1 −iπz_2 −iπz_3 } =−((iπ)/8){ z_0 +z_1 −(z_2 +z_3 )} but z_0 +z_1 =e^((iπ)/4) +e^(i(π−(π/4))) =e^((iπ)/4) −e^(−((iπ)/4)) =2i sin((π/4)) =2i((√2)/2) =i(√2) z_2 +z_3 =e^(i((5π)/4)) +e^((i7π)/4) =e^(i(π+(π/4))) +e^(i(2π−(π/4))) =−e^((iπ)/4) +e^(−((iπ)/4)) =−(e^((iπ)/4) −e^(−((iπ)/4)) ) =−2isin((π/4)) =−2i((√2)/2) =−i(√2) ⇒ ∫_0 ^∞ (dx/(1+x^4 )) =−((iπ)/8){i(√2)+i(√2)} =(π/8)(2(√2)) =((π(√2))/4)](https://www.tinkutara.com/question/Q65682.png)
$$\left.\mathrm{2}\right)\:{let}\:{decompose}\:{F}\left({x}\right)=\frac{\mathrm{1}}{{x}^{\mathrm{4}} \:+\mathrm{1}} \\ $$$${x}^{\mathrm{4}} +\mathrm{1}\:=\mathrm{0}\:\Rightarrow{x}^{\mathrm{4}} =−\mathrm{1}\:\Rightarrow\left({re}^{{i}\theta} \right)^{\mathrm{4}} \:={e}^{{i}\left(\mathrm{2}{k}+\mathrm{1}\right)\pi} \:\Rightarrow{r}\:=\mathrm{1}\:{and}\:\theta\:=\left(\mathrm{2}{k}+\mathrm{1}\right)\frac{\pi}{\mathrm{4}} \\ $$$${so}\:{the}\:{roots}\:{are}\:{z}_{{k}} ={e}^{{i}\left(\mathrm{2}{k}+\mathrm{1}\right)\frac{\pi}{\mathrm{4}}} \:\:\:{k}\in\left[\left[\mathrm{0},\mathrm{3}\right]\right] \\ $$$${z}_{\mathrm{0}} ={e}^{\frac{{i}\pi}{\mathrm{4}}} \:,\:{z}_{\mathrm{1}} ={e}^{{i}\frac{\mathrm{3}\pi}{\mathrm{4}}} \:\:,{z}_{\mathrm{2}} ={e}^{{i}\left(\frac{\mathrm{5}\pi}{\mathrm{4}}\right)} \:\:,{z}_{\mathrm{3}} ={e}^{{i}\left(\frac{\mathrm{7}\pi}{\mathrm{4}}\right)} \\ $$$${F}\left({x}\right)\:=\sum_{{i}=\mathrm{0}} ^{\mathrm{3}} \:\frac{\lambda_{{i}} }{{x}−{z}_{{i}} }\:\:\:\:\:{and}\:\:\lambda_{{i}} =\frac{\mathrm{1}}{\mathrm{4}{z}_{{i}} ^{\mathrm{3}} }\:=−\frac{\mathrm{1}}{\mathrm{4}}{z}_{{i}} \:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{+\infty} \:\frac{{dx}}{{x}^{\mathrm{4}} \:+\mathrm{1}}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \:{F}\left({x}\right){dx}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \sum_{{i}=\mathrm{0}} ^{\mathrm{3}} \:\:\left(−\frac{\mathrm{1}}{\mathrm{4}}\right)\frac{{z}_{{i}} }{{x}−{z}_{{i}} }{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{8}}\:\sum_{{i}=\mathrm{0}} ^{\mathrm{3}} \:\:\int_{−\infty} ^{+\infty} \:\:\frac{{z}_{{i}} }{{x}−{z}_{{i}} }{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{8}}\left\{\:\:{z}_{\mathrm{0}} \:\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{{x}−{z}_{\mathrm{0}} }\:+{z}_{\mathrm{1}} \int_{−\infty} ^{+\infty} \:\frac{{dx}}{{x}−{z}_{\mathrm{1}} }\:+{z}_{\mathrm{2}} \:\int_{−\infty} ^{+\infty} \:\frac{{dx}}{{x}−{z}_{\mathrm{2}} }\:+{z}_{\mathrm{3}} \int_{−\infty} ^{+\infty} \:\frac{{dx}}{{x}−{z}_{\mathrm{3}} }\right\} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{8}}\left\{\:{i}\pi\left({z}_{\mathrm{0}} \right)+{i}\pi\:{z}_{\mathrm{1}} −{i}\pi{z}_{\mathrm{2}} −{i}\pi{z}_{\mathrm{3}} \right\} \\ $$$$=−\frac{{i}\pi}{\mathrm{8}}\left\{\:{z}_{\mathrm{0}} +{z}_{\mathrm{1}} −\left({z}_{\mathrm{2}} +{z}_{\mathrm{3}} \right)\right\}\:\:{but}\:\:{z}_{\mathrm{0}} \:+{z}_{\mathrm{1}} ={e}^{\frac{{i}\pi}{\mathrm{4}}} \:+{e}^{{i}\left(\pi−\frac{\pi}{\mathrm{4}}\right)} \\ $$$$={e}^{\frac{{i}\pi}{\mathrm{4}}} −{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:=\mathrm{2}{i}\:{sin}\left(\frac{\pi}{\mathrm{4}}\right)\:=\mathrm{2}{i}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:={i}\sqrt{\mathrm{2}} \\ $$$${z}_{\mathrm{2}} \:+{z}_{\mathrm{3}} ={e}^{{i}\frac{\mathrm{5}\pi}{\mathrm{4}}} \:+{e}^{\frac{{i}\mathrm{7}\pi}{\mathrm{4}}} \:={e}^{{i}\left(\pi+\frac{\pi}{\mathrm{4}}\right)} \:+{e}^{{i}\left(\mathrm{2}\pi−\frac{\pi}{\mathrm{4}}\right)} =−{e}^{\frac{{i}\pi}{\mathrm{4}}} \:\:+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \\ $$$$=−\left({e}^{\frac{{i}\pi}{\mathrm{4}}} −{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\:=−\mathrm{2}{isin}\left(\frac{\pi}{\mathrm{4}}\right)\:=−\mathrm{2}{i}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:=−{i}\sqrt{\mathrm{2}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{4}} }\:=−\frac{{i}\pi}{\mathrm{8}}\left\{{i}\sqrt{\mathrm{2}}+{i}\sqrt{\mathrm{2}}\right\}\:=\frac{\pi}{\mathrm{8}}\left(\mathrm{2}\sqrt{\mathrm{2}}\right)\:=\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$$$ \\ $$