Question Number 40007 by math khazana by abdo last updated on 15/Jul/18
$${find}\:{thevalue}\:{of}\:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{\mathrm{1}\:+{x}^{\mathrm{6}} }\:\:{by}\:{using}\:{the} \\ $$$${value}\:{of}\:\:\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{{x}−{z}}\:{with}\:{z}\:\in\:{C}\:{and}\:{Im}\left({z}\right)\neq\mathrm{0} \\ $$$$ \\ $$
Commented by prof Abdo imad last updated on 16/Jul/18
![let I = ∫_0 ^∞ (dx/(1+x^6 )) 2I = ∫_(−∞) ^(+∞) (dx/(1+x^6 )) let drcompose inside C(x) F(x)= (1/(x^6 +1)) poles of F? x^6 =−1 ⇒ x_k = e^(i((2k+1)π)/6)) with k∈[[0,5]] x_0 = e^((iπ)/6) , x_1 = e^(i(π/2)) , x_2 = e^((i5π)/6) = −e^(−((iπ)/6)) x_3 = e^(i((7π)/6)) =−e^((iπ)/6) , x_4 =e^((i9π)/6) =e^((i3π)/2) , x_5 = e^((i11π)/6) =e^(−((iπ)/6)) we have λ_i = (1/(6x_i ^5 )) =−(1/6) x_i ⇒ F(x)=−(1/6){ (x_0 /(x−x_0 )) +(x_1 /(x−x_1 )) +(x_2 /(x−x_2 )) +(x_3 /(x−x_3 )) + (x_4 /(x−x_4 )) +(x_5 /(x−x_5 ))} ⇒ ∫_(−∞) ^(+∞) F(x)dx =−(1/6){ x_0 ∫_(−∞) ^(+∞) (dx/(x−x_0 )) +x_1 ∫_(−∞) ^(+∞) (dx/(x−x_1 )) +x_2 ∫_(−∞) ^(+∞) (dx/(x−x_2 )) + x_3 ∫_(−∞) ^(+∞) (dx/(x−x_3 )) + x_4 ∫_(−∞) ^(+∞) (dx/(x−x_4 )) +x_5 ∫_(−∞) ^(+∞) (dx/(x−x_5 ))} but we have proved that ∫_(−∞) ^(+∞) (dx/(x−z)) =iπ if Im(z)>0 and −iπ if Im(z)<0 ∫_(−∞) ^(+∞) F(x)dx =−(1/6){x_0 (iπ) +x_1 (iπ) +x_2 (iπ) +x_3 (−iπ)+x_4 (−iπ) +x_5 (−iπ)} =−((iπ)/6){ x_0 +x_1 +x_2 −x_3 −x_4 −x_5 } =−((iπ)/6){ e^((iπ)/6) + i −e^(−((iπ)/6)) +e^((iπ)/6) −e^((i3π)/2) −e^(−((iπ)/6)) } =−((iπ)/6){ 2i +2(e^((iπ)/6) −e^(−((iπ)/6)) )} =(π/3) −((iπ)/3) 2i (1/2) =(π/3) +(π/3) =((2π)/3) 2I =((2π)/3) ⇒ I =(π/3) .](https://www.tinkutara.com/question/Q40164.png)
$${let}\:{I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{6}} } \\ $$$$\mathrm{2}{I}\:=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{6}} }\:{let}\:{drcompose}\:{inside}\:{C}\left({x}\right) \\ $$$${F}\left({x}\right)=\:\frac{\mathrm{1}}{{x}^{\mathrm{6}} +\mathrm{1}}\:\:{poles}\:{of}\:{F}? \\ $$$${x}^{\mathrm{6}} \:=−\mathrm{1}\:\:\Rightarrow\:{x}_{{k}} =\:{e}^{{i}\frac{\left.\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{6}}} \:\:{with}\:{k}\in\left[\left[\mathrm{0},\mathrm{5}\right]\right] \\ $$$${x}_{\mathrm{0}} =\:{e}^{\frac{{i}\pi}{\mathrm{6}}} \:\:,\:{x}_{\mathrm{1}} =\:{e}^{{i}\frac{\pi}{\mathrm{2}}} \:\:\:,\:{x}_{\mathrm{2}} =\:{e}^{\frac{{i}\mathrm{5}\pi}{\mathrm{6}}} \:=\:−{e}^{−\frac{{i}\pi}{\mathrm{6}}} \\ $$$${x}_{\mathrm{3}} =\:{e}^{{i}\frac{\mathrm{7}\pi}{\mathrm{6}}} \:=−{e}^{\frac{{i}\pi}{\mathrm{6}}} \:,\:{x}_{\mathrm{4}} ={e}^{\frac{{i}\mathrm{9}\pi}{\mathrm{6}}} \:={e}^{\frac{{i}\mathrm{3}\pi}{\mathrm{2}}} \:\:,\:{x}_{\mathrm{5}} =\:{e}^{\frac{{i}\mathrm{11}\pi}{\mathrm{6}}} \:={e}^{−\frac{{i}\pi}{\mathrm{6}}} \\ $$$${we}\:{have}\:\lambda_{{i}} =\:\frac{\mathrm{1}}{\mathrm{6}{x}_{{i}} ^{\mathrm{5}} }\:=−\frac{\mathrm{1}}{\mathrm{6}}\:{x}_{{i}} \:\Rightarrow \\ $$$${F}\left({x}\right)=−\frac{\mathrm{1}}{\mathrm{6}}\left\{\:\:\frac{{x}_{\mathrm{0}} }{{x}−{x}_{\mathrm{0}} }\:+\frac{{x}_{\mathrm{1}} }{{x}−{x}_{\mathrm{1}} }\:+\frac{{x}_{\mathrm{2}} }{{x}−{x}_{\mathrm{2}} }\:+\frac{{x}_{\mathrm{3}} }{{x}−{x}_{\mathrm{3}} }\right. \\ $$$$\left.+\:\frac{{x}_{\mathrm{4}} }{{x}−{x}_{\mathrm{4}} }\:+\frac{{x}_{\mathrm{5}} }{{x}−{x}_{\mathrm{5}} }\right\}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:{F}\left({x}\right){dx}\:=−\frac{\mathrm{1}}{\mathrm{6}}\left\{\:{x}_{\mathrm{0}} \int_{−\infty} ^{+\infty} \:\frac{{dx}}{{x}−{x}_{\mathrm{0}} }\:+{x}_{\mathrm{1}} \:\int_{−\infty} ^{+\infty} \:\frac{{dx}}{{x}−{x}_{\mathrm{1}} }\right. \\ $$$$+{x}_{\mathrm{2}} \:\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{{x}−{x}_{\mathrm{2}} }\:+\:{x}_{\mathrm{3}} \:\:\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{{x}−{x}_{\mathrm{3}} }\:+\:{x}_{\mathrm{4}} \:\:\int_{−\infty} ^{+\infty} \:\frac{{dx}}{{x}−{x}_{\mathrm{4}} } \\ $$$$\left.+{x}_{\mathrm{5}} \:\int_{−\infty} ^{+\infty} \:\:\:\frac{{dx}}{{x}−{x}_{\mathrm{5}} }\right\}\:{but}\:{we}\:{have}\:{proved}\:{that} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\:\frac{{dx}}{{x}−{z}}\:\:={i}\pi\:\:{if}\:{Im}\left({z}\right)>\mathrm{0}\:{and}\:−{i}\pi\:{if}\:{Im}\left({z}\right)<\mathrm{0} \\ $$$$\int_{−\infty} ^{+\infty} \:{F}\left({x}\right){dx}\:=−\frac{\mathrm{1}}{\mathrm{6}}\left\{{x}_{\mathrm{0}} \left({i}\pi\right)\:+{x}_{\mathrm{1}} \left({i}\pi\right)\:+{x}_{\mathrm{2}} \left({i}\pi\right)\right. \\ $$$$\left.+{x}_{\mathrm{3}} \left(−{i}\pi\right)+{x}_{\mathrm{4}} \left(−{i}\pi\right)\:+{x}_{\mathrm{5}} \left(−{i}\pi\right)\right\} \\ $$$$=−\frac{{i}\pi}{\mathrm{6}}\left\{\:{x}_{\mathrm{0}} \:+{x}_{\mathrm{1}} \:+{x}_{\mathrm{2}} \:−{x}_{\mathrm{3}} \:−{x}_{\mathrm{4}} −{x}_{\mathrm{5}} \right\} \\ $$$$=−\frac{{i}\pi}{\mathrm{6}}\left\{\:{e}^{\frac{{i}\pi}{\mathrm{6}}} \:+\:{i}\:\:−{e}^{−\frac{{i}\pi}{\mathrm{6}}} \:+{e}^{\frac{{i}\pi}{\mathrm{6}}} \:−{e}^{\frac{{i}\mathrm{3}\pi}{\mathrm{2}}} \:−{e}^{−\frac{{i}\pi}{\mathrm{6}}} \right\} \\ $$$$=−\frac{{i}\pi}{\mathrm{6}}\left\{\:\:\mathrm{2}{i}\:\:+\mathrm{2}\left({e}^{\frac{{i}\pi}{\mathrm{6}}} \:−{e}^{−\frac{{i}\pi}{\mathrm{6}}} \right)\right\} \\ $$$$=\frac{\pi}{\mathrm{3}}\:−\frac{{i}\pi}{\mathrm{3}}\:\mathrm{2}{i}\:\frac{\mathrm{1}}{\mathrm{2}}\:=\frac{\pi}{\mathrm{3}}\:+\frac{\pi}{\mathrm{3}}\:=\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$$\mathrm{2}{I}\:=\frac{\mathrm{2}\pi}{\mathrm{3}}\:\Rightarrow\:{I}\:=\frac{\pi}{\mathrm{3}}\:. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$