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Question Number 93633 by abdomathmax last updated on 14/May/20
calvulate ∫_0 ^∞ ((cos(πx))/(1+x^4 ))dx
$${calvulate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left(\pi{x}\right)}{\mathrm{1}+{x}^{\mathrm{4}} }{dx} \\ $$
Commented by mathmax by abdo last updated on 15/May/20
A =∫_0 ^∞ ((cos(πx))/(x^4 +1))dx ⇒2A =∫_(−∞) ^(+∞) ((cos(πx))/(x^4 +1))dx =Re(∫_(−∞) ^(+∞) (e^(iπx) /(x^4 +1))dx) let ϕ(z) =(e^(iπz) /(z^4 +1)) poles of ϕ? ϕ(z) =(e^(iπz) /((z^2 −i)(z^2 +i))) =(e^(iπz) /((z−(√i))(z+(√i))(z−(√(−i)))(z+(√(−i))))) =(e^(iπz) /((z−e^((iπ)/4) )(z+e^((iπ)/4) )(z−e^(−((iπ)/4)) )(z+e^(−((iπ)/4)) ))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{Res(ϕ,e^((iπ)/4) )+Res(ϕ,−e^(−((iπ)/4)) )} Res(ϕ,e^((iπ)/4) ) =(e^(iπ(e^((iπ)/4) )) /(2e^((iπ)/4) (2i))) =(1/(4i))e^(−((iπ)/4)) ×e^(iπ((1/( (√2)))+(i/( (√2))))) =(1/(4i))e^(−((iπ)/4)) ×e^(−(π/( (√2)))) ×e^((iπ)/( (√2))) =(e^(−(π/( (√2)))) /(4i)) e^(i((π/( (√2)))−(π/4))) Res(ϕ,−e^(−i(π/4)) ) = (e^(iπ (−e^(−((iπ)/4)) )) /(−2e^(−((iπ)/4)) (−2i))) =(1/(4i))e^((iπ)/4) e^(−iπ((1/( (√2)))−(i/( (√2))))) =(1/(4i)) e^((iπ)/4) ×e^(π/( (√2))) e^(−((iπ)/( (√2)))) =(e^(π/( (√2))) /(4i)) ×e^(i((π/4)−(π/( (√2))))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =(π/2){ e^(−(π/2)) ×e^(i((π/( (√2)))−(π/4))) +e^(π/2) × e^(i((π/4)−(π/( (√2))))) } =((πe^(−(π/2)) )/2)( cos((π/( (√2)))−(π/4))+isin((π/( (√2)))−(π/4)))+((πe^(π/2) )/2)(cos((π/4)−(π/( (√2))))+isin((π/4)−(π/( (√2))))) =((π/2)e^(−(π/2)) +(π/2)e^(π/2) )cos((π/( (√2)))−(π/4)) +i(.....) ⇒ 2A =π ch((π/2))cos((π/( (√2)))−(π/4)) ⇒ A =(π/2)ch((π/2))cos((π/( (√2)))−(π/4))
$${A}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{cos}\left(\pi{x}\right)}{{x}^{\mathrm{4}} \:+\mathrm{1}}{dx}\:\Rightarrow\mathrm{2}{A}\:=\int_{−\infty} ^{+\infty} \:\frac{{cos}\left(\pi{x}\right)}{{x}^{\mathrm{4}} +\mathrm{1}}{dx}\:={Re}\left(\int_{−\infty} ^{+\infty} \:\frac{{e}^{{i}\pi{x}} }{{x}^{\mathrm{4}} \:+\mathrm{1}}{dx}\right) \\ $$$${let}\:\varphi\left({z}\right)\:=\frac{{e}^{{i}\pi{z}} }{{z}^{\mathrm{4}} \:+\mathrm{1}}\:\:\:\:{poles}\:{of}\:\varphi? \\ $$$$\varphi\left({z}\right)\:=\frac{{e}^{{i}\pi{z}} }{\left({z}^{\mathrm{2}} −{i}\right)\left({z}^{\mathrm{2}} \:+{i}\right)}\:=\frac{{e}^{{i}\pi{z}} }{\left({z}−\sqrt{{i}}\right)\left({z}+\sqrt{{i}}\right)\left({z}−\sqrt{−{i}}\right)\left({z}+\sqrt{−{i}}\right)} \\ $$$$=\frac{{e}^{{i}\pi{z}} }{\left({z}−{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{{Res}\left(\varphi,{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)+{Res}\left(\varphi,−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\right\} \\ $$$${Res}\left(\varphi,{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\:=\frac{{e}^{{i}\pi\left({e}^{\frac{{i}\pi}{\mathrm{4}}} \right)} }{\mathrm{2}{e}^{\frac{{i}\pi}{\mathrm{4}}} \left(\mathrm{2}{i}\right)}\:=\frac{\mathrm{1}}{\mathrm{4}{i}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:×{e}^{{i}\pi\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{{i}}{\:\sqrt{\mathrm{2}}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}{i}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:×{e}^{−\frac{\pi}{\:\sqrt{\mathrm{2}}}} ×{e}^{\frac{{i}\pi}{\:\sqrt{\mathrm{2}}}} \:=\frac{{e}^{−\frac{\pi}{\:\sqrt{\mathrm{2}}}} }{\mathrm{4}{i}}\:{e}^{{i}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}−\frac{\pi}{\mathrm{4}}\right)} \\ $$$${Res}\left(\varphi,−{e}^{−{i}\frac{\pi}{\mathrm{4}}} \right)\:=\:\frac{{e}^{{i}\pi\:\left(−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)} }{−\mathrm{2}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \left(−\mathrm{2}{i}\right)}\:=\frac{\mathrm{1}}{\mathrm{4}{i}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \:{e}^{−{i}\pi\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{{i}}{\:\sqrt{\mathrm{2}}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}{i}}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \:×{e}^{\frac{\pi}{\:\sqrt{\mathrm{2}}}} \:{e}^{−\frac{{i}\pi}{\:\sqrt{\mathrm{2}}}} \:\:=\frac{{e}^{\frac{\pi}{\:\sqrt{\mathrm{2}}}} }{\mathrm{4}{i}}\:\:×{e}^{{i}\left(\frac{\pi}{\mathrm{4}}−\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)} \:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\frac{\pi}{\mathrm{2}}\left\{\:\:{e}^{−\frac{\pi}{\mathrm{2}}} \:×{e}^{{i}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}−\frac{\pi}{\mathrm{4}}\right)} \:+{e}^{\frac{\pi}{\mathrm{2}}} \:×\:{e}^{{i}\left(\frac{\pi}{\mathrm{4}}−\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)} \right\} \\ $$$$=\frac{\pi{e}^{−\frac{\pi}{\mathrm{2}}} }{\mathrm{2}}\left(\:{cos}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}−\frac{\pi}{\mathrm{4}}\right)+{isin}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}−\frac{\pi}{\mathrm{4}}\right)\right)+\frac{\pi{e}^{\frac{\pi}{\mathrm{2}}} }{\mathrm{2}}\left({cos}\left(\frac{\pi}{\mathrm{4}}−\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)+{isin}\left(\frac{\pi}{\mathrm{4}}−\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)\right) \\ $$$$=\left(\frac{\pi}{\mathrm{2}}{e}^{−\frac{\pi}{\mathrm{2}}} \:\:\:+\frac{\pi}{\mathrm{2}}{e}^{\frac{\pi}{\mathrm{2}}} \right){cos}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}−\frac{\pi}{\mathrm{4}}\right)\:+{i}\left(…..\right)\:\Rightarrow \\ $$$$\mathrm{2}{A}\:=\pi\:{ch}\left(\frac{\pi}{\mathrm{2}}\right){cos}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}−\frac{\pi}{\mathrm{4}}\right)\:\Rightarrow\:{A}\:=\frac{\pi}{\mathrm{2}}{ch}\left(\frac{\pi}{\mathrm{2}}\right){cos}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}−\frac{\pi}{\mathrm{4}}\right) \\ $$

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