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Question Number 43808 by maxmathsup by imad last updated on 15/Sep/18
let I = ∫_0 ^∞   (dx/(x^4 −i)) and J = ∫_0 ^∞     (dx/(x^4  +i))  1) find  values of I and J  2) calculate I +J  3) calculate ∫_0 ^∞    (dx/(x^(8 )  +1))
$${let}\:{I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{{x}^{\mathrm{4}} −{i}}\:{and}\:{J}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{{x}^{\mathrm{4}} \:+{i}} \\ $$$$\left.\mathrm{1}\right)\:{find}\:\:{values}\:{of}\:{I}\:{and}\:{J} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{I}\:+{J} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{{x}^{\mathrm{8}\:} \:+\mathrm{1}} \\ $$
Commented by maxmathsup by imad last updated on 21/Sep/18
1) we have i =(e^((iπ)/8) )^4  changement   x =t e^((iπ)/8)   give  J= ∫_0 ^∞   ((e^((iπ)/8)  dt)/(i t^4  +i)) =−i e^((iπ)/8)   ∫_0 ^∞    (dt/(1+t^4 )) also cha7gement t =u^(1/4)  ⇒  ∫_0 ^∞   (dt/(1+t^4 )) =∫_0 ^∞    (1/(1+u)) (1/4) u^((1/4)−1) du =(1/4) (π/(sin((π/4)))) =(π/4) (1/((√2)/2)) =(π/(2(√2))) ⇒  J =−((iπ)/(2(√2))) e^((iπ)/8)   and I =J^−  =((iπ)/(2(√2))) e^(−((iπ)/8))   2) we have I +J =((iπ)/(2(√2))) e^(−((iπ)/8))  −((iπ)/(2(√2))) e^((iπ)/8)  =−((iπ)/(2(√2))){e^((iπ)/8)  −e^(−i(π/8)) }  =−((iπ)/(2(√2)))(2i sin((π/8))) =(π/( (√2))) ((√(2−(√2)))/2) =(π/(2(√2)))(√(2−(√2))).  3) we have I −J =∫_0 ^∞   ((1/(x^4 −i)) −(1/(x^4  +i)))dx  =∫_0 ^∞   ((2i)/(x^8  +1)) dx  ⇒∫_0 ^∞   (dx/(1+x^8 )) =(1/(2i)){I −J} but   I −J = ((iπ)/(2(√2))) e^(−((iπ)/8))   +((iπ)/(2(√2))) e^((iπ)/8)  =((iπ)/(2(√2))) (2cos((π/8))} =((iπ)/( (√2))) ((√(2+(√2)))/2) ⇒  ∫_0 ^∞    (dx/(1+x^8 )) =(1/(2i)) ((iπ)/(2(√2)))(√(2+(√2)))=(π/(4(√2))) (√(2+(√2))).  remark we have I +J =∫_0 ^∞  ((1/(x^4 −i)) +(1/(x^4  +i)))dx  = ∫_0 ^∞   ((2x^4 )/(1+x^8 )) dx =(π/(2(√2)))(√(2−(√2))) ⇒∫_0 ^∞     (x^4 /(1+x^8 ))dx =(π/(4(√2)))(√(2−(√2))).
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{i}\:=\left({e}^{\frac{{i}\pi}{\mathrm{8}}} \right)^{\mathrm{4}} \:{changement}\:\:\:{x}\:={t}\:{e}^{\frac{{i}\pi}{\mathrm{8}}} \:\:{give} \\ $$$${J}=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{\frac{{i}\pi}{\mathrm{8}}} \:{dt}}{{i}\:{t}^{\mathrm{4}} \:+{i}}\:=−{i}\:{e}^{\frac{{i}\pi}{\mathrm{8}}} \:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{4}} }\:{also}\:{cha}\mathrm{7}{gement}\:{t}\:={u}^{\frac{\mathrm{1}}{\mathrm{4}}} \:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{4}} }\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}}{\mathrm{1}+{u}}\:\frac{\mathrm{1}}{\mathrm{4}}\:{u}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} {du}\:=\frac{\mathrm{1}}{\mathrm{4}}\:\frac{\pi}{{sin}\left(\frac{\pi}{\mathrm{4}}\right)}\:=\frac{\pi}{\mathrm{4}}\:\frac{\mathrm{1}}{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:\Rightarrow \\ $$$${J}\:=−\frac{{i}\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:{e}^{\frac{{i}\pi}{\mathrm{8}}} \:\:{and}\:{I}\:=\overset{−} {{J}}\:=\frac{{i}\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:{e}^{−\frac{{i}\pi}{\mathrm{8}}} \\ $$$$\left.\mathrm{2}\right)\:{we}\:{have}\:{I}\:+{J}\:=\frac{{i}\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:{e}^{−\frac{{i}\pi}{\mathrm{8}}} \:−\frac{{i}\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:{e}^{\frac{{i}\pi}{\mathrm{8}}} \:=−\frac{{i}\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\left\{{e}^{\frac{{i}\pi}{\mathrm{8}}} \:−{e}^{−{i}\frac{\pi}{\mathrm{8}}} \right\} \\ $$$$=−\frac{{i}\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\left(\mathrm{2}{i}\:{sin}\left(\frac{\pi}{\mathrm{8}}\right)\right)\:=\frac{\pi}{\:\sqrt{\mathrm{2}}}\:\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}}{\mathrm{2}}\:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}. \\ $$$$\left.\mathrm{3}\right)\:{we}\:{have}\:{I}\:−{J}\:=\int_{\mathrm{0}} ^{\infty} \:\:\left(\frac{\mathrm{1}}{{x}^{\mathrm{4}} −{i}}\:−\frac{\mathrm{1}}{{x}^{\mathrm{4}} \:+{i}}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2}{i}}{{x}^{\mathrm{8}} \:+\mathrm{1}}\:{dx}\:\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{8}} }\:=\frac{\mathrm{1}}{\mathrm{2}{i}}\left\{{I}\:−{J}\right\}\:{but}\: \\ $$$${I}\:−{J}\:=\:\frac{{i}\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:{e}^{−\frac{{i}\pi}{\mathrm{8}}} \:\:+\frac{{i}\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:{e}^{\frac{{i}\pi}{\mathrm{8}}} \:=\frac{{i}\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:\left(\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{8}}\right)\right\}\:=\frac{{i}\pi}{\:\sqrt{\mathrm{2}}}\:\frac{\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}{\mathrm{2}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{8}} }\:=\frac{\mathrm{1}}{\mathrm{2}{i}}\:\frac{{i}\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}=\frac{\pi}{\mathrm{4}\sqrt{\mathrm{2}}}\:\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}. \\ $$$${remark}\:{we}\:{have}\:{I}\:+{J}\:=\int_{\mathrm{0}} ^{\infty} \:\left(\frac{\mathrm{1}}{{x}^{\mathrm{4}} −{i}}\:+\frac{\mathrm{1}}{{x}^{\mathrm{4}} \:+{i}}\right){dx} \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2}{x}^{\mathrm{4}} }{\mathrm{1}+{x}^{\mathrm{8}} }\:{dx}\:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{x}^{\mathrm{4}} }{\mathrm{1}+{x}^{\mathrm{8}} }{dx}\:=\frac{\pi}{\mathrm{4}\sqrt{\mathrm{2}}}\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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