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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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