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Question Number 121492 by mathmax by abdo last updated on 08/Nov/20

calculate ∫_0 ^(+∞) (dx/((x^4 +1)^3 ))

$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{+\infty} \:\:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{3}} } \\ $$

Answered by mnjuly1970 last updated on 09/Nov/20

solution: Θ=^(x^4 =y) (1/4)∫_0 ^( ∞) ((y^((((−3)/4)+1)−1) dy)/((y+1)^3 ))=(1/4)β((1/4) ,3−(1/4)) =(1/4) ((Γ((1/4))Γ(3−(1/4)))/(Γ(3)))=(1/4) ∗(((7/4)Γ((7/4))Γ((1/4)))/2) =(1/4)∗(((7/4)∗(3/4)Γ((3/4))Γ((1/4)))/2) =((21)/(128)) ∗(π/(sin((π/4))))=((42π)/(128(√2))) =((21π)/(64(√2)))

$${solution}: \\ $$$$\:\:\Theta\overset{{x}^{\mathrm{4}} ={y}} {=}\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\:\infty} \:\frac{{y}^{\left(\frac{−\mathrm{3}}{\mathrm{4}}+\mathrm{1}\right)−\mathrm{1}} {dy}}{\left({y}+\mathrm{1}\right)^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{4}}\beta\left(\frac{\mathrm{1}}{\mathrm{4}}\:,\mathrm{3}−\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\Gamma\left(\mathrm{3}−\frac{\mathrm{1}}{\mathrm{4}}\right)}{\Gamma\left(\mathrm{3}\right)}=\frac{\mathrm{1}}{\mathrm{4}}\:\ast\frac{\frac{\mathrm{7}}{\mathrm{4}}\Gamma\left(\frac{\mathrm{7}}{\mathrm{4}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\mathrm{2}} \\ $$$$\:=\frac{\mathrm{1}}{\mathrm{4}}\ast\frac{\frac{\mathrm{7}}{\mathrm{4}}\ast\frac{\mathrm{3}}{\mathrm{4}}\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\mathrm{2}} \\ $$$$=\frac{\mathrm{21}}{\mathrm{128}}\:\ast\frac{\pi}{{sin}\left(\frac{\pi}{\mathrm{4}}\right)}=\frac{\mathrm{42}\pi}{\mathrm{128}\sqrt{\mathrm{2}}} \\ $$$$\:\:=\frac{\mathrm{21}\pi}{\mathrm{64}\sqrt{\mathrm{2}}} \\ $$

Answered by MJS_new last updated on 09/Nov/20

∫(dx/((x^4 +1)^3 ))= [Ostrogradski] =((x(7x^4 +11))/(32(x^4 +1)^2 ))+((21)/(32))∫(dx/(x^4 +1))= =((x(7x^4 +11))/(32(x^4 +1)^2 ))+((21(√2))/(256))ln ((x^2 +(√2)x+1)/(x^2 −(√2)x+1)) +((21(√2))/(128))(tan^(−1) ((√2)x−1) +tan^(−1) ((√2)x+1)) +C ⇒ answer is ((21(√2))/(128))π

$$\int\frac{{dx}}{\left({x}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{3}} }= \\ $$$$\:\:\:\:\:\left[\mathrm{Ostrogradski}\right] \\ $$$$=\frac{{x}\left(\mathrm{7}{x}^{\mathrm{4}} +\mathrm{11}\right)}{\mathrm{32}\left({x}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{21}}{\mathrm{32}}\int\frac{{dx}}{{x}^{\mathrm{4}} +\mathrm{1}}= \\ $$$$=\frac{{x}\left(\mathrm{7}{x}^{\mathrm{4}} +\mathrm{11}\right)}{\mathrm{32}\left({x}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{21}\sqrt{\mathrm{2}}}{\mathrm{256}}\mathrm{ln}\:\frac{{x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}+\mathrm{1}}{{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{1}}\:+\frac{\mathrm{21}\sqrt{\mathrm{2}}}{\mathrm{128}}\left(\mathrm{tan}^{−\mathrm{1}} \:\left(\sqrt{\mathrm{2}}{x}−\mathrm{1}\right)\:+\mathrm{tan}^{−\mathrm{1}} \:\left(\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)\right)\:+{C} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\frac{\mathrm{21}\sqrt{\mathrm{2}}}{\mathrm{128}}\pi \\ $$

Answered by mathmax by abdo last updated on 09/Nov/20

A =∫_0 ^∞ (dx/((x^4 +1)^3 )) ⇒2A =∫_(−∞) ^(+∞) (dx/((x^4 +1)^3 )) =∫_(−∞) ^(+∞) (dx/((x^2 −i)^2 (x^2 +i)^2 )) let ϕ(z) =(1/((z^2 −i)^2 (z^2 +i)^2 )) ⇒ϕ(z)=(1/((z−(√i))^2 (z+(√i))^2 (z−(√(−i)))^2 (z+(√(−i)))^2 )) =(1/((z−e^((iπ)/4) )^2 (z+e^((iπ)/4) )^2 (z−e^(−((iπ)/4)) )^2 (z+e^(−((iπ)/4)) )^2 )) the poles of ϕ are +^− e^((iπ)/4) and +^− e^(−((iπ)/4)) (doubles) so ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ { Res(ϕ,e^((iπ)/4) ) +Res(ϕ,−e^(−((iπ)/4)) )} Res(ϕ,e^((iπ)/4) ) =lim_(z→e^((iπ)/4) ) (1/((2−1)!)){(z−e^((iπ)/4) )^2 ϕ(z)}^((1)) =lim_(z→e^((iπ)/4) ) {(1/((z+e^((iπ)/4) )^2 (z^2 +i)^2 ))}^((1)) =−lim_(z→e^((iπ)/4) ) ((2(z+e^((iπ)/4) )(z^2 +i)^2 +4z(z^2 +i)(z+e^((iπ)/4) )^2 )/((z+e^((iπ)/4) )^4 (z^2 +i)^4 )) =−lim_(z→e^((iπ)/4) ) ((2(z^2 +i)+4z(z+e^((iπ)/4) ))/((z+e^((iπ)/4) )^3 (z^2 +i)^3 )) =−((2(2i)+4e^((iπ)/4) ×2e^((iπ)/4) )/((2e^((iπ)/4) )^3 (2i)^3 )) =−((4i+8i)/(2^3 (−8i)))e^(−((i3π)/4)) =((12)/(64)) e^(−((i3π)/4)) rest to find Res(ϕ,−e^(−((iπ)/4)) )....be continued....

$$\mathrm{A}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{4}} \:+\mathrm{1}\right)^{\mathrm{3}} }\:\Rightarrow\mathrm{2A}\:=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{3}} }\:=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{i}\right)^{\mathrm{2}} \left(\mathrm{x}^{\mathrm{2}} +\mathrm{i}\right)^{\mathrm{2}} } \\ $$$$\mathrm{let}\:\varphi\left(\mathrm{z}\right)\:=\frac{\mathrm{1}}{\left(\mathrm{z}^{\mathrm{2}} −\mathrm{i}\right)^{\mathrm{2}} \left(\mathrm{z}^{\mathrm{2}} +\mathrm{i}\right)^{\mathrm{2}} }\:\Rightarrow\varphi\left(\mathrm{z}\right)=\frac{\mathrm{1}}{\left(\mathrm{z}−\sqrt{\mathrm{i}}\right)^{\mathrm{2}} \left(\mathrm{z}+\sqrt{\mathrm{i}}\right)^{\mathrm{2}} \left(\mathrm{z}−\sqrt{−\mathrm{i}}\right)^{\mathrm{2}} \left(\mathrm{z}+\sqrt{−\mathrm{i}}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{z}−\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} \left(\mathrm{z}+\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} \left(\mathrm{z}−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} \left(\mathrm{z}+\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} } \\ $$$$\mathrm{the}\:\mathrm{poles}\:\mathrm{of}\:\varphi\:\mathrm{are}\:\overset{−} {+}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \:\mathrm{and}\:\overset{−} {+}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \:\left(\mathrm{doubles}\right)\:\mathrm{so} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\:\left\{\:\mathrm{Res}\left(\varphi,\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\:+\mathrm{Res}\left(\varphi,−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\right\} \\ $$$$\mathrm{Res}\left(\varphi,\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\:=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} } \:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\left(\mathrm{z}−\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} \varphi\left(\mathrm{z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} } \:\:\:\left\{\frac{\mathrm{1}}{\left(\mathrm{z}+\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} \left(\mathrm{z}^{\mathrm{2}} +\mathrm{i}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$=−\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} } \:\:\:\:\:\frac{\mathrm{2}\left(\mathrm{z}+\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\left(\mathrm{z}^{\mathrm{2}} +\mathrm{i}\right)^{\mathrm{2}} +\mathrm{4z}\left(\mathrm{z}^{\mathrm{2}} +\mathrm{i}\right)\left(\mathrm{z}+\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} }{\left(\mathrm{z}+\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{4}} \left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{i}\right)^{\mathrm{4}} } \\ $$$$=−\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} } \:\:\:\:\frac{\mathrm{2}\left(\mathrm{z}^{\mathrm{2}} +\mathrm{i}\right)+\mathrm{4z}\left(\mathrm{z}+\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)}{\left(\mathrm{z}+\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{3}} \left(\mathrm{z}^{\mathrm{2}} +\mathrm{i}\right)^{\mathrm{3}} } \\ $$$$=−\frac{\mathrm{2}\left(\mathrm{2i}\right)+\mathrm{4e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} ×\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} }{\left(\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{3}} \left(\mathrm{2i}\right)^{\mathrm{3}} }\:=−\frac{\mathrm{4i}+\mathrm{8i}}{\mathrm{2}^{\mathrm{3}} \:\left(−\mathrm{8i}\right)}\mathrm{e}^{−\frac{\mathrm{i3}\pi}{\mathrm{4}}} \:=\frac{\mathrm{12}}{\mathrm{64}}\:\mathrm{e}^{−\frac{\mathrm{i3}\pi}{\mathrm{4}}} \\ $$$$\mathrm{rest}\:\mathrm{to}\:\mathrm{find}\:\:\mathrm{Res}\left(\varphi,−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)....\mathrm{be}\:\mathrm{continued}.... \\ $$

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