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Question Number 142425 by Mathspace last updated on 31/May/21

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

$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{log}^{\mathrm{3}} {x}}{\mathrm{1}+{x}^{\mathrm{3}} }{dx} \\ $$

Answered by mathmax by abdo last updated on 02/Jun/21

Φ=∫_0 ^∞ ((log^3 x)/(1+x^3 ))dx changement x^3 =t give Φ=∫_0 ^∞ (((log(t^(1/3) ))^3 )/(1+t))(1/3)t^((1/3)−1) dt =(1/3^4 )∫_0 ^∞ t^((1/3)−1) ((log^3 t)/(1+t))dt let f(a)=∫_0 ^∞ (t^(a−1) /(1+t))dt with 0<a<1 we have f(a)=(π/(sin(πa))) and f^′ (a) =∫_0 ^∞ t^(a−1) ((logt)/(1+t))dt ⇒f^(,,) (a)=∫_0 ^∞ ((t^(a−1) log^2 t)/(1+t))dt ⇒ f^((3)) (a)=∫_0 ^∞ ((t^(a−1) log^3 t)/(1+t))dt ⇒f^((3)) ((1/3))=∫_0 ^∞ ((t^((1/3)−1) log^3 t)/(1+t))dt =3^4 Φ but f(a)=(π/(sin(πa))) ⇒f^′ (a)=−π^2 ((cos(πa))/(sin^2 (πa))) ⇒ f^((2)) (a)=−π^2 .((−πsin(πa)−2πsin(πa)cos(πa))/(sin^4 (πa))) =π^3 .((1+2cos(πa))/(sin^3 (πa))) ⇒ f^((3)) (a)=π^3 .((−2πsin(πa).sin^3 (πa)−3πsin^2 (πa)cos(πa))/(sin^6 (πa))) =π^4 .((−2sin^2 (πa)−3 cos(πa))/(sin^4 (πa))) ⇒ f^((3)) ((1/3))=−π^4 ×((2(((√3)/2))^2 +3.(1/2))/((((√3)/2))^4 ))=−π^4 .(((3/2)+(3/2))/(9/(16))) =(−π^4 ).(((16)/9))(3) =−((16π^4 )/3) ⇒Φ=−((16π^4 )/3^5 )

$$\Phi=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{log}^{\mathrm{3}} \mathrm{x}}{\mathrm{1}+\mathrm{x}^{\mathrm{3}} }\mathrm{dx}\:\mathrm{changement}\:\mathrm{x}^{\mathrm{3}} \:=\mathrm{t}\:\mathrm{give} \\ $$$$\Phi=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\left(\mathrm{log}\left(\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)\right)^{\mathrm{3}} }{\mathrm{1}+\mathrm{t}}\frac{\mathrm{1}}{\mathrm{3}}\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}} \:\mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{4}} }\int_{\mathrm{0}} ^{\infty} \mathrm{t}^{\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}} \:\:\frac{\mathrm{log}^{\mathrm{3}} \mathrm{t}}{\mathrm{1}+\mathrm{t}}\mathrm{dt} \\ $$$$\mathrm{let}\:\mathrm{f}\left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{t}^{\mathrm{a}−\mathrm{1}} }{\mathrm{1}+\mathrm{t}}\mathrm{dt}\:\mathrm{with}\:\:\mathrm{0}<\mathrm{a}<\mathrm{1}\:\mathrm{we}\:\mathrm{have}\:\mathrm{f}\left(\mathrm{a}\right)=\frac{\pi}{\mathrm{sin}\left(\pi\mathrm{a}\right)} \\ $$$$\mathrm{and}\:\mathrm{f}^{'} \left(\mathrm{a}\right)\:=\int_{\mathrm{0}} ^{\infty} \mathrm{t}^{\mathrm{a}−\mathrm{1}} \:\frac{\mathrm{logt}}{\mathrm{1}+\mathrm{t}}\mathrm{dt}\:\Rightarrow\mathrm{f}^{,,} \left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{t}^{\mathrm{a}−\mathrm{1}} \:\mathrm{log}^{\mathrm{2}} \mathrm{t}}{\mathrm{1}+\mathrm{t}}\mathrm{dt}\:\Rightarrow \\ $$$$\mathrm{f}^{\left(\mathrm{3}\right)} \left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{t}^{\mathrm{a}−\mathrm{1}} \:\mathrm{log}^{\mathrm{3}} \mathrm{t}}{\mathrm{1}+\mathrm{t}}\mathrm{dt}\:\Rightarrow\mathrm{f}^{\left(\mathrm{3}\right)} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}} \:\mathrm{log}^{\mathrm{3}} \mathrm{t}}{\mathrm{1}+\mathrm{t}}\mathrm{dt} \\ $$$$=\mathrm{3}^{\mathrm{4}} \:\Phi\:\mathrm{but}\:\mathrm{f}\left(\mathrm{a}\right)=\frac{\pi}{\mathrm{sin}\left(\pi\mathrm{a}\right)}\:\Rightarrow\mathrm{f}^{'} \left(\mathrm{a}\right)=−\pi^{\mathrm{2}} \frac{\mathrm{cos}\left(\pi\mathrm{a}\right)}{\mathrm{sin}^{\mathrm{2}} \left(\pi\mathrm{a}\right)}\:\Rightarrow \\ $$$$\mathrm{f}^{\left(\mathrm{2}\right)} \left(\mathrm{a}\right)=−\pi^{\mathrm{2}} .\frac{−\pi\mathrm{sin}\left(\pi\mathrm{a}\right)−\mathrm{2}\pi\mathrm{sin}\left(\pi\mathrm{a}\right)\mathrm{cos}\left(\pi\mathrm{a}\right)}{\mathrm{sin}^{\mathrm{4}} \left(\pi\mathrm{a}\right)} \\ $$$$=\pi^{\mathrm{3}} \:.\frac{\mathrm{1}+\mathrm{2cos}\left(\pi\mathrm{a}\right)}{\mathrm{sin}^{\mathrm{3}} \left(\pi\mathrm{a}\right)}\:\Rightarrow \\ $$$$\mathrm{f}^{\left(\mathrm{3}\right)} \left(\mathrm{a}\right)=\pi^{\mathrm{3}} .\frac{−\mathrm{2}\pi\mathrm{sin}\left(\pi\mathrm{a}\right).\mathrm{sin}^{\mathrm{3}} \left(\pi\mathrm{a}\right)−\mathrm{3}\pi\mathrm{sin}^{\mathrm{2}} \left(\pi\mathrm{a}\right)\mathrm{cos}\left(\pi\mathrm{a}\right)}{\mathrm{sin}^{\mathrm{6}} \left(\pi\mathrm{a}\right)} \\ $$$$=\pi^{\mathrm{4}} .\frac{−\mathrm{2sin}^{\mathrm{2}} \left(\pi\mathrm{a}\right)−\mathrm{3}\:\mathrm{cos}\left(\pi\mathrm{a}\right)}{\mathrm{sin}^{\mathrm{4}} \left(\pi\mathrm{a}\right)}\:\Rightarrow \\ $$$$\mathrm{f}^{\left(\mathrm{3}\right)} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)=−\pi^{\mathrm{4}} ×\frac{\mathrm{2}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{3}.\frac{\mathrm{1}}{\mathrm{2}}}{\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{4}} }=−\pi^{\mathrm{4}} .\frac{\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{2}}}{\frac{\mathrm{9}}{\mathrm{16}}} \\ $$$$=\left(−\pi^{\mathrm{4}} \right).\left(\frac{\mathrm{16}}{\mathrm{9}}\right)\left(\mathrm{3}\right)\:=−\frac{\mathrm{16}\pi^{\mathrm{4}} }{\mathrm{3}}\:\Rightarrow\Phi=−\frac{\mathrm{16}\pi^{\mathrm{4}} }{\mathrm{3}^{\mathrm{5}} } \\ $$$$ \\ $$

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