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Question Number 141319 by mnjuly1970 last updated on 18/May/21
prove:: Ω:=∫_0 ^( 1) ((ln^2 (x))/(1−x^4 ))dx =(π^3 /(32))+(7/8)ζ(3)..
$$\:\:\: \\ $$$$\:\:\:\:\:{prove}:: \\ $$$$\:\:\:\:\:\:\:\Omega:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left({x}\right)}{\mathrm{1}−{x}^{\mathrm{4}} }{dx}\:=\frac{\pi^{\mathrm{3}} }{\mathrm{32}}+\frac{\mathrm{7}}{\mathrm{8}}\zeta\left(\mathrm{3}\right).. \\ $$
Answered by qaz last updated on 17/May/21
∫_0 ^∞ ((ln^2 (x))/(1−x^4 ))dx =∫_0 ^1 ((ln^2 (x))/(1−x^4 ))dx+∫_1 ^∞ ((ln^2 (x))/(1−x^4 ))dx =∫_0 ^1 ((ln^2 x)/(1−x^4 ))dx+∫_0 ^1 ((x^2 ln^2 x)/(x^4 −1))dx =∫_0 ^1 ((ln^2 x)/(1+x^2 ))dx =∫_0 ^∞ ((y^2 e^(−y) )/(1+e^(−2y) ))dy =Σ_(n=0) ^∞ (−1)^n ∫_0 ^∞ y^2 e^(−(2n+1)y) dy =2Σ_(n=0) ^∞ (((−1)^n )/((2n+1)^3 )) =(π^3 /(16))
$$\int_{\mathrm{0}} ^{\infty} \frac{{ln}^{\mathrm{2}} \left({x}\right)}{\mathrm{1}−{x}^{\mathrm{4}} }{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left({x}\right)}{\mathrm{1}−{x}^{\mathrm{4}} }{dx}+\int_{\mathrm{1}} ^{\infty} \frac{{ln}^{\mathrm{2}} \left({x}\right)}{\mathrm{1}−{x}^{\mathrm{4}} }{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}^{\mathrm{2}} {x}}{\mathrm{1}−{x}^{\mathrm{4}} }{dx}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} {ln}^{\mathrm{2}} {x}}{{x}^{\mathrm{4}} −\mathrm{1}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}^{\mathrm{2}} {x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{{y}^{\mathrm{2}} {e}^{−{y}} }{\mathrm{1}+{e}^{−\mathrm{2}{y}} }{dy} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\infty} {y}^{\mathrm{2}} {e}^{−\left(\mathrm{2}{n}+\mathrm{1}\right){y}} {dy} \\ $$$$=\mathrm{2}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\frac{\pi^{\mathrm{3}} }{\mathrm{16}} \\ $$
Answered by Dwaipayan Shikari last updated on 17/May/21
∫_0 ^∞ ((log^2 (x))/(1−x^4 ))dx =∫_1 ^∞ ((log^2 (x))/(1−x^4 ))dx+∫_0 ^1 ((log^2 (x))/(1−x^4 ))dx (1/x)=i =∫_0 ^1 ((log^2 (u)u^2 )/(u^4 −1))+∫_0 ^1 ((log^2 (x))/(1−x^4 ))dx x^4 =m u^4 =k =−(1/(64))∫_0 ^1 ((log^2 (k)k^(−1/4) )/(1−k))+(1/(64))∫_0 ^1 ((log^2 (m)m^(−3/4) )/(1−m))dm =(1/(64))(ψ′′((3/4))−ψ′′((1/4)))=(π^3 /(16)) ψ(1−a)−ψ(a)=πcot(πa) ψ′(1−a)+ψ′(a)=π^2 csc^2 (πa) ψ′′(1−a)−ψ′′(a)=2π^3 csc^2 (πa)cot(πa) ψ′′((3/4))−ψ′′((1/4))=4π^3
$$\int_{\mathrm{0}} ^{\infty} \frac{{log}^{\mathrm{2}} \left({x}\right)}{\mathrm{1}−{x}^{\mathrm{4}} }{dx} \\ $$$$=\int_{\mathrm{1}} ^{\infty} \frac{{log}^{\mathrm{2}} \left({x}\right)}{\mathrm{1}−{x}^{\mathrm{4}} }{dx}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}^{\mathrm{2}} \left({x}\right)}{\mathrm{1}−{x}^{\mathrm{4}} }{dx}\:\:\:\frac{\mathrm{1}}{{x}}={i} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}^{\mathrm{2}} \left({u}\right){u}^{\mathrm{2}} }{{u}^{\mathrm{4}} −\mathrm{1}}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}^{\mathrm{2}} \left({x}\right)}{\mathrm{1}−{x}^{\mathrm{4}} }{dx}\:\:\:\:{x}^{\mathrm{4}} ={m}\:\:\:{u}^{\mathrm{4}} ={k} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{64}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}^{\mathrm{2}} \left({k}\right){k}^{−\mathrm{1}/\mathrm{4}} }{\mathrm{1}−{k}}+\frac{\mathrm{1}}{\mathrm{64}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}^{\mathrm{2}} \left({m}\right){m}^{−\mathrm{3}/\mathrm{4}} }{\mathrm{1}−{m}}{dm} \\ $$$$=\frac{\mathrm{1}}{\mathrm{64}}\left(\psi''\left(\frac{\mathrm{3}}{\mathrm{4}}\right)−\psi''\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\right)=\frac{\pi^{\mathrm{3}} }{\mathrm{16}} \\ $$$$\psi\left(\mathrm{1}−{a}\right)−\psi\left({a}\right)=\pi{cot}\left(\pi{a}\right) \\ $$$$\psi'\left(\mathrm{1}−{a}\right)+\psi'\left({a}\right)=\pi^{\mathrm{2}} {csc}^{\mathrm{2}} \left(\pi{a}\right) \\ $$$$\psi''\left(\mathrm{1}−{a}\right)−\psi''\left({a}\right)=\mathrm{2}\pi^{\mathrm{3}} {csc}^{\mathrm{2}} \left(\pi{a}\right){cot}\left(\pi{a}\right) \\ $$$$\psi''\left(\frac{\mathrm{3}}{\mathrm{4}}\right)−\psi''\left(\frac{\mathrm{1}}{\mathrm{4}}\right)=\mathrm{4}\pi^{\mathrm{3}} \\ $$

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