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Question Number 158293 by mnjuly1970 last updated on 02/Nov/21
question# If , Ω =∫_0 ^( 1) ((ln^( 2) (1−x^( 4) ))/x) dx= a ζ b) find the value of , a , b .
$$ \\ $$$$\:\:\:\:\:{question}# \\ $$$$\left.\mathrm{If}\:,\:\:\Omega\:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}^{\:\mathrm{2}} \left(\mathrm{1}−{x}^{\:\mathrm{4}} \right)}{{x}}\:{dx}=\:{a}\:\zeta\:{b}\right) \\ $$$$\:\:\:\:\:\:{find}\:{the}\:{value}\:{of}\:,\:\:\:\:\:{a}\:\:,\:{b}\:\:. \\ $$$$ \\ $$$$ \\ $$
Answered by qaz last updated on 02/Nov/21
∫_0 ^1 ((ln^2 (1−x^4 ))/x)dx =(1/4)∫_0 ^1 ((ln^2 (1−x))/x)dx =(1/4)Σ_(n=0) ^∞ ∫_0 ^1 x^n ln^2 xdx =(1/4)Σ_(n=0) ^∞ ((2!)/((n+1)^3 )) =(1/2)ζ(3)
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{x}^{\mathrm{4}} \right)}{\mathrm{x}}\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{x}\right)}{\mathrm{x}}\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}^{\mathrm{n}} \mathrm{ln}^{\mathrm{2}} \mathrm{xdx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{2}!}{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\zeta\left(\mathrm{3}\right) \\ $$
Commented by mnjuly1970 last updated on 02/Nov/21
grateful sir qaz excellent as always
$${grateful}\:{sir}\:{qaz} \\ $$$${excellent}\:{as}\:{always} \\ $$
Answered by mindispower last updated on 02/Nov/21
=∫_0 ^1 ((ln^2 (1−x^4 )x^3 )/x^4 )dx =∫_0 ^1 ((ln^2 (1−t))/t)(dt/3) ln(1−t)=−u⇒dt=e^(−u) du =(1/3)∫_0 ^∞ ((u^2 e^(−u) )/(1−e^(−u) ))du=(1/3)Γ(3)ζ(3)=(2/3)ζ(3)
$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{4}} \right){x}^{\mathrm{3}} }{{x}^{\mathrm{4}} }{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left(\mathrm{1}−{t}\right)}{{t}}\frac{{dt}}{\mathrm{3}} \\ $$$${ln}\left(\mathrm{1}−{t}\right)=−{u}\Rightarrow{dt}={e}^{−{u}} {du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} \frac{{u}^{\mathrm{2}} {e}^{−{u}} }{\mathrm{1}−{e}^{−{u}} }{du}=\frac{\mathrm{1}}{\mathrm{3}}\Gamma\left(\mathrm{3}\right)\zeta\left(\mathrm{3}\right)=\frac{\mathrm{2}}{\mathrm{3}}\zeta\left(\mathrm{3}\right) \\ $$
Commented by mnjuly1970 last updated on 03/Nov/21
thx sir power
$${thx}\:{sir}\:{power} \\ $$
Answered by Ar Brandon last updated on 02/Nov/21
Ω=∫_0 ^1 ((ln^2 (1−x^4 ))/x)dx, x=u^(1/4) ⇒dx=(1/4)u^(−(3/4)) du =(1/4)∫_0 ^1 ((ln^2 (1−u))/u)du=(1/4)∫_0 ^1 ((ln^2 u)/(1−u))du =−(1/4)ψ^((2)) (1)=(2/4)ζ(3)=(1/2)ζ(3)
$$\Omega=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{4}} \right)}{{x}}{dx},\:{x}={u}^{\frac{\mathrm{1}}{\mathrm{4}}} \Rightarrow{dx}=\frac{\mathrm{1}}{\mathrm{4}}{u}^{−\frac{\mathrm{3}}{\mathrm{4}}} {du} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}−{u}\right)}{{u}}{du}=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{2}} {u}}{\mathrm{1}−{u}}{du} \\ $$$$\:\:\:\:=−\frac{\mathrm{1}}{\mathrm{4}}\psi^{\left(\mathrm{2}\right)} \left(\mathrm{1}\right)=\frac{\mathrm{2}}{\mathrm{4}}\zeta\left(\mathrm{3}\right)=\frac{\mathrm{1}}{\mathrm{2}}\zeta\left(\mathrm{3}\right) \\ $$
Commented by mnjuly1970 last updated on 03/Nov/21
mercey mr brandon
$${mercey}\:{mr}\:{brandon} \\ $$

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