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Question Number 69233 by ~ À ® @ 237 ~ last updated on 21/Sep/19
Prove that B=∫_0 ^1 [ln(−lnu)]^2 du = γ^2 + ζ(2)
$${Prove}\:{that}\:\:{B}=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\left[{ln}\left(−{lnu}\right)\right]^{\mathrm{2}} \:{du}\:=\:\gamma^{\mathrm{2}} +\:\zeta\left(\mathrm{2}\right)\:\: \\ $$
Commented by mathmax by abdo last updated on 22/Sep/19
B=∫_0 ^1 {ln(−lnx)}^2 dx changement −lnx =t give x =e^(−t) B =−∫_0 ^∞ (ln(t)^2 (−e^(−t) )dt =∫_0 ^∞ (ln(t))^2 e^(−t) dt we have Γ(x)=∫_0 ^∞ t^(x−1) e^(−t) dt with x>0 ⇒Γ(x)=∫_0 ^∞ e^((x−1)lnt) e^(−t) dt ⇒Γ^′ (x) =∫_0 ^∞ ln(t) t^(x−1) e^(−t) dt ⇒Γ^((2)) (x)=∫_0 ^∞ (ln(t))^2 t^(x−1) e^(−t) dt ⇒ Γ^((2)) (1) =∫_0 ^∞ (ln(t))^(2 ) e^(−t) dt but we have the formula ((Γ^′ (x))/(Γ(x))) =−γ−(1/x) +Σ_(n=1) ^∞ ((1/n)−(1/(n+x))) by derivation we get ((Γ^((2)) (x)Γ(x)−(Γ^′ (x))^2 )/((Γ(x))^2 )) =(1/x^2 ) +Σ_(n=1) ^∞ (1/((n+x)^2 )) ⇒ ((Γ^((2)) (1)Γ(1)−(Γ^′ (1))^2 )/((Γ(1))^2 )) =1+Σ_(n=1) ^∞ (1/((n+1)^2 )) =Σ_(n=0) ^∞ (1/((n+1)^2 )) =(π^2 /6) Γ(1)=1 and Γ^′ (1) =∫_0 ^∞ e^(−t) ln(t)dt =−γ (this result is proved) ⇒ Γ^((2)) (1) −γ^2 =(π^2 /6) ⇒Γ^((2)) (1) =γ^2 +(π^2 /6) =B
$${B}=\int_{\mathrm{0}} ^{\mathrm{1}} \left\{{ln}\left(−{lnx}\right)\right\}^{\mathrm{2}} {dx}\:\:\:{changement}\:−{lnx}\:={t}\:{give}\:{x}\:={e}^{−{t}} \\ $$$${B}\:=−\int_{\mathrm{0}} ^{\infty} \left({ln}\left({t}\right)^{\mathrm{2}} \:\:\left(−{e}^{−{t}} \right){dt}\:=\int_{\mathrm{0}} ^{\infty} \:\:\left({ln}\left({t}\right)\right)^{\mathrm{2}} {e}^{−{t}} {dt}\right. \\ $$$${we}\:{have}\:\Gamma\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:{t}^{{x}−\mathrm{1}} {e}^{−{t}} \:{dt}\:\:{with}\:{x}>\mathrm{0}\:\Rightarrow\Gamma\left({x}\right)=\int_{\mathrm{0}} ^{\infty} {e}^{\left({x}−\mathrm{1}\right){lnt}} {e}^{−{t}} {dt} \\ $$$$\Rightarrow\Gamma^{'} \left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} {ln}\left({t}\right)\:{t}^{{x}−\mathrm{1}} \:{e}^{−{t}} \:{dt}\:\Rightarrow\Gamma^{\left(\mathrm{2}\right)} \left({x}\right)=\int_{\mathrm{0}} ^{\infty} \left({ln}\left({t}\right)\right)^{\mathrm{2}} {t}^{{x}−\mathrm{1}} \:{e}^{−{t}} \:{dt}\:\Rightarrow \\ $$$$\Gamma^{\left(\mathrm{2}\right)} \left(\mathrm{1}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\left({ln}\left({t}\right)\right)^{\mathrm{2}\:} {e}^{−{t}} \:{dt}\:\:{but}\:{we}\:{have}\:{the}\:{formula} \\ $$$$\frac{\Gamma^{'} \left({x}\right)}{\Gamma\left({x}\right)}\:=−\gamma−\frac{\mathrm{1}}{{x}}\:+\sum_{{n}=\mathrm{1}} ^{\infty} \left(\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+{x}}\right)\:\:{by}\:{derivation}\:\:{we}\:{get} \\ $$$$\frac{\Gamma^{\left(\mathrm{2}\right)} \left({x}\right)\Gamma\left({x}\right)−\left(\Gamma^{'} \left({x}\right)\right)^{\mathrm{2}} }{\left(\Gamma\left({x}\right)\right)^{\mathrm{2}} }\:=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{\left({n}+{x}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\frac{\Gamma^{\left(\mathrm{2}\right)} \left(\mathrm{1}\right)\Gamma\left(\mathrm{1}\right)−\left(\Gamma^{'} \left(\mathrm{1}\right)\right)^{\mathrm{2}} }{\left(\Gamma\left(\mathrm{1}\right)\right)^{\mathrm{2}} }\:=\mathrm{1}+\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\Gamma\left(\mathrm{1}\right)=\mathrm{1}\:\:\:{and}\:\Gamma^{'} \left(\mathrm{1}\right)\:=\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}} {ln}\left({t}\right){dt}\:=−\gamma\:\left({this}\:{result}\:{is}\:{proved}\right)\:\Rightarrow \\ $$$$\Gamma^{\left(\mathrm{2}\right)} \left(\mathrm{1}\right)\:−\gamma^{\mathrm{2}} \:=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:\Rightarrow\Gamma^{\left(\mathrm{2}\right)} \left(\mathrm{1}\right)\:=\gamma^{\mathrm{2}} \:+\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:={B}\: \\ $$
Answered by mind is power last updated on 21/Sep/19
−ln(u)=t ⇒∫_0 ^(+∞) [ln(t)]^2 e^(−t) dt Γ(z)=∫_0 ^(+∞) t^(z−1) e^(−t) dt ⇒Γ′(z)=∫_0 ^(+∞) (d/dz)e^((z−1)ln(t)) e^(−t) dt ⇒Γ′(z)=∫_0 ^(+∞) ln(t)t^(z−1) e^(−t) dt ⇒Γ′′(z)=∫_0 ^(+∞) (ln(t))^2 t^(z−1) e^(−t) dt ⇒Γ′′(1)=∫_0 ^(+∞) (ln(t))^2 e^(−t) dt we have ψ(z)=((Γ′(z))/(Γ(z)))=−γ−(1/z)+Σ_(n≥1) (1/n)−(1/(n+z))...psi function to prove it tack ln(Γ) Γ′(1)=−γ (((Γ′(z))/(Γ(z))))^′ =((Γ′′(z)Γ(z)−(Γ′(z))^2 )/(Γ(z)^2 ))=(1/z^2 )+Σ_(n≥1) (1/((n+z)^2 )) ⇒((Γ′′(1)Γ(1)−(Γ′(1))^2 )/(Γ(1)^2 ))=1+Σ_(n≥1) (1/((1+n)^2 ))=Σ_(n≥1) (1/n^2 )=ζ(2) Γ(1)=1 .Γ′(1)=−γ ⇒Γ′′(1)−γ^2 =ζ(2)⇒Γ′′(1)=γ^2 +ζ(2)
$$−{ln}\left({u}\right)={t} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{+\infty} \left[{ln}\left({t}\right)\right]^{\mathrm{2}} {e}^{−{t}} {dt} \\ $$$$\Gamma\left({z}\right)=\int_{\mathrm{0}} ^{+\infty} {t}^{{z}−\mathrm{1}} {e}^{−{t}} {dt} \\ $$$$\Rightarrow\Gamma'\left({z}\right)=\int_{\mathrm{0}} ^{+\infty} \frac{{d}}{{dz}}{e}^{\left({z}−\mathrm{1}\right){ln}\left({t}\right)} {e}^{−{t}} {dt} \\ $$$$\Rightarrow\Gamma'\left({z}\right)=\int_{\mathrm{0}} ^{+\infty} {ln}\left({t}\right){t}^{{z}−\mathrm{1}} {e}^{−{t}} {dt} \\ $$$$\Rightarrow\Gamma''\left({z}\right)=\int_{\mathrm{0}} ^{+\infty} \left({ln}\left({t}\right)\right)^{\mathrm{2}} {t}^{{z}−\mathrm{1}} {e}^{−{t}} {dt} \\ $$$$\Rightarrow\Gamma''\left(\mathrm{1}\right)=\int_{\mathrm{0}} ^{+\infty} \left({ln}\left({t}\right)\right)^{\mathrm{2}} {e}^{−{t}} {dt} \\ $$$${we}\:{have} \\ $$$$\psi\left({z}\right)=\frac{\Gamma'\left({z}\right)}{\Gamma\left({z}\right)}=−\gamma−\frac{\mathrm{1}}{{z}}+\sum_{{n}\geqslant\mathrm{1}} \frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+{z}}…{psi}\:{function}\:{to}\:{prove}\:{it}\:{tack}\:{ln}\left(\Gamma\right) \\ $$$$\Gamma'\left(\mathrm{1}\right)=−\gamma \\ $$$$\left(\frac{\Gamma'\left({z}\right)}{\Gamma\left({z}\right)}\overset{'} {\right)}=\frac{\Gamma''\left({z}\right)\Gamma\left({z}\right)−\left(\Gamma'\left({z}\right)\right)^{\mathrm{2}} }{\Gamma\left({z}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{{z}^{\mathrm{2}} }+\sum_{{n}\geqslant\mathrm{1}} \frac{\mathrm{1}}{\left({n}+{z}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{\Gamma''\left(\mathrm{1}\right)\Gamma\left(\mathrm{1}\right)−\left(\Gamma'\left(\mathrm{1}\right)\right)^{\mathrm{2}} }{\Gamma\left(\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{1}+\sum_{{n}\geqslant\mathrm{1}} \frac{\mathrm{1}}{\left(\mathrm{1}+{n}\right)^{\mathrm{2}} }=\sum_{{n}\geqslant\mathrm{1}} \frac{\mathrm{1}}{{n}^{\mathrm{2}} }=\zeta\left(\mathrm{2}\right) \\ $$$$\Gamma\left(\mathrm{1}\right)=\mathrm{1}\:\:\:\:.\Gamma'\left(\mathrm{1}\right)=−\gamma \\ $$$$\Rightarrow\Gamma''\left(\mathrm{1}\right)−\gamma^{\mathrm{2}} =\zeta\left(\mathrm{2}\right)\Rightarrow\Gamma''\left(\mathrm{1}\right)=\gamma^{\mathrm{2}} +\zeta\left(\mathrm{2}\right) \\ $$
Commented by ~ À ® @ 237 ~ last updated on 21/Sep/19
Thanks you sir ! have you prove that Γ′(1)=−γ ?
$${Thanks}\:{you}\:{sir}\:\:!\:\:{have}\:{you}\:{prove}\:{that}\:\Gamma'\left(\mathrm{1}\right)=−\gamma\:? \\ $$$$ \\ $$
Commented by mind is power last updated on 21/Sep/19
ψ(1)=((Γ′(1))/(Γ(1)))=−γ−(1/1)+Σ_(n≥1) (1/n)−(1/(n+1))=−γ ⇒Γ′(1)=−γ
$$\psi\left(\mathrm{1}\right)=\frac{\Gamma'\left(\mathrm{1}\right)}{\Gamma\left(\mathrm{1}\right)}=−\gamma−\frac{\mathrm{1}}{\mathrm{1}}+\sum_{{n}\geqslant\mathrm{1}} \frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}=−\gamma \\ $$$$\Rightarrow\Gamma'\left(\mathrm{1}\right)=−\gamma \\ $$
Commented by ~ À ® @ 237 ~ last updated on 21/Sep/19
you′re solving a question by another : else thanks!
$${you}'{re}\:{solving}\:{a}\:{question}\:{by}\:{another}\::\:{else}\:{thanks}! \\ $$
Commented by mind is power last updated on 21/Sep/19
Γ(z)=(e^(−γz) /z).∐_(k≥1) (e^(z/k) /(1+(z/k))) ⇒ln(Γ(z))=−γz−ln(z)+Σ_(k≥1) (z/k)−ln(1+(z/k)) ⇒((Γ′(z))/(Γ(z)))=−γ−(1/z)+Σ_(k≥1) (1/k)−(1/k).(k/(k+z)) ⇒((Γ′(z))/(Γ(z)))=−γ−(1/z)+Σ_(k≥1) (1/k)−(1/(k+z))
$$\Gamma\left({z}\right)=\frac{{e}^{−\gamma{z}} }{{z}}.\coprod_{{k}\geqslant\mathrm{1}} \frac{{e}^{\frac{{z}}{{k}}} }{\mathrm{1}+\frac{{z}}{{k}}} \\ $$$$\Rightarrow{ln}\left(\Gamma\left({z}\right)\right)=−\gamma{z}−{ln}\left({z}\right)+\sum_{{k}\geqslant\mathrm{1}} \frac{{z}}{{k}}−{ln}\left(\mathrm{1}+\frac{{z}}{{k}}\right) \\ $$$$\Rightarrow\frac{\Gamma'\left({z}\right)}{\Gamma\left({z}\right)}=−\gamma−\frac{\mathrm{1}}{{z}}+\sum_{{k}\geqslant\mathrm{1}} \frac{\mathrm{1}}{{k}}−\frac{\mathrm{1}}{{k}}.\frac{{k}}{{k}+{z}} \\ $$$$\Rightarrow\frac{\Gamma'\left({z}\right)}{\Gamma\left({z}\right)}=−\gamma−\frac{\mathrm{1}}{{z}}+\sum_{{k}\geqslant\mathrm{1}} \frac{\mathrm{1}}{{k}}−\frac{\mathrm{1}}{{k}+{z}} \\ $$

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