Prove that B=∫_0 ^1 [ln(−lnu)]^2 du = γ^2 + ζ(2)


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Question Number 69233 by ~ À ® @ 237 ~ last updated on 21/Sep/19

$P r o v e t h a t B = \int_{0}^{1} {[l n (- l n u)]}^{2} d u = γ^{2} + ζ (2)$
Commented by mathmax by abdo last updated on 22/Sep/19
$B=∫_0 ^1 {ln(−lnx)}^2 dx changement −lnx =t give x =e^(−t) B =−∫_0 ^∞ (ln(t)^2 (−e^(−t) )dt =∫_0 ^∞ (ln(t))^2 e^(−t) dt we have Γ(x)=∫_0 ^∞ t^(x−1) e^(−t) dt with x>0 ⇒Γ(x)=∫_0 ^∞ e^((x−1)lnt) e^(−t) dt ⇒Γ^′ (x) =∫_0 ^∞ ln(t) t^(x−1) e^(−t) dt ⇒Γ^((2)) (x)=∫_0 ^∞ (ln(t))^2 t^(x−1) e^(−t) dt ⇒ Γ^((2)) (1) =∫_0 ^∞ (ln(t))^(2 ) e^(−t) dt but we have the formula ((Γ^′ (x))/(Γ(x))) =−γ−(1/x) +Σ_(n=1) ^∞ ((1/n)−(1/(n+x))) by derivation we get ((Γ^((2)) (x)Γ(x)−(Γ^′ (x))^2 )/((Γ(x))^2 )) =(1/x^2 ) +Σ_(n=1) ^∞ (1/((n+x)^2 )) ⇒ ((Γ^((2)) (1)Γ(1)−(Γ^′ (1))^2 )/((Γ(1))^2 )) =1+Σ_(n=1) ^∞ (1/((n+1)^2 )) =Σ_(n=0) ^∞ (1/((n+1)^2 )) =(π^2 /6) Γ(1)=1 and Γ^′ (1) =∫_0 ^∞ e^(−t) ln(t)dt =−γ (this result is proved) ⇒ Γ^((2)) (1) −γ^2 =(π^2 /6) ⇒Γ^((2)) (1) =γ^2 +(π^2 /6) =B$
$B = \int_{0}^{1} {l n (- l n x)}^{2} d x c h a n g e m e n t - l n x = t g i v e x = e^{- t}$ $B = - \int_{0}^{\infty} (l n {(t)}^{2} (- e^{- t}) d t = \int_{0}^{\infty} {(l n (t))}^{2} e^{- t} d t$ $w e h a v e Γ (x) = \int_{0}^{\infty} t^{x - 1} e^{- t} d t w i t h x > 0 \Rightarrow Γ (x) = \int_{0}^{\infty} e^{(x - 1) l n t} e^{- t} d t$ $\Rightarrow Γ^{^{'}} (x) = \int_{0}^{\infty} l n (t) t^{x - 1} e^{- t} d t \Rightarrow Γ^{(2)} (x) = \int_{0}^{\infty} {(l n (t))}^{2} t^{x - 1} e^{- t} d t \Rightarrow$ $Γ^{(2)} (1) = \int_{0}^{\infty} {(l n (t))}^{2} e^{- t} d t b u t w e h a v e t h e f o r m u l a$ $\frac{Γ^{^{'}} (x)}{Γ (x)} = - γ - \frac{1}{x} + \sum_{n = 1}^{\infty} (\frac{1}{n} - \frac{1}{n + x}) b y d e r i v a t i o n w e g e t$ $\frac{Γ^{(2)} (x) Γ (x) - {(Γ^{^{'}} (x))}^{2}}{{(Γ (x))}^{2}} = \frac{1}{x^{2}} + \sum_{n = 1}^{\infty} \frac{1}{{(n + x)}^{2}} \Rightarrow$ $\frac{Γ^{(2)} (1) Γ (1) - {(Γ^{^{'}} (1))}^{2}}{{(Γ (1))}^{2}} = 1 + \sum_{n = 1}^{\infty} \frac{1}{{(n + 1)}^{2}} = \sum_{n = 0}^{\infty} \frac{1}{{(n + 1)}^{2}} = \frac{π^{2}}{6}$ $Γ (1) = 1 a n d Γ^{^{'}} (1) = \int_{0}^{\infty} e^{- t} l n (t) d t = - γ (t h i s r e s u l t i s p r o v e d) \Rightarrow$ $Γ^{(2)} (1) - γ^{2} = \frac{π^{2}}{6} \Rightarrow Γ^{(2)} (1) = γ^{2} + \frac{π^{2}}{6} = B$
	Answered by mind is power last updated on 21/Sep/19

	$- l n (u) = t$ $\Rightarrow \int_{0}^{+ \infty} {[l n (t)]}^{2} e^{- t} d t$ $Γ (z) = \int_{0}^{+ \infty} t^{z - 1} e^{- t} d t$ $\Rightarrow Γ^{'} (z) = \int_{0}^{+ \infty} \frac{d}{d z} e^{(z - 1) l n (t)} e^{- t} d t$ $\Rightarrow Γ^{'} (z) = \int_{0}^{+ \infty} l n (t) t^{z - 1} e^{- t} d t$ $\Rightarrow Γ^{″} (z) = \int_{0}^{+ \infty} {(l n (t))}^{2} t^{z - 1} e^{- t} d t$ $\Rightarrow Γ^{″} (1) = \int_{0}^{+ \infty} {(l n (t))}^{2} e^{- t} d t$ $w e h a v e$ $ψ (z) = \frac{Γ^{'} (z)}{Γ (z)} = - γ - \frac{1}{z} + \sum_{n ⩾ 1} \frac{1}{n} - \frac{1}{n + z} . . . p s i f u n c t i o n t o p r o v e i t t a c k l n (Γ)$ $Γ^{'} (1) = - γ$ $Missing \left or extra \right$ $\Rightarrow \frac{Γ^{″} (1) Γ (1) - {(Γ^{'} (1))}^{2}}{Γ {(1)}^{2}} = 1 + \sum_{n ⩾ 1} \frac{1}{{(1 + n)}^{2}} = \sum_{n ⩾ 1} \frac{1}{n^{2}} = ζ (2)$ $Γ (1) = 1 . Γ^{'} (1) = - γ$ $\Rightarrow Γ^{″} (1) - γ^{2} = ζ (2) \Rightarrow Γ^{″} (1) = γ^{2} + ζ (2)$
	Commented by ~ À ® @ 237 ~ last updated on 21/Sep/19

	$T h a n k s y o u s i r! h a v e y o u p r o v e t h a t Γ^{'} (1) = - γ ?$
	Commented by mind is power last updated on 21/Sep/19

	$ψ (1) = \frac{Γ^{'} (1)}{Γ (1)} = - γ - \frac{1}{1} + \sum_{n ⩾ 1} \frac{1}{n} - \frac{1}{n + 1} = - γ$ $\Rightarrow Γ^{'} (1) = - γ$
	Commented by ~ À ® @ 237 ~ last updated on 21/Sep/19

	${y o u}^{'} r e s o l v i n g a q u e s t i o n b y a n o t h e r : e l s e t h a n k s!$
	Commented by mind is power last updated on 21/Sep/19

	$Γ (z) = \frac{e^{- γ z}}{z} . \underset{k ⩾ 1}{∐} \frac{e^{\frac{z}{k}}}{1 + \frac{z}{k}}$ $\Rightarrow l n (Γ (z)) = - γ z - l n (z) + \sum_{k ⩾ 1} \frac{z}{k} - l n (1 + \frac{z}{k})$ $\Rightarrow \frac{Γ^{'} (z)}{Γ (z)} = - γ - \frac{1}{z} + \sum_{k ⩾ 1} \frac{1}{k} - \frac{1}{k} . \frac{k}{k + z}$ $\Rightarrow \frac{Γ^{'} (z)}{Γ (z)} = - γ - \frac{1}{z} + \sum_{k ⩾ 1} \frac{1}{k} - \frac{1}{k + z}$
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