question proposed by A8;15: ∫_0 ^1 ((ln(ln(1/x)))/(1+x))dx


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Question Number 112642 by mathdave last updated on 09/Sep/20

$q u e s t i o n p r o p o s e d b y A 8; 15 :$ $\int_{0}^{1} \frac{\ln (\ln \frac{1}{x})}{1 + x} d x$
	Answered by mathdave last updated on 09/Sep/20

	$m y s o l u t i o n g o e s$ $f i r s t w e n e e d t o c o n v e r t f r o m$ ${m a l s t e n}^{'} s i n t e g r a l t o {v e r d i}^{'} s i n t e g r a l$ $l e t I = \int_{0}^{1} \frac{\ln (\ln \frac{1}{x})}{1 + x} d x l e t t = \ln (\frac{1}{x})$ $x = e^{- t} a n d d x = - e^{- t}$ $I = \int_{\infty}^{0} \frac{\ln t}{1 + e^{- t}} \times - e^{- t} d t = \int_{0}^{\infty} \frac{\ln t}{1 + e^{- t}} e^{- t} d t$ $I = \int_{0}^{\infty} \frac{\ln t}{1 + e^{t}} d t$ $I = \frac{\partial}{\partial a} ∣_{a = 0} \int_{0}^{\infty} \frac{t^{a}}{1 + e^{t}} d t$ $b u t n o t e \int_{0}^{\infty} \frac{t^{s}}{1 + e^{t}} d t = η (s + 1) Γ (s + 1)$ $I = \frac{\partial}{\partial a} ∣_{a = 0} [η (a + 1) Γ (a + 1)]$ $I = {[η^{'} (a + 1) Γ (a + 1) + η (a + 1) Γ^{^{'}} (a + 1)]}_{a = 0}$ $b u t Γ (a + 1) = Γ (a + 1) ψ (a + 1)$ $I = {[η^{^{'}} (a + 1) Γ (a + 1) + η (a + 1) Γ (a + 1) ψ (a + 1)]}_{a = 0}$ $I = [η^{^{'}} (1) Γ (1) + η (1) Γ (1) ψ (1)]$ $b u t$ $η^{^{'}} (1) = - γ \ln 2 - \frac{1}{2} \ln^{2} (2), Γ (1) = 1, η (1) = \ln 2, ψ (1) = γ$ $I = - γ \ln 2 - \frac{1}{2} \ln^{2} (2) + γ \ln 2 = - \frac{1}{2} \ln^{2} (2)$ $∵ \int_{0}^{1} \frac{\ln (\ln \frac{1}{x})}{1 + x} d x = - \frac{1}{2} \ln^{2} (2)$ $b y m a t h d a v e (09 / 08 / 2020)$
	Commented by mnjuly1970 last updated on 09/Sep/20

	$v e r y n i c e m r b a t h d a v e . . .$
	Commented by Tawa11 last updated on 06/Sep/21

	$great sir$
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