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Question Number 133938 by Algoritm last updated on 25/Feb/21

	Answered by mathmax by abdo last updated on 25/Feb/21
	$I =∫_0 ^1 ((ln(−lnx))/(1+x^2 ))dx changement −lnx=t give x=e^(−t) ⇒ I=∫_0 ^∞ ((ln(t))/(1 +e^(−2t) ))e^(−t) dt =∫_0 ^∞ e^(−t) ln(t)Σ_(n=0) ^∞ (−1)^n e^(−2nt) dt =Σ_(n=0) ^∞ (−1)^n ∫_0 ^∞ e^(−(2n+1)t) ln(t)dt =_((2n+1)t=u) Σ_(n=0) ^∞ (−1)^n ∫_0 ^∞ e^(−u) ln((u/(2n+1)))(du/(2n+1)) =Σ_(n=0) ^∞ (((−1)^n )/(2n+1)) ∫_0 ^∞ e^(−u) {lnu−ln(2n+1)}du =Σ_(n=0) ^∞ (((−1)^n )/(2n+1))∫_0 ^∞ e^(−u) lnudu −Σ_(n=0) ^∞ (((−1)^n ln(2n+1))/(2n+1)) =−(π/4)γ −Σ_(n=0) ^∞ (((−1)^n )/(2n+1))ln(2n+1) rest to find the value of this serie ....be continued...$
	$I = \int_{0}^{1} \frac{\ln (- lnx)}{1 + x^{2}} dx changement - lnx = t give x = e^{- t} \Rightarrow$ $I = \int_{0}^{\infty} \frac{\ln (t)}{1 + e^{- 2 t}} e^{- t} dt = \int_{0}^{\infty} e^{- t} \ln (t) \sum_{n = 0}^{\infty} {(- 1)}^{n} e^{- 2 nt} dt$ $= \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{\infty} e^{- (2 n + 1) t} \ln (t) dt =_{(2 n + 1) t = u} \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{\infty} e^{- u} \ln (\frac{u}{2 n + 1}) \frac{du}{2 n + 1}$ $= \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1} \int_{0}^{\infty} e^{- u} {lnu - \ln (2 n + 1)} du$ $= \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1} \int_{0}^{\infty} e^{- u} lnudu - \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} \ln (2 n + 1)}{2 n + 1}$ $= - \frac{π}{4} γ - \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1} \ln (2 n + 1) rest to find the value of this serie$ $. . . . be continued . . .$
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