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Question Number 107328 by mathdave last updated on 10/Aug/20

Commented by mnjuly1970 last updated on 10/Aug/20

Commented by mnjuly1970 last updated on 10/Aug/20

Commented by mathdave last updated on 10/Aug/20

$p l s h w u g o t a l l d s$
Commented by mathdave last updated on 10/Aug/20

$d s i s n t a c t u a l l y s o l v e b y u$
Commented by mnjuly1970 last updated on 10/Aug/20

$p l e a s e : n o w, s o l v e i t$ $i n y o u r o w n w a y!$
Commented by mathmax by abdo last updated on 10/Aug/20

$can you show how you get the value of Σ \frac{{(- 1)}^{n}}{2 n + 1} \ln (2 n + 1) ?$
	Answered by mathmax by abdo last updated on 10/Aug/20
	$I =∫_0 ^∞ ((lnx)/(e^x +e^(−x) )) dx ⇒I =∫_0 ^∞ ((e^(−x) ln(x))/(1+e^(−2x) ))dx =∫_0 ^∞ e^(−x) ln(x)Σ_(n=0) ^∞ (−1)^n e^(−2nx) dx =Σ_(n=0) ^∞ (−1)^n ∫_0 ^∞ e^(−(2n+1)x) lnx dx changement (2n+1)x =t give ∫_0 ^∞ e^(−(2n+1)x) ln(x)dx =∫_0 ^∞ e^(−t) ln((t/(2n+1)))(dt/(2n+1)) =(1/(2n+1))∫_0 ^∞ e^(−t) {ln(t)−ln(2n+1)}dt =(1/(2n+1))∫_0 ^∞ e^(−t) ln(t)dt−((ln(2n+1))/(2n+1)) ∫_0 ^∞ e^(−t) dt =−(γ/(2n+1))−((ln(2n+1))/(2n+1)) ⇒I =−γ Σ_(n=0) ^∞ (((−1)^n )/(2n+1))−Σ_(n=0) ^∞ (((−1)^n ln(2n+1))/(2n+1)) =−((πγ)/4) −Σ_(n=0) ^∞ (((−1)^n )/(2n+1))ln(2n+1) rest to find this sum ...be continued..$
	$I = \int_{0}^{\infty} \frac{lnx}{e^{x} + e^{- x}} dx \Rightarrow I = \int_{0}^{\infty} \frac{e^{- x} \ln (x)}{1 + e^{- 2 x}} dx$ $= \int_{0}^{\infty} e^{- x} \ln (x) \sum_{n = 0}^{\infty} {(- 1)}^{n} e^{- 2 nx} dx$ $= \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{\infty} e^{- (2 n + 1) x} lnx dx changement (2 n + 1) x = t give$ $\int_{0}^{\infty} e^{- (2 n + 1) x} \ln (x) dx = \int_{0}^{\infty} e^{- t} \ln (\frac{t}{2 n + 1}) \frac{dt}{2 n + 1}$ $= \frac{1}{2 n + 1} \int_{0}^{\infty} e^{- t} {\ln (t) - \ln (2 n + 1)} dt$ $= \frac{1}{2 n + 1} \int_{0}^{\infty} e^{- t} \ln (t) dt - \frac{\ln (2 n + 1)}{2 n + 1} \int_{0}^{\infty} e^{- t} dt$ $= - \frac{γ}{2 n + 1} - \frac{\ln (2 n + 1)}{2 n + 1} \Rightarrow I = - γ \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1} - \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} \ln (2 n + 1)}{2 n + 1}$ $= - \frac{π γ}{4} - \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1} \ln (2 n + 1) rest to find this sum . . . be continued . .$
	Commented by mnjuly1970 last updated on 10/Aug/20

	$p e r f e c t . m e r c e y$
	Commented by mathdave last updated on 10/Aug/20

	$u h a d r e a l l y d o n e w e l l o o o w e l l d o n e s i r$
	Commented by mathmax by abdo last updated on 10/Aug/20

	$you are welcome$
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