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Question Number 168707 by safojontoshtemirov last updated on 16/Apr/22

Commented by safojontoshtemirov last updated on 16/Apr/22

$h e l p m e$
	Answered by mindispower last updated on 19/Apr/22
	$sin(2x)=((2tg(x))/(1+tg^2 (x))) ⇔∫_0 ^1 ((2xln(1+x))/(1+x^2 ))−∫_0 ^1 ((xln(1+x^2 ))/(1+x^2 )) A−B A=∫_0 ^1 ((ln(1+x))/(x+i))+((ln(1+x))/(x−i))dx=2Re{∫_0 ^1 ((ln(1+x))/(x+i))d(3/)(3/))x}) =2Re{ln(2)ln(1+i)−∫_0 ^1 ((ln(x+i))/(1+x))dx} =ln^2 (2)−2Re∫_1 ^2 ((ln(u−1+i))/u)du u=(1−i)t A=ln^2 (2)−2Re∫_((1+i)/2) ^((1+i)) ((ln((i−1)(1−t))/t)dt ln(i−1)=ln((√2))+((3i𝛑)/4) A=ln^2 (2)−2Re(ln((√2))+((3iπ)/4))ln(2) +2Re(−∫_((1+i)/2) ^(1+i) ((ln(1−t))/t)dt) =2Re(Li_2 (1+i)−Li_2 (((1+i)/2))) =Li_2 (1+i)−Li_2 (((1+i)/2))−Li_2 (((1−i)/2))+Li_2 (1−i) Li_2 (z)+li_2 (1−z)=ζ(2)−ln(z)ln(1−z) z=((1+i)/2)⇒li_2 (((1+i)/2))+li_2 (((1−i)/2))=(π^2 /6)−ln((e^(i(π/4)) /( (√2))))ln((e^(−((iπ)/4)) /( (√2)))) =ζ(2)−(ln((1/( (√2))))+((iπ)/4))(ln((1/( (√2))))−i(π/4)) =ζ(2)−((ln^2 (2))/4)−(π^2 /(16)) li_2 (1−z)+Li_2 (1−(1/z))=−((ln^2 (z))/2) li_2 (1−i)+Li_2 (1+i)=(π^2 /8) (π^2 /8)−(π^2 /6)+(π^2 /(16))+((ln^2 (2))/4)=(π^2 /(48))+((ln^2 (2))/4) B=(1/2)∫_0 ^1 ((2x)/(1+x^2 ))ln(1+x^2 )dx=(1/4)ln^2 (2) we Get ∫_0 ^(π/4) tg(x)ln(1+sin(2x))dx=A−B=(π^2 /(48))$
	$s i n (2 x) = \frac{2 t g (x)}{1 + {t g}^{2} (x)}$ $\Leftrightarrow \int_{0}^{1} \frac{2 x l n (1 + x)}{1 + x^{2}} - \int_{0}^{1} \frac{x l n (1 + x^{2})}{1 + x^{2}}$ $A - B$ $A = \int_{0}^{1} \frac{l n (1 + x)}{x + i} + \frac{l n (1 + x)}{x - i} d x = 2 R e {\int_{0}^{1} \frac{l n (1 + x)}{x + i} d \frac{3}{} \frac{3}{}) x})$ $= 2 R e {l n (2) l n (1 + i) - \int_{0}^{1} \frac{l n (x + i)}{1 + x} d x}$ $= {l n}^{2} (2) - 2 R e \int_{1}^{2} \frac{l n (u - 1 + i)}{u} d u$ $u = (1 - i) t$ $A = {l n}^{2} (2) - 2 R e \int_{\frac{1 + i}{2}}^{(1 + i)} \frac{l n ((i - 1) (1 - t)}{t} d t$ $l n (i - 1) = l n (\sqrt{2}) + \frac{3 i π}{4}$ $A = {l n}^{2} (2) - 2 R e (l n (\sqrt{2}) + \frac{3 i π}{4}) l n (2)$ $+ 2 R e (- \int_{\frac{1 + i}{2}}^{1 + i} \frac{l n (1 - t)}{t} d t)$ $= 2 R e ({L i}_{2} (1 + i) - L i_{2} (\frac{1 + i}{2}))$ $= {L i}_{2} (1 + i) - {L i}_{2} (\frac{1 + i}{2}) - {L i}_{2} (\frac{1 - i}{2}) + {L i}_{2} (1 - i)$ ${L i}_{2} (z) + {l i}_{2} (1 - z) = ζ (2) - l n (z) l n (1 - z)$ $z = \frac{1 + i}{2} \Rightarrow {l i}_{2} (\frac{1 + i}{2}) + {li}_{2} (\frac{1 - i}{2}) = \frac{π^{2}}{6} - l n (\frac{e^{i \frac{π}{4}}}{\sqrt{2}}) l n (\frac{e^{- \frac{i π}{4}}}{\sqrt{2}})$ $= ζ (2) - (l n (\frac{1}{\sqrt{2}}) + \frac{i π}{4}) (l n (\frac{1}{\sqrt{2}}) - i \frac{π}{4})$ $= ζ (2) - \frac{{l n}^{2} (2)}{4} - \frac{π^{2}}{16}$ ${l i}_{2} (1 - z) + {L i}_{2} (1 - \frac{1}{z}) = - \frac{{l n}^{2} (z)}{2}$ ${l i}_{2} (1 - i) + {L i}_{2} (1 + i) = \frac{π^{2}}{8}$ $\frac{π^{2}}{8} - \frac{π^{2}}{6} + \frac{π^{2}}{16} + \frac{{l n}^{2} (2)}{4} = \frac{π^{2}}{48} + \frac{{l n}^{2} (2)}{4}$ $B = \frac{1}{2} \int_{0}^{1} \frac{2 x}{1 + x^{2}} l n (1 + x^{2}) d x = \frac{1}{4} {l n}^{2} (2)$ $w e G e t$ $\int_{0}^{\frac{π}{4}} t g (x) l n (1 + s i n (2 x)) d x = A - B = \frac{π^{2}}{48}$
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