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

Commented by safojontoshtemirov last updated on 16/Apr/22

help me

$${help}\:{me} \\ $$

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))

$${sin}\left(\mathrm{2}{x}\right)=\frac{\mathrm{2}{tg}\left({x}\right)}{\mathrm{1}+{tg}^{\mathrm{2}} \left({x}\right)} \\ $$$$\Leftrightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{xln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{xln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$${A}−{B} \\ $$$$\left.{A}\left.=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}+{i}}+\frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}−{i}}{dx}=\mathrm{2}{Re}\left\{\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}+{i}}{d}\frac{\mathrm{3}}{}\frac{\mathrm{3}}{}\right){x}\right\}\right) \\ $$$$=\mathrm{2}{Re}\left\{{ln}\left(\mathrm{2}\right){ln}\left(\mathrm{1}+{i}\right)−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({x}+{i}\right)}{\mathrm{1}+{x}}{dx}\right\} \\ $$$$={ln}^{\mathrm{2}} \left(\mathrm{2}\right)−\mathrm{2}{Re}\int_{\mathrm{1}} ^{\mathrm{2}} \frac{{ln}\left({u}−\mathrm{1}+{i}\right)}{{u}}{du} \\ $$$${u}=\left(\mathrm{1}−{i}\right){t} \\ $$$${A}={ln}^{\mathrm{2}} \left(\mathrm{2}\right)−\mathrm{2}{Re}\int_{\frac{\mathrm{1}+{i}}{\mathrm{2}}} ^{\left(\mathrm{1}+{i}\right)} \frac{{ln}\left(\left({i}−\mathrm{1}\right)\left(\mathrm{1}−\boldsymbol{{t}}\right)\right.}{\boldsymbol{{t}}}\boldsymbol{{dt}} \\ $$$$\boldsymbol{{ln}}\left(\boldsymbol{{i}}−\mathrm{1}\right)=\boldsymbol{{ln}}\left(\sqrt{\mathrm{2}}\right)+\frac{\mathrm{3}\boldsymbol{{i}\pi}}{\mathrm{4}} \\ $$$${A}={ln}^{\mathrm{2}} \left(\mathrm{2}\right)−\mathrm{2}{Re}\left({ln}\left(\sqrt{\mathrm{2}}\right)+\frac{\mathrm{3}{i}\pi}{\mathrm{4}}\right){ln}\left(\mathrm{2}\right) \\ $$$$+\mathrm{2}{Re}\left(−\int_{\frac{\mathrm{1}+{i}}{\mathrm{2}}} ^{\mathrm{1}+{i}} \frac{{ln}\left(\mathrm{1}−{t}\right)}{{t}}{dt}\right) \\ $$$$=\mathrm{2}{Re}\left({Li}_{\mathrm{2}} \left(\mathrm{1}+{i}\right)−\boldsymbol{{L}}{i}_{\mathrm{2}} \left(\frac{\mathrm{1}+{i}}{\mathrm{2}}\right)\right) \\ $$$$={Li}_{\mathrm{2}} \left(\mathrm{1}+{i}\right)−{Li}_{\mathrm{2}} \left(\frac{\mathrm{1}+{i}}{\mathrm{2}}\right)−{Li}_{\mathrm{2}} \left(\frac{\mathrm{1}−{i}}{\mathrm{2}}\right)+{Li}_{\mathrm{2}} \left(\mathrm{1}−{i}\right) \\ $$$${Li}_{\mathrm{2}} \left({z}\right)+{li}_{\mathrm{2}} \left(\mathrm{1}−{z}\right)=\zeta\left(\mathrm{2}\right)−{ln}\left({z}\right){ln}\left(\mathrm{1}−{z}\right) \\ $$$${z}=\frac{\mathrm{1}+{i}}{\mathrm{2}}\Rightarrow{li}_{\mathrm{2}} \left(\frac{\mathrm{1}+{i}}{\mathrm{2}}\right)+\mathrm{li}_{\mathrm{2}} \left(\frac{\mathrm{1}−\mathrm{i}}{\mathrm{2}}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−{ln}\left(\frac{{e}^{{i}\frac{\pi}{\mathrm{4}}} }{\:\sqrt{\mathrm{2}}}\right){ln}\left(\frac{{e}^{−\frac{{i}\pi}{\mathrm{4}}} }{\:\sqrt{\mathrm{2}}}\right) \\ $$$$=\zeta\left(\mathrm{2}\right)−\left({ln}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)+\frac{{i}\pi}{\mathrm{4}}\right)\left({ln}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)−{i}\frac{\pi}{\mathrm{4}}\right) \\ $$$$=\zeta\left(\mathrm{2}\right)−\frac{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{4}}−\frac{\pi^{\mathrm{2}} }{\mathrm{16}} \\ $$$${li}_{\mathrm{2}} \left(\mathrm{1}−{z}\right)+{Li}_{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{1}}{{z}}\right)=−\frac{{ln}^{\mathrm{2}} \left({z}\right)}{\mathrm{2}} \\ $$$${li}_{\mathrm{2}} \left(\mathrm{1}−{i}\right)+{Li}_{\mathrm{2}} \left(\mathrm{1}+{i}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$$$\frac{\pi^{\mathrm{2}} }{\mathrm{8}}−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}+\frac{\pi^{\mathrm{2}} }{\mathrm{16}}+\frac{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{4}}=\frac{\pi^{\mathrm{2}} }{\mathrm{48}}+\frac{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{4}} \\ $$$${B}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right){dx}=\frac{\mathrm{1}}{\mathrm{4}}{ln}^{\mathrm{2}} \left(\mathrm{2}\right) \\ $$$${we}\:{Get} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {tg}\left({x}\right){ln}\left(\mathrm{1}+{sin}\left(\mathrm{2}{x}\right)\right){dx}={A}−{B}=\frac{\pi^{\mathrm{2}} }{\mathrm{48}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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