Question Number 156923 by MathSh last updated on 17/Oct/21
$$\boldsymbol{\Omega}\:=\underset{\:\mathrm{0}} {\overset{\:\frac{\boldsymbol{\pi}}{\mathrm{4}}} {\int}}\mathrm{x}\:\mathrm{log}\:\left(\mathrm{1}\:+\:\mathrm{tan}\boldsymbol{\mathrm{x}}\right)\:\mathrm{dx}\:=\:? \\ $$
Answered by mindispower last updated on 18/Oct/21
![β«_0 ^(Ο/4) xln(cos(x))dx=a...? ln(cos(x))=βΞ£_(kβ₯1) (((β1)^k cos(2kx))/k)βln(2)..fourie sere a=βΞ£_(kβ₯1) (((β1)^k )/k)β«_0 ^(Ο/4) xcos(2kx)dxβln(2)β«_0 ^(Ο/4) xdx =βΞ£_(kβ₯1) (((β1)^k )/k){[((xsin(2kx))/(2k))]_0 ^(Ο/4) +(1/(4k^2 ))[cos(2kx)]_0 ^(Ο/4) β(Ο^2 /(32))ln(2) =β(Ξ£_(kβ₯1) (((β1)^k sin(((kΟ)/2)))/(2k^2 ))+(((β1)^k cos(((kΟ)/2)))/(4k^3 ))) spit k=2m,k=2m+1we get withe sin(Οm)=0 sin(Οm+(Ο/2))=(β1)^m ,cos(mΟ)=(β1)^m ,cos(mΟ+(Ο/2))=(β1)^m =Ξ£_(mβ₯0) (((β1)^m )/(2(2m+1)^2 ))βΞ£_(mβ₯1) (((β1)^m )/(32m^3 )) =(1/2)π(β1)β(1/(32))Li_3 (β1)=(G/2)+(3/(4.32))ΞΆ(3)β((Ο^2 ln(2))/(32)) β«_0 ^(Ο/4) ln(cos(x))dx,β«_0 ^(Ο/4) ln(tg(x))=G β«_0 ^(Ο/2) ln(sinx)dx=β(Ο/2)ln(2) cos(x)=(1/2)ln(((sin(x)cos(x))/(sin(x)))cos(x)),xβ[0,(Ο/2)] =(1/2)ln(sin(2x))β((ln(tg(x)))/2) ββ«_0 ^(Ο/4) ln(cos(x))=(Ο/8)ln(2)β(G/2) Ξ©=β«_0 ^(Ο/4) xln((((β2)sin(x+(Ο/4)))/(cos(x))))dx =((ln(2))/2)β«_0 ^(Ο/4) xdx+β«_0 ^(Ο/4) xln(sin(x+(Ο/4)))dxββ«_0 ^(Ο/4) xln(cos(x))dx =((ln(2))/(32))Ο^2 +β«_0 ^(Ο/4) (Ο/4)ln(cos(x))dxβ2β«_0 ^(Ο/4) xln(cos(x))dx =((ln(2))/(32))Ο^2 +(Ο/4)(((Οln2)/8)β(G/2))β2((G/2)+((3ΞΆ(1))/(128))β((Ο^2 ln(2))/(32))) =β((Ο/8)+1)G+((Ο^2 ln(2))/8)β(3/(64))ΞΆ(3)](https://www.tinkutara.com/question/Q157032.png)
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {xln}\left({cos}\left({x}\right)\right){dx}={a}…? \\ $$$${ln}\left({cos}\left({x}\right)\right)=β\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(β\mathrm{1}\right)^{{k}} {cos}\left(\mathrm{2}{kx}\right)}{{k}}β{ln}\left(\mathrm{2}\right)..{fourie}\:{sere} \\ $$$${a}=β\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(β\mathrm{1}\right)^{{k}} }{{k}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {xcos}\left(\mathrm{2}{kx}\right){dx}β{ln}\left(\mathrm{2}\right)\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {xdx} \\ $$$$=β\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(β\mathrm{1}\right)^{{k}} }{{k}}\left\{\left[\frac{{xsin}\left(\mathrm{2}{kx}\right)}{\mathrm{2}{k}}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} +\frac{\mathrm{1}}{\mathrm{4}{k}^{\mathrm{2}} }\left[{cos}\left(\mathrm{2}{kx}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} β\frac{\pi^{\mathrm{2}} }{\mathrm{32}}{ln}\left(\mathrm{2}\right)\right. \\ $$$$=β\left(\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(β\mathrm{1}\right)^{{k}} {sin}\left(\frac{{k}\pi}{\mathrm{2}}\right)}{\mathrm{2}{k}^{\mathrm{2}} }+\frac{\left(β\mathrm{1}\right)^{{k}} {cos}\left(\frac{{k}\pi}{\mathrm{2}}\right)}{\mathrm{4}{k}^{\mathrm{3}} }\right) \\ $$$${spit}\:{k}=\mathrm{2}{m},{k}=\mathrm{2}{m}+\mathrm{1}{we}\:{get}\:{withe}\:{sin}\left(\pi{m}\right)=\mathrm{0} \\ $$$${sin}\left(\pi{m}+\frac{\pi}{\mathrm{2}}\right)=\left(β\mathrm{1}\right)^{{m}} ,{cos}\left({m}\pi\right)=\left(β\mathrm{1}\right)^{{m}} ,{cos}\left({m}\pi+\frac{\pi}{\mathrm{2}}\right)=\left(β\mathrm{1}\right)^{{m}} \\ $$$$=\underset{{m}\geqslant\mathrm{0}} {\sum}\frac{\left(β\mathrm{1}\right)^{{m}} }{\mathrm{2}\left(\mathrm{2}{m}+\mathrm{1}\right)^{\mathrm{2}} }β\underset{{m}\geqslant\mathrm{1}} {\sum}\frac{\left(β\mathrm{1}\right)^{\boldsymbol{{m}}} }{\mathrm{32}\boldsymbol{{m}}^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\beta}\left(β\mathrm{1}\right)β\frac{\mathrm{1}}{\mathrm{32}}{Li}_{\mathrm{3}} \left(β\mathrm{1}\right)=\frac{{G}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{4}.\mathrm{32}}\zeta\left(\mathrm{3}\right)β\frac{\pi^{\mathrm{2}} {ln}\left(\mathrm{2}\right)}{\mathrm{32}} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({cos}\left({x}\right)\right){dx},\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({tg}\left({x}\right)\right)={G} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sinx}\right){dx}=β\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right) \\ $$$${cos}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{{sin}\left({x}\right){cos}\left({x}\right)}{{sin}\left({x}\right)}{cos}\left({x}\right)\right),{x}\in\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({sin}\left(\mathrm{2}{x}\right)\right)β\frac{{ln}\left({tg}\left({x}\right)\right)}{\mathrm{2}} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({cos}\left({x}\right)\right)=\frac{\pi}{\mathrm{8}}{ln}\left(\mathrm{2}\right)β\frac{{G}}{\mathrm{2}} \\ $$$$\Omega=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {xln}\left(\frac{\sqrt{\mathrm{2}}{sin}\left({x}+\frac{\pi}{\mathrm{4}}\right)}{{cos}\left({x}\right)}\right){dx} \\ $$$$=\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {xdx}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {xln}\left({sin}\left({x}+\frac{\pi}{\mathrm{4}}\right)\right){dx}β\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {xln}\left({cos}\left({x}\right)\right){dx} \\ $$$$=\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{32}}\pi^{\mathrm{2}} +\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\pi}{\mathrm{4}}{ln}\left({cos}\left({x}\right)\right){dx}β\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {xln}\left({cos}\left({x}\right)\right){dx} \\ $$$$=\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{32}}\pi^{\mathrm{2}} +\frac{\pi}{\mathrm{4}}\left(\frac{\pi{ln}\mathrm{2}}{\mathrm{8}}β\frac{{G}}{\mathrm{2}}\right)β\mathrm{2}\left(\frac{{G}}{\mathrm{2}}+\frac{\mathrm{3}\zeta\left(\mathrm{1}\right)}{\mathrm{128}}β\frac{\pi^{\mathrm{2}} {ln}\left(\mathrm{2}\right)}{\mathrm{32}}\right) \\ $$$$=β\left(\frac{\pi}{\mathrm{8}}+\mathrm{1}\right){G}+\frac{\pi^{\mathrm{2}} {ln}\left(\mathrm{2}\right)}{\mathrm{8}}β\frac{\mathrm{3}}{\mathrm{64}}\zeta\left(\mathrm{3}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by MathSh last updated on 18/Oct/21
$$\mathrm{Perfect}\:\mathrm{dear}\:\mathrm{Ser},\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$
Commented by mindispower last updated on 19/Oct/21
$${withe}\:{Pleasur} \\ $$