0-pi-2-ln-sinx-ln-cosx-dx-

Question Number 160358 by Ar Brandon last updated on 28/Nov/21

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{sin}{x}\right)\mathrm{ln}\left(\mathrm{cos}{x}\right){dx} \\ $$

Answered by amin96 last updated on 28/Nov/21

sin(x)=t (dt/dx)=cos(x)=(√(1−t^2 )) 𝛀=∫_0 ^1 ((ln(t)ln((√(1−t^2 ))))/( (√(1−t^2 ))))dt=(1/2)∫_0 ^1 ((ln(t)ln(1−t^2 ))/( (√(1−t^2 ))))dt t^2 =x (dx/dt)=2t=2(√x) 𝛀=(1/4)∫_0 ^1 ((ln((√x))ln(1−x))/( (√x) (√(1−x))))dx=(1/8)∫_0 ^1 ((((√(1−x)))ln(x)ln(1−x))/( (√x)×(1−x)))dx= =−(1/8)Σ_(n=1) ^∞ H_n ∫_0 ^1 x^(n−(1/2)) (1−x)^(1/2) ln(x)dx= =−(1/8)ΣH_n (Σ(((−1)^(n+1) )/n))∫_0 ^1 x^(n−(1/2)) (1−x)^(1/2) (x−1)^n dx= =(1/8)Σ_(n=1) ^∞ H_n {Σ_(n=1) ^∞ (1/n)∫_0 ^1 x^(n−(1/2)) (1−x)^(n+(1/2)) dx}= =(1/8)Σ_(n=1) ^∞ H_n {Σ_(n=1) ^∞ (1/n)∫_0 ^1 x^((n+(1/2))−1) (1−x)^((n+(3/2))−1) dx}= =(1/8)Σ_(n=1) ^∞ H_n {Σ_(n=1) ^∞ (1/n)𝛃(n+(1/2); n+(3/2))}

$$\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{x}}\right)=\boldsymbol{\mathrm{t}}\:\:\:\frac{\boldsymbol{\mathrm{dt}}}{\boldsymbol{\mathrm{dx}}}=\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\mathrm{x}}\right)=\sqrt{\mathrm{1}−\boldsymbol{\mathrm{t}}^{\mathrm{2}} } \\ $$$$\boldsymbol{\Omega}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{t}}\right)\boldsymbol{\mathrm{ln}}\left(\sqrt{\mathrm{1}−\boldsymbol{\mathrm{t}}^{\mathrm{2}} }\right)}{\:\sqrt{\mathrm{1}−\boldsymbol{\mathrm{t}}^{\mathrm{2}} }}\boldsymbol{\mathrm{dt}}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{t}}\right)\boldsymbol{\mathrm{ln}}\left(\mathrm{1}−\boldsymbol{\mathrm{t}}^{\mathrm{2}} \right)}{\:\sqrt{\mathrm{1}−\boldsymbol{\mathrm{t}}^{\mathrm{2}} }}\boldsymbol{\mathrm{dt}} \\ $$$$\boldsymbol{\mathrm{t}}^{\mathrm{2}} =\boldsymbol{\mathrm{x}}\:\:\:\:\frac{\boldsymbol{\mathrm{dx}}}{\boldsymbol{\mathrm{dt}}}=\mathrm{2}\boldsymbol{\mathrm{t}}=\mathrm{2}\sqrt{\boldsymbol{\mathrm{x}}}\:\:\: \\ $$$$\boldsymbol{\Omega}=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\boldsymbol{\mathrm{ln}}\left(\sqrt{\boldsymbol{\mathrm{x}}}\right)\boldsymbol{\mathrm{ln}}\left(\mathrm{1}−\boldsymbol{\mathrm{x}}\right)}{\:\sqrt{\boldsymbol{\mathrm{x}}}\:\sqrt{\mathrm{1}−\boldsymbol{\mathrm{x}}}}\boldsymbol{\mathrm{dx}}=\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\sqrt{\mathrm{1}−\boldsymbol{\mathrm{x}}}\right)\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{ln}}\left(\mathrm{1}−\boldsymbol{\mathrm{x}}\right)}{\:\sqrt{\boldsymbol{\mathrm{x}}}×\left(\mathrm{1}−\boldsymbol{\mathrm{x}}\right)}\boldsymbol{\mathrm{dx}}= \\ $$$$=−\frac{\mathrm{1}}{\mathrm{8}}\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\boldsymbol{\mathrm{H}}_{\boldsymbol{\mathrm{n}}} \int_{\mathrm{0}} ^{\mathrm{1}} \boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{n}}−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−\boldsymbol{\mathrm{x}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{dx}}= \\ $$$$=−\frac{\mathrm{1}}{\mathrm{8}}\Sigma\boldsymbol{\mathrm{H}}_{\boldsymbol{\mathrm{n}}} \left(\Sigma\frac{\left(−\mathrm{1}\right)^{\boldsymbol{\mathrm{n}}+\mathrm{1}} }{\boldsymbol{\mathrm{n}}}\right)\int_{\mathrm{0}} ^{\mathrm{1}} \boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{n}}−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−\boldsymbol{\mathrm{x}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\boldsymbol{\mathrm{x}}−\mathrm{1}\right)^{\boldsymbol{\mathrm{n}}} \boldsymbol{\mathrm{dx}}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\boldsymbol{\mathrm{H}}_{\boldsymbol{\mathrm{n}}} \left\{\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\boldsymbol{\mathrm{n}}}\int_{\mathrm{0}} ^{\mathrm{1}} \boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{n}}−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−\boldsymbol{\mathrm{x}}\right)^{\boldsymbol{\mathrm{n}}+\frac{\mathrm{1}}{\mathrm{2}}} \boldsymbol{\mathrm{dx}}\right\}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\boldsymbol{\mathrm{H}}_{\boldsymbol{\mathrm{n}}} \left\{\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\boldsymbol{\mathrm{n}}}\int_{\mathrm{0}} ^{\mathrm{1}} \boldsymbol{\mathrm{x}}^{\left(\boldsymbol{\mathrm{n}}+\frac{\mathrm{1}}{\mathrm{2}}\right)−\mathrm{1}} \left(\mathrm{1}−\boldsymbol{\mathrm{x}}\right)^{\left(\boldsymbol{\mathrm{n}}+\frac{\mathrm{3}}{\mathrm{2}}\right)−\mathrm{1}} \boldsymbol{\mathrm{dx}}\right\}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\boldsymbol{\mathrm{H}}_{\boldsymbol{\mathrm{n}}} \left\{\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\boldsymbol{\mathrm{n}}}\boldsymbol{\beta}\left(\boldsymbol{\mathrm{n}}+\frac{\mathrm{1}}{\mathrm{2}};\:\boldsymbol{\mathrm{n}}+\frac{\mathrm{3}}{\mathrm{2}}\right)\right\} \\ $$

Commented by Ar Brandon last updated on 28/Nov/21

$$\mathrm{I}\:\mathrm{wish}\:\mathrm{it}\:\mathrm{is}\:\mathrm{further}\:\mathrm{simplified}. \\ $$

Commented by amin96 last updated on 28/Nov/21

$$ \\ $$ do you know the answer?

Commented by Ar Brandon last updated on 29/Nov/21

$$\mathrm{Yeah} \\ $$

Commented by Ar Brandon last updated on 29/Nov/21

$$\frac{\pi}{\mathrm{2}}\mathrm{ln}^{\mathrm{2}} \mathrm{2}−\frac{\pi^{\mathrm{3}} }{\mathrm{48}} \\ $$

Commented by amin96 last updated on 29/Nov/21

∫_0 ^(π/2) xln(sinx)ln(cosx)dx The answer to this integral

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \boldsymbol{\mathrm{xln}}\left(\boldsymbol{\mathrm{sinx}}\right)\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{cosx}}\right)\boldsymbol{\mathrm{dx}}\: \\ $$$$ \\ $$The answer to this integral

Commented by Ar Brandon last updated on 29/Nov/21

∫_0 ^(π/2) xln(sinx)ln(cosx)dx=(π^2 /8)ln^2 2−(π^4 /(192)) ∫_0 ^(π/2) xln(sinx)ln(cosx)dx=(π/4)∫_0 ^(π/2) ln(sinx)ln(cosx)dx

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {x}\mathrm{ln}\left(\mathrm{sin}{x}\right)\mathrm{ln}\left(\mathrm{cos}{x}\right){dx}=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\mathrm{ln}^{\mathrm{2}} \mathrm{2}−\frac{\pi^{\mathrm{4}} }{\mathrm{192}} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {x}\mathrm{ln}\left(\mathrm{sin}{x}\right)\mathrm{ln}\left(\mathrm{cos}{x}\right){dx}=\frac{\pi}{\mathrm{4}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{sin}{x}\right)\mathrm{ln}\left(\mathrm{cos}{x}\right){dx} \\ $$

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