## Integral with Logarithms

∫π/20log(cos(x))log(sin(x)) dx One solution: Consider : F(m,n)=∫π/20sin2m−1xcos2n−1xdx To solve this put sin2x=t to get our integral as : F(m,n)=12∫10tm−1(1−t)n−1dt=β(m,n)2 Where β(m,n) is the beta function. F(m,n)=Γ(m)Γ(n)2Γ(m+n) Hence we have : Γ(m)Γ(n)Γ(m+n)=2∫π/20sin2m−1xcos2n−1xdx First differentiating both sides with respect to m we have : Γ(n)(Γ(m+n))2(Γ′(m)Γ(m+n)−Γ(m)Γ′(m+n))=4∫π/20log(sin(x))sin2m−1xcos2n−1xdx Better written as : Γ(m)Γ(n)Γ(m+n)(ψ(m)−ψ(m+n))=4∫π/20log(sin(x))sin2m−1xcos2n−1xdx where ψ(x) is the digamma function. Now differentiate … Read more