Integral involging log-sine
We prove the following integral using the famous Feynman's trick ∫π20ln2(sinx)dx=πln222+π324
Proof
∫π20ln2(sinx)dx=∫π20d2dt2|t=0+sint(x)dx=d2dt2|t=0+∫π20sint(x)dx(w↦sinx)=d2dt2|t=0+∫10wt(1−w2)−12dw(y↦w2)=d2dt2|t=0+12∫10yt2−12(1−y)−12dw=d2dt2|t=0+12B(t+12,12)=d2dt2|t=0+√π2Γ(1+t2)Γ(1+t2) Then ddt|t=0+[ddt√π2Γ(1+t2)Γ(1+t2)]=ddt|t=0+√πΓ(1+t2)(ψ0(1+t2)−ψ0(t2+1))4Γ(t2+1)=limt→0+√πΓ(1+t2)8Γ(t2+1)[ψ0(t2+1)2−2ψ0(t+12)ψ0(t2+1)+ψ0(t+12)2−ψ1(t2+1)+ψ1(t+12)] Taking the limit as t→0+ ∫π20ln2(sinx)dx=πln222+π324
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