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Thursday, July 1, 2021

Feynman's trick III

Powers of logsin

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(wsinx)=d2dt2|t=0+10wt(1w2)12dw(yw2)=d2dt2|t=0+1210yt212(1y)12dw=d2dt2|t=0+12B(t+12,12)=d2dt2|t=0+π2Γ(1+t2)Γ(1+t2) Then ddt|t=0+[ddtπ2Γ(1+t2)Γ(1+t2)]=ddt|t=0+πΓ(1+t2)(ψ0(1+t2)ψ0(t2+1))4Γ(t2+1)=limt0+πΓ(1+t2)8Γ(t2+1)[ψ0(t2+1)22ψ0(t+12)ψ0(t2+1)+ψ0(t+12)2ψ1(t2+1)+ψ1(t+12)] Taking the limit as t0+ π20ln2(sinx)dx=πln222+π324

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