xを1/2-xと置き換えると、
I=∫(1/2,0) log(sin(π/2-πx))・(-dx)=∫(0,1/2) log(cosπx)dx
よって、
2I=∫(0,1/2) log(sinπx)dx + ∫(0,1/2) log(cosπx)dx
=∫(0,1/2) log(1/2・sin2πx)dx
=1/2・log(1/2) + ∫(0,1/2) log(sin2πx)dx
2x=tとおくと
2I=-1/2・log2 + ∫(0,1) log(sinπt)・(1/2)dt
=-1/2・log2 + (1/2){∫(0,1/2) log(sinπt)dt + ∫(1/2,1) log(sinπt)dt}
=-1/2・log2 + (1/2){I + ∫(1/2,1) log(sinπt)dt}
t=1 - sとおくと
2I=-1/2・log2 + (1/2){I + ∫(1/2,0) log(sin(π-πs))・(-ds)}
=-1/2・log2 + (1/2){I + ∫(0,1/2) log(sinπs)ds}
=-1/2・log2 + I
すなわち　I=-1/2・log2　（積分収束判定は省略）