∫[0,π/2]log(sinx)dx の値
∫[0,π/2]log(sinx)dx を求めるにあたってS=∫[0,π/2]log(sinx)dx とおく S を求めるにあたって，J=∫[0,π]log(sinx)dx を考える．J=∫[0,π]log(sinx)dx=∫[0,π/2]log(sinx)dx +∫[π/2,π]log(sinx)dx=S+∫[π/2,π]log(sinx)dxここで，∫[π/2,π]log(sinx)dx についてx=π-t と置換-∫[π/2,0]log(sin(π-t))dt=∫[0,π/2]log(sint)dtよってJ=∫[0,π/2]log(sinx)dx +∫[0,π/2]log(sinx)dx=2∫[0,π/2]log(sinx)dx=2S次に J=∫[0,π]log(sinx)dx を別な視点から考察するsinx=2sin(x/2)cos(x/2) を使ってJ=∫[0,π]log(2sin(x/2)cos(x/2) )dx=∫[0,π]log(2)dx+∫[0,π]log(sin(x/2))dx+∫[0,π]log(cos(x/2))dx=πlog2+∫[0,π]log(sin(x/2))dx+∫[0,π]log(cos(x/2))dxここで∫[0,π]log(sin(x/2))dx は, x/2=t として2∫[0,π/2]log(sint)dt =2Sまた，∫[0,π]log(cos(x/2))dx は，x/2=t として2∫[0,π/2]log(cost)dtここで，t=π/2-x とすると∫[0,π/2]log(cost)dt=-∫[π/2,0]log(cos(π/2-x))dx=∫[0,π/2]log(sinx)dx=S以上よりJ=∫[0,π]log(2sin(x/2)cos(x/2) )dx2S=πlog2+∫[0,π]log(sin(x/2))dx+∫[0,π]log(cos(x/2))dx2S=πlog2+2S+2S2S=-πlog2S=-π(1/2)log2∫[0,π/2]log(sinx)dx=-(π/2)log2続編をみてください。https://ameblo.jp/titchmarsh/entry-12617677318.html＜別解＞2021年4月21日追加S=∫[0,π/2]log(sinx)dx とおくS=∫[0,π/2]log(sinx)dx=∫[0,π/4]log(sinx)dx +∫[π/4,π/2]log(sinx)dxここで，∫[π/4,π/2]log(sinx)dx についてx=π/2-t と置換-∫[π/4,0]log(sin(π/2-t))dt=∫[0,π/4]log(cost)dtよってS=∫[0,π/4]log(sinx)dx +∫[π/4,π/2]log(sinx)dx=∫[0,π/4]log(sinx)dx +∫[0,π/4]log(cost)dt=∫[0,π/4](log(sinx)+log(cosx))dx=∫[0,π/4](log(sinxcosx))dx=∫[0,π/4](log((1/2)sin(2x)))dx=∫[0,π/4](log(1/2)+log(sin(2x)))dx=∫[0,π/4](log(1/2)dx +∫[0,π/4](log(sin(2x)))dx=-(π/4 )log2+∫[0,π/4](log(sin(2x)))dx2x=t として 第２項の積分を置換するS=-(π/4)log2+(½)∫[0,π/2](log(sint))dtS=-(π/4 )log2+(½)S(½)S=-(π/4 )log2S=-(π/2 )log2よって S=∫[0,π/2]log(sinx)dx=-(π/2 )log2
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