integral from x to y of ln(sin(t))

解

\int_{x}^{y} ln (sin (t)) d t

\frac{i y ^{2} - i y ln ( 1 - ( - 1 ) ^{\frac{2 y}{π}} ) + i y ln ( sin ( y ) ) - i x ^{2} + i x ln ( 1 - ( - 1 ) ^{\frac{2 x}{π}} ) - i x ln ( sin ( x ) ) + i Li _{2} ( e ^{2 i y} ) - i Li _{2} ( e ^{2 i x} )}{2}

解答ステップ

\int_{x}^{y} ln (sin (t)) d t

非初等積分を使用する:

\int ln (sin (x)) d x = i \frac{1}{2} (x^{2} + Li_{2} (e^{2 i x}) - x ln (1 - (- 1)^{\frac{2 x}{π}}) + x ln (sin (x)))

= [i \frac{1}{2} (t^{2} + Li_{2} (e^{2 i t}) - t ln (1 - (- 1)^{\frac{2 t}{π}}) + t ln (sin (t)))]_{x}^{y}

限度を計算する:

i \frac{1}{2} (y^{2} + Li_{2} (e^{2 i y}) - y ln (1 - (- 1)^{\frac{2 y}{π}}) + y ln (sin (y))) - i \frac{1}{2} (x^{2} + Li_{2} (e^{2 i x}) - x ln (1 - (- 1)^{\frac{2 x}{π}}) + x ln (sin (x)))

= i \frac{1}{2} (y^{2} + Li_{2} (e^{2 i y}) - y ln (1 - (- 1)^{\frac{2 y}{π}}) + y ln (sin (y))) - i \frac{1}{2} (x^{2} + Li_{2} (e^{2 i x}) - x ln (1 - (- 1)^{\frac{2 x}{π}}) + x ln (sin (x)))

簡素化

= \frac{i y ^{2} - i y ln ( 1 - ( - 1 ) ^{\frac{2 y}{π}} ) + i y ln ( sin ( y ) ) - i x ^{2} + i x ln ( 1 - ( - 1 ) ^{\frac{2 x}{π}} ) - i x ln ( sin ( x ) ) + i Li _{2} ( e ^{2 i y} ) - i Li _{2} ( e ^{2 i x} )}{2}