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

解答

\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}

求解步骤

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

这是非初等函数积分:

\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}

带入上下限计算后得:

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)))

化简

= \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}

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