integral from 0 to 1 of pi(-ln(x))^2

解答

\int_{0}^{1} π (- ln (x))^{2} d x

π 2

十进制

6.28318 \dots

求解步骤

\int_{0}^{1} π (- ln (x))^{2} d x

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= π \cdot \int_{0}^{1} (- ln (x))^{2} d x

化简

(- ln (x))^{2} : ln^{2} (x)

= π \cdot \int_{0}^{1} ln^{2} (x) d x

使用分布积分法

= π [x ln^{2} (x) - \int 2 ln (x) d x]_{0}^{1}

\int 2 ln (x) d x = 2 (x ln (x) - x)

= π [x ln^{2} (x) - 2 (x ln (x) - x)]_{0}^{1}

带入上下限计算后得:

2

= π 2

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