integral from 0 to 1 of x^{-1}ln(x)

解答

\int_{0}^{1} x^{- 1} ln (x) d x

发散

求解步骤

\int_{0}^{1} x^{- 1} ln (x) d x

化简

x^{- 1} ln (x) : \frac{ln ( x )}{x}

= \int_{0}^{1} \frac{ln ( x )}{x} d x

使用分布积分法

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

\int \frac{1 - ln ( x )}{x} d x = - \frac{( 1 - ln ( x ) ) ^{2}}{2}

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

化简

[ln (x) - (- \frac{( 1 - ln ( x ) ) ^{2}}{2})]_{0}^{1} : [\frac{1}{2} (ln^{2} (x) + 1)]_{0}^{1}

= [\frac{1}{2} (ln^{2} (x) + 1)]_{0}^{1}

带入上下限计算后得:

发散

= 发散

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