integral from 0 to 1 of (e^{-1/x})/(x^2)

解答

\int_{0}^{1} \frac{e ^{- \frac{1}{x}}}{x ^{2}} d x

\frac{1}{e}

十进制

0.36787 \dots

求解步骤

\int_{0}^{1} \frac{e ^{- \frac{1}{x}}}{x ^{2}} d x

使用分布积分法

= [\frac{e ^{- \frac{1}{x}}}{x} - \int \frac{e ^{- \frac{1}{x}} - 2 e ^{- \frac{1}{x}} x}{x ^{3}} d x]_{0}^{1}

\int \frac{e ^{- \frac{1}{x}} - 2 e ^{- \frac{1}{x}} x}{x ^{3}} d x = \frac{e ^{- \frac{1}{x}}}{x} - e^{- \frac{1}{x}}

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

化简

= [e^{- \frac{1}{x}}]_{0}^{1}

带入上下限计算后得:

\frac{1}{e}

= \frac{1}{e}

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