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

解答

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

- ln (2) + ln (1 + \frac{1}{e}) + 1

十进制

0.62011 \dots

求解步骤

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

\frac{1}{1 + e ^{- x}} = - \frac{e ^{- x}}{1 + e ^{- x}} + 1

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

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int_{0}^{1} \frac{e ^{- x}}{1 + e ^{- x}} d x = ln (2) - ln (1 + \frac{1}{e})

\int_{0}^{1} 1 d x = 1

= - (ln (2) - ln (1 + \frac{1}{e})) + 1

化简

= - ln (2) + ln (1 + \frac{1}{e}) + 1

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