integral from 0 to 1 of e^{-(1+j^2pin)t}

解答

\int_{0}^{1} e^{- (1 + j^{2} πn) t} d t

\frac{e ^{- 1 - π j^{2} n} - 1}{- 1 - π j ^{2} n}

求解步骤

\int_{0}^{1} e^{- (1 + j^{2} πn) t} d t

使用换元积分法

= \int_{0}^{- 1 - π j^{2} n} \frac{e ^{u}}{- 1 - π j ^{2} n} d u

提出常数:

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

= \frac{1}{- 1 - π j ^{2} n} \cdot \int_{0}^{- 1 - π j^{2} n} e^{u} d u

使用积分常见定则:

\int e^{u} d u = e^{u}

= \frac{1}{- 1 - π j ^{2} n} [e^{u}]_{0}^{- 1 - π j^{2} n}

化简

\frac{1}{- 1 - π j ^{2} n} [e^{u}]_{0}^{- 1 - π j^{2} n} : \frac{[ e ^{u} ] _{0}^{- 1 - π j^{2} n}}{- 1 - π j ^{2} n}

= \frac{[ e ^{u} ] _{0}^{- 1 - π j^{2} n}}{- 1 - π j ^{2} n}

带入上下限计算后得:

e^{- 1 - π j^{2} n} - 1

= \frac{e ^{- 1 - π j^{2} n} - 1}{- 1 - π j ^{2} n}

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