integral from-1 to-0.2588 of e^{-ipi^2ws}

解答

\int_{- 1}^{- 0.2588} e^{- i π^{2} w s} d w

\frac{- 0.10132 \dots sin ( 2.55425 \dots s ) + 0.10132 \dots sin ( 9.86960 \dots s )}{s} + i \frac{0.10132 \dots cos ( 2.55425 \dots s ) - 0.10132 \dots cos ( 9.86960 \dots s )}{s}

求解步骤

\int_{- 1}^{- 0.2588} e^{- i π^{2} w s} d w

使用换元积分法

= \int_{π^{2} i s}^{2.55425 \dots i s} \frac{i e ^{u}}{π ^{2} s} d u

提出常数:

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

= \frac{i}{π ^{2} s} \cdot \int_{π^{2} i s}^{2.55425 \dots i s} e^{u} d u

使用积分常见定则:

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

= \frac{i}{π ^{2} s} [e^{u}]_{π^{2} i s}^{2.55425 \dots i s}

化简

\frac{i}{π ^{2} s} [e^{u}]_{π^{2} i s}^{2.55425 \dots i s} : \frac{0.10132 \dots i [ e ^{u} ] _{9.86960 \dots i s}^{2.55425 \dots i s}}{s}

= \frac{0.10132 \dots i [ e ^{u} ] _{9.86960 \dots i s}^{2.55425 \dots i s}}{s}

带入上下限计算后得:

(cos (2.55425 \dots s) - cos (9.86960 \dots s)) + i (sin (2.55425 \dots s) - sin (9.86960 \dots s))

= \frac{0.10132 \dots i ( cos ( 2.55425 \dots s ) - cos ( 9.86960 \dots s ) + i ( sin ( 2.55425 \dots s ) - sin ( 9.86960 \dots s ) ) )}{s}

化简

= \frac{- 0.10132 \dots sin ( 2.55425 \dots s ) + 0.10132 \dots sin ( 9.86960 \dots s )}{s} + i \frac{0.10132 \dots cos ( 2.55425 \dots s ) - 0.10132 \dots cos ( 9.86960 \dots s )}{s}

k = 2 \sum \infty k^{k}

a re a 7 x e^{- x^{2}}, y = 0, 0 \leq x \leq 1

d er i v a t i v e ln (2 x - 1)

\frac{d y}{d x} + 4 y = e^{- x}

n = 1 \sum \infty \frac{n}{e ^{2 n}}