- 应用泰勒级数求下列积分：
(1) 菲涅尔积分 s ( z ) = ∫ 0 z sin z 2 d z \displaystyle s(z)=\int_{0}^{z}\sin z^{2}\mathrm{d}z s(z)=∫0zsinz2dz； c ( z ) = ∫ 0 z cos z 2 d z \displaystyle c(z)=\int_{0}^{z}\cos z^{2}\mathrm{d}z c(z)=∫0zcosz2dz；
解1：
s ( z ) = ∫ 0 z ∑ k = 0 ∞ ( − 1 ) k z 2 ( 2 k + 1 ) ( 2 k + 1 ) ! d z = ∑ k = 0 ∞ ∫ 0 z ( − 1 ) k z 2 ( 2 k + 1 ) ( 2 k + 1 ) ! d z \displaystyle s(z)=\int_{0}^{z}\sum_{k=0}^{\infty}\frac{(-1)^{k}z^{2(2k+1)}}{(2k+1)!}\mathrm{d}z =\sum_{k=0}^{\infty}\int_{0}^{z}\frac{(-1)^{k}z^{2(2k+1)}}{(2k+1)!}\mathrm{d}z s(z)=∫0zk=0∑∞(2k+1)!(−1)kz2(2k+1)dz=k=0∑∞∫0z(2k+1)!(−1)kz2(2k+1)dz
= ∑ k = 0 ∞ ( − 1 ) k z 4 k + 4 ( 4 k + 3 ) ( 2 k + 1 ) ! \displaystyle =\sum_{k=0}^{\infty}\frac{(-1)^{k}z^{4k+4}}{(4k+3)(2k+1)!} =k=0∑∞(4k+3)(2k+1)!(−1)kz4k+4
c ( z ) = ∫ 0 z ∑ k = 0 ∞ ( − 1 ) k z 4 k ( 2 k ) ! d z = ∑ k = 0 ∞ ∫ 0 z ( − 1 ) k z 4 k ( 2 k ) ! d z \displaystyle c(z)=\int_{0}^{z}\sum_{k=0}^{\infty}\frac{(-1)^{k}z^{4k}}{(2k)!}\mathrm{d}z =\sum_{k=0}^{\infty}\int_{0}^{z}\frac{(-1)^{k}z^{4k}}{(2k)!}\mathrm{d}z c(z)=∫0zk=0∑∞(2k)!(−1)kz4kdz=k=0∑∞∫0z(2k)!(−1)kz4kdz
= ∑ k = 0 ∞ ( − 1 ) k z 4 k + 1 ( 4 k + 1 ) ( 2 k ) ! \displaystyle =\sum_{k=0}^{\infty}\frac{(-1)^{k}z^{4k+1}}{(4k+1)(2k)!} =k=0∑∞(4k+1)(2k)!(−1)kz4k+1
菲涅尔积分的基础解法2:
∫ 0 ∞ sin x 2 d x \displaystyle \int_{0}^{\infty} \sin x^{2} \mathrm{d}x ∫0∞sinx2dx
解：令 x 2 = t x^{2}=t x2=t， ∫ 0 ∞ sin x 2 d x = ∫ 0 ∞ sin t ⋅ 1 2 t d t \displaystyle \int_{0}^{\infty} \sin x^{2} \mathrm{d}x = \int_{0}^{\infty} \sin t \cdot \frac{1}{2\sqrt{t}} \mathrm{d}t ∫0∞sinx2dx=∫0∞sint⋅2t 1dt
∵ ∫ 0 ∞ e − t u 2 d u = 1 t ∫ 0 ∞ e − ( t u ) 2 d ( t u ) = π 2 ⋅ 1 t \displaystyle \because \int_{0}^{\infty} e^{-tu^{2}} \mathrm{d}u = \frac{1}{\sqrt{t}} \int_{0}^{\infty} e^{-(\sqrt{t}u)^{2}} \mathrm{d}(\sqrt{t}u) = \frac{\sqrt{\pi}}{2} \cdot \frac{1}{\sqrt{t}} ∵∫0∞e−tu2du=t 1∫0∞e−(t u)2d(t u)=2π ⋅t 1（高斯积分）
∴ 1 t = 2 π ∫ 0 ∞ e − t u 2 d u \displaystyle \therefore \frac{1}{\sqrt{t}} = \frac{2}{\sqrt{\pi}} \int_{0}^{\infty} e^{-tu^{2}} \mathrm{d}u ∴t 1=π 2∫0∞e−tu2du
∴ 1 2 ∫ 0 ∞ sin t ⋅ 1 t d t = 1 2 ∫ 0 ∞ sin t ⋅ 2 π ∫ 0 ∞ e − t u 2 d u d t \displaystyle \therefore \frac{1}{2}\int_{0}^{\infty} \sin t \cdot \frac{1}{\sqrt{t}} \mathrm{d}t = \frac{1}{2} \int_{0}^{\infty} \sin t \cdot \frac{2}{\sqrt{\pi}} \int_{0}^{\infty} e^{-tu^{2}} \mathrm{d}u \mathrm{d}t ∴21∫0∞sint⋅t 1dt=21∫0∞sint⋅π 2∫0∞e−tu2dudt（交换积分次序）
= 1 π ∫ 0 ∞ ∫ 0 ∞ sin t ⋅ e − t u 2 d t d u ⋯ ⋯ ① . \displaystyle = \frac{1}{\sqrt{\pi}} \int_{0}^{\infty} \int_{0}^{\infty} \sin t \cdot e^{-tu^{2}} \mathrm{d}t \mathrm{d}u \quad \cdots\cdots ①. =π 1∫0∞∫0∞sint⋅e−tu2dtdu⋯⋯①.
令 I = ∫ 0 ∞ sin t ⋅ e − t u 2 d t = − ∫ 0 ∞ e − t u 2 d cos t \displaystyle I=\int_{0}^{\infty} \sin t \cdot e^{-tu^{2}} \mathrm{d}t = -\int_{0}^{\infty} e^{-tu^{2}} \mathrm{d}\cos t I=∫0∞sint⋅e−tu2dt=−∫0∞e−tu2dcost（分部积分）
= − [ e − t u 2 cos t ∣ 0 ∞ − ∫ 0 ∞ cos t ⋅ e − t u 2 ( − u 2 ) d t ] \displaystyle = -\left[ e^{-tu^{2}} \cos t \bigg|{0}^{\infty} - \int{0}^{\infty} \cos t \cdot e^{-tu^{2}} (-u^{2}) \mathrm{d}t \right] =−[e−tu2cost 0∞−∫0∞cost⋅e−tu2(−u2)dt]
= − [ − 1 + u 2 ∫ 0 ∞ e − t u 2 d sin t ] \displaystyle = -\left[ -1 + u^{2} \int_{0}^{\infty} e^{-tu^{2}} \mathrm{d}\sin t \right] =−[−1+u2∫0∞e−tu2dsint]（再分部积分）
= − [ − 1 + u 2 ( e − t u 2 sin t ∣ 0 ∞ − ∫ 0 ∞ e − t u 2 ⋅ ( − u 2 ) sin t d t ) ] \displaystyle = -\left[ -1 + u^{2} \left( e^{-tu^{2}} \sin t \bigg|{0}^{\infty} - \int{0}^{\infty} e^{-tu^{2}} \cdot (-u^{2}) \sin t \mathrm{d}t \right) \right] =−[−1+u2(e−tu2sint 0∞−∫0∞e−tu2⋅(−u2)sintdt)]
= 1 − u 4 ∫ 0 ∞ e − t u 2 sin t d t = 1 − u 4 ⋅ I \displaystyle = 1 - u^{4} \int_{0}^{\infty} e^{-tu^{2}} \sin t \mathrm{d}t = 1 - u^{4} \cdot I =1−u4∫0∞e−tu2sintdt=1−u4⋅I
∴ I = 1 1 + u 4 \displaystyle \therefore I = \frac{1}{1+u^{4}} ∴I=1+u41，代入①得.
I 2 = 1 π ∫ 0 ∞ 1 1 + u 4 d u = 1 π ∫ 0 ∞ 1 u 2 1 u 2 + u 2 d u \displaystyle I_{2}= \frac{1}{\sqrt{\pi}} \int_{0}^{\infty} \frac{1}{1+u^{4}} \mathrm{d}u = \frac{1}{\sqrt{\pi}} \int_{0}^{\infty} \frac{\frac{1}{u^{2}}}{\frac{1}{u^{2}}+u^{2}} \mathrm{d}u I2=π 1∫0∞1+u41du=π 1∫0∞u21+u2u21du, 令 u = 1 t u=\frac{1}{t} u=t1
= 1 π ∫ ∞ 0 t 2 t 2 + 1 t 2 ( − 1 ) 1 t 2 d t = 1 π ∫ 0 ∞ t 2 t 4 + 1 d t \displaystyle = \frac{1}{\sqrt{\pi}} \int_{\infty}^{0} \frac{t^{2}}{t^{2}+\frac{1}{t^{2}}} (-1) \frac{1}{t^{2}} \mathrm{d}t = \frac{1}{\sqrt{\pi}} \int_{0}^{\infty} \frac{t^{2}}{t^{4}+1} \mathrm{d}t =π 1∫∞0t2+t21t2(−1)t21dt=π 1∫0∞t4+1t2dt
∴ 2 I 2 = 1 π ∫ 0 ∞ ( 1 1 + u 2 + u 2 1 + u 4 ) d u = 1 π ∫ 0 ∞ 1 u 2 + 1 1 u 2 + 2 + u 2 − 2 d u \displaystyle \therefore 2I_{2}= \frac{1}{\sqrt{\pi}} \int_{0}^{\infty} \left( \frac{1}{1+u^{2}} + \frac{u^{2}}{1+u^{4}} \right) \mathrm{d}u = \frac{1}{\sqrt{\pi}} \int_{0}^{\infty} \frac{\frac{1}{u^{2}}+1}{\frac{1}{u^{2}}+2+u^{2}-2} \mathrm{d}u ∴2I2=π 1∫0∞(1+u21+1+u4u2)du=π 1∫0∞u21+2+u2−2u21+1du
= 1 π ∫ 0 ∞ d ( u − 1 u ) ( u − 1 u ) 2 + 2 = 1 π 2 ∫ 0 ∞ d ( u − 1 u ) 1 2 1 + [ ( u − 1 u ) ⋅ 1 2 ] 2 = 1 π 2 arctan [ 1 2 ( u − 1 u ) ] ∣ 0 ∞ \displaystyle =\frac{1}{\sqrt{\pi}}\int_{0}^{\infty}\frac{\mathrm{d}\left(u-\frac{1}{u}\right)}{\left(u-\frac{1}{u}\right)^{2}+2} =\frac{1}{\sqrt{\pi}\sqrt{2}}\int_{0}^{\infty}\frac{\mathrm{d}\left(u-\frac{1}{u}\right)\frac{1}{\sqrt{2}}}{1+\left[\left(u-\frac{1}{u}\right)\cdot\frac{1}{\sqrt{2}}\right]^{2}} =\frac{1}{\sqrt{\pi}\sqrt{2}}\arctan\left[\frac{1}{\sqrt{2}}\left(u-\frac{1}{u}\right)\right]\bigg|_{0}^{\infty} =π 1∫0∞(u−u1)2+2d(u−u1)=π 2 1∫0∞1+[(u−u1)⋅2 1]2d(u−u1)2 1=π 2 1arctan[2 1(u−u1)] 0∞
= 1 π ⋅ π 2 = 2 2 π . ∴ I 2 = 2 4 π \displaystyle =\frac{1}{\sqrt{\pi}}\cdot\frac{\pi}{2} =\frac{\sqrt{2}}{2}\sqrt{\pi}. \quad\therefore I_{2}=\frac{\sqrt{2}}{4}\pi =π 1⋅2π=22 π .∴I2=42 π
(2) 误差函数 e r f z = 2 π ∫ 0 z e − ζ 2 d ζ \displaystyle \mathrm{erf}z=\frac{2}{\sqrt{\pi}}\int_{0}^{z}\mathrm{e}^{-\zeta^{2}}\mathrm{d}\zeta erfz=π 2∫0ze−ζ2dζ；
解： e r f z = 2 π ∫ 0 z ∑ k = 0 ∞ ( − 1 ) k ζ 2 k ( k ) ! d ζ \displaystyle\mathrm{erf}z = \frac{2}{\sqrt{\pi}}\int_{0}^{z}\sum_{k=0}^{\infty}\frac{(-1)^{k}\zeta^{2k}}{(k)!}\mathrm{d}\zeta erfz=π 2∫0zk=0∑∞(k)!(−1)kζ2kdζ
= 2 π ∑ k = 0 ∞ ∫ 0 z ( − 1 ) k ζ 2 k ( k ) ! d ζ \displaystyle= \frac{2}{\sqrt{\pi}}\sum_{k=0}^{\infty}\int_{0}^{z}\frac{(-1)^{k}\zeta^{2k}}{(k)!}\mathrm{d}\zeta =π 2k=0∑∞∫0z(k)!(−1)kζ2kdζ
= 2 π ∑ k = 0 ∞ ( − 1 ) k z 2 k + 1 ( 2 k + 1 ) ( k ) ! \displaystyle= \frac{2}{\sqrt{\pi}}\sum_{k=0}^{\infty}\frac{(-1)^{k}z^{2k+1}}{(2k+1)(k)!} =π 2k=0∑∞(2k+1)(k)!(−1)kz2k+1
(3) 积分正弦 S i z = ∫ 0 z sin t t d t \displaystyle \mathrm{Si}z=\int_{0}^{z}\frac{\sin t}{t}\mathrm{d}t Siz=∫0ztsintdt。
解： S i z = ∫ 0 z 1 t ∑ k = 0 ∞ ( − 1 ) k t 2 k + 1 ( 2 k + 1 ) ! d t \displaystyle\mathrm{Si}z = \int_{0}^{z}\frac{1}{t}\sum_{k=0}^{\infty}\frac{(-1)^{k}t^{2k+1}}{(2k+1)!}\mathrm{d}t Siz=∫0zt1k=0∑∞(2k+1)!(−1)kt2k+1dt
= ∑ k = 0 ∞ ∫ 0 z ( − 1 ) k t 2 k ( 2 k + 1 ) ! d t \displaystyle= \sum_{k=0}^{\infty}\int_{0}^{z}\frac{(-1)^{k}t^{2k}}{(2k+1)!}\mathrm{d}t =k=0∑∞∫0z(2k+1)!(−1)kt2kdt
= ∑ k = 0 ∞ ( − 1 ) k z 2 k + 1 ( 2 k + 1 ) ( 2 k + 1 ) ! \displaystyle= \sum_{k=0}^{\infty}\frac{(-1)^{k}z^{2k+1}}{(2k+1)(2k+1)!} =k=0∑∞(2k+1)(2k+1)!(−1)kz2k+1