8.1Fourier 变换及其性质习题参考解答

zhangzujin

# Fourier 变换及其性质习题参考解答 --- 1、 按定义求下列函数的 Fourier 变换: (1)、 $\displaystyle f(x)=\left\\{\begin\{array\}\{llllllllllll\}0,&|x| > a,\\\\ |x|,&|x|\leq a;\end\{array\}\right.$ [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / \begin\{aligned\} \hat f(\xi)&=\frac\{1\}\{\sqrt\{2\pi\}\}\int\_\{-a\}^a |x|\mathrm\{e\}^\{-\mathrm\{ i\} x\xi\}\mathrm\{ d\} x =\frac\{1\}\{\sqrt\{2\pi\}\}\int\_\{-a\}^a |x|\left(\cos x\xi-\mathrm\{ i\} \sin x\xi\right)\mathrm\{ d\} x\\\\ &\xlongequal\{\tiny\mbox\{对称性\}\} \frac\{2\}\{\sqrt\{2\pi\}\}\int\_0^a x\cos x\xi \mathrm\{ d\} x \xlongequal[\tiny\mbox\{积分\}]\{\tiny\mbox\{分部\}\} \sqrt\{\frac\{2\}\{\pi\}\}\frac\{\cos a\xi-1-a\xi \sin a\xi\}\{\xi^2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (2)、 $\displaystyle f(x)=\left\\{\begin\{array\}\{llllllllllll\}0,&|x| > a,\\\\ 1-\frac\{|x|\}\{a\},&|x|\leq a;\end\{array\}\right.$ [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / \begin\{aligned\} \hat f(\xi)&=\frac\{1\}\{\sqrt\{2\pi\}\}\int\_\{-a\}^a \left(1-\frac\{|x|\}\{a\}\right) \mathrm\{e\}^\{-\mathrm\{ i\} x\xi\}\mathrm\{ d\} x\\\\ &\xlongequal\{\tiny\mbox\{对称性\}\} \frac\{2\}\{\sqrt\{2\pi\}\}\int\_\{-a\}^a \left(1-\frac\{|x|\}\{a\}\right)\mathrm\{e\}^\{-\mathrm\{ i\} x\xi\}\mathrm\{ d\} x\\\\ &\xlongequal[\tiny\mbox\{积分\}]\{\tiny\mbox\{分部\}\} \sqrt\{\frac\{2\}\{\pi\}\}\frac\{1-\cos a\xi\}\{a\xi^2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (3)、 $\displaystyle f(x)=\left\\{\begin\{array\}\{llllllllllll\}\sin bx,&|x| < a,\\\\ 0,&|x| > a,\end\{array\}\right.$ $\displaystyle b$ 为常数; [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / \begin\{aligned\} \hat f(\xi)&=\frac\{1\}\{\sqrt\{2\pi\}\} \int\_\{-a\}^a \sin bx\left(\cos x\xi -\mathrm\{ i\} \sin x\xi\right)\mathrm\{ d\} x\\\\ &\xlongequal\{\tiny\mbox\{对称性\}\} \frac\{2\mathrm\{ i\}\}\{\sqrt\{2\pi\}\}\int\_0^a \sin bx\sin x\xi\mathrm\{ d\} x\\\\ &=\frac\{\mathrm\{ i\} \}\{\sqrt\{2\pi\}\}\int\_0^a \left\[\cos (b-\xi)x-\cos(b+\xi)x\right\]\mathrm\{ d\} x\\\\ &=\frac\{\mathrm\{ i\}\}\{\sqrt\{2\pi\}\} \left\[\frac\{\sin(b-\xi)a\}\{b-\xi\} -\frac\{\sin (b+\xi)a\}\{b+\xi\}\right\]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (4)、 $\displaystyle f(x)=\mathrm\{e\}^\{-a|x|\}$, $\displaystyle a$ 为正常数; [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由书第 200 页例 8.3 知 \begin\{aligned\} &f\_1(x)=\mathrm\{e\}^\{-|x|\}\Rightarrow f(x)=f\_1(ax)\\\\ \Rightarrow& \hat f(\xi)=\frac\{1\}\{a\}\hat\{f\}\_1\left(\frac\{\xi\}\{a\}\right) =\frac\{1\}\{a\} \frac\{1\}\{\sqrt\{2\pi\}\}\frac\{2\}\{1+\left(\frac\{\xi\}\{a\}\right)^2\} =\frac\{2a\}\{\sqrt\{2\pi\}(\xi^2+a^2)\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (5)、 $\displaystyle f(x)=\mathrm\{e\}^\{-a|x|\}\cos x$, $\displaystyle a$ 为正常数; [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / \begin\{aligned\} \hat f(\xi)&=\frac\{1\}\{\sqrt\{2\pi\}\}\int\_\{\mathbb\{R\}\} \mathrm\{e\}^\{-a|x|\} \cos x \mathrm\{e\}^\{-\mathrm\{ i\} x\xi\}\mathrm\{ d\} x \xlongequal\{\tiny\mbox\{对称性\}\} \frac\{2\}\{\sqrt\{2\pi\}\} \int\_0^\infty \mathrm\{e\}^\{-ax\}\cos x\cos x\xi\mathrm\{ d\} x\\\\ &=\frac\{1\}\{\sqrt\{2\pi\}\}\int\_0^\infty \mathrm\{e\}^\{-ax\}\left\[\cos (1+\xi)x+\cos(1-\xi)x\right\]\mathrm\{ d\} x\\\\ &\xlongequal[\tiny\mbox\{积分\}]\{\tiny\mbox\{分部\}\} \frac\{a\}\{\sqrt\{2\pi\}\} \left\[\frac\{1\}\{a^2+(\xi+1)^2\} +\frac\{1\}\{a^2+(\xi-1)^2\}\right\]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (6)、 $\displaystyle f(x)=\mathrm\{e\}^\{-a|x|\}\sin^2 x$, $\displaystyle a$ 为正常数. [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / \begin\{aligned\} \hat f(\xi)&=\frac\{1\}\{\sqrt\{2\pi\}\} \int\_\{\mathbb\{R\}\}\mathrm\{e\}^\{-a|x|\}\sin^2x \cdot \cos x \xi\mathrm\{ d\} x\\\\ &\xlongequal\{\tiny\mbox\{对称性\}\} \frac\{2\}\{\sqrt\{2\pi\}\}\int\_0^\infty \mathrm\{e\}^\{-ax\} \frac\{1-\cos 2x\}\{2\}\cos x\xi\mathrm\{ d\} x\\\\ &=2a\sqrt\{\frac\{2\}\{\pi\}\} \frac\{4+a^2-3\xi^2\}\{ [a^2+(\xi-2)^2] (a^2+\xi^2)[a^2+(\xi+2)^2]\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 2、 利用 Fourier 变换的性质求下列函数的 Fourier 变换: (1)、 $\displaystyle f(x)=\left\\{\begin\{array\}\{llllllllllll\}x^2,&|x| < a,\\\\ 0,&|x| > a;\end\{array\}\right.$ [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由书第 199 页例 8.1 知 \begin\{aligned\} f\_1(x)=\left\\{\begin\{array\}\{llllllllllll\}1,&|x|\leq A, \\\\ 0,&|x| > A\end\{array\}\right.\Rightarrow \hat f\_1(\xi)=\sqrt\{\frac\{2\}\{\pi\}\} \frac\{\sin A\xi\}\{\xi\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 \begin\{aligned\} \hat f&=(x^2f\_1)^\wedge =\mathrm\{ i\}^2 \frac\{\mathrm\{ d\}^2\}\{\mathrm\{ d\} \xi^2\}\hat f\_1 =-\sqrt\{\frac\{2\}\{\pi\}\} \frac\{\mathrm\{ d\}^2\}\{\mathrm\{ d\} \xi^2\}\left(\frac\{\sin a\xi\}\{\xi\}\right)\\\\ &=\sqrt\{\frac\{2\}\{\pi\}\}\frac\{2a\xi \cos a\xi-2\sin a\xi+a^2\xi^2\sin a\xi\}\{\xi^3\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (2)、 $\displaystyle f(x)=x\mathrm\{e\}^\{-a|x|\}$, $\displaystyle a$ 为正常数; [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由题 1 (4) 知若 $\displaystyle f\_1(x)=\mathrm\{e\}^\{-a|x|\}$, 则 \begin\{aligned\} \hat f\_1(\xi)=\frac\{2a\}\{\sqrt\{2\pi\}(\xi^2+a^2)\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 \begin\{aligned\} f(x)=xf\_1(x)\Rightarrow \hat f=\mathrm\{ i\} \frac\{\mathrm\{ d\} \}\{\mathrm\{ d\} \xi\}\left\[\frac\{2a\}\{\sqrt\{2\pi\}(\xi^2+a^2)\}\right\] =-\sqrt\{\frac\{2\}\{\pi\}\}\frac\{2a\mathrm\{ i\} t\}\{(a^2+t^2)^2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (3)、 $\displaystyle f(x)=\left\\{\begin\{array\}\{llllllllllll\}\mathrm\{e\}^\{b x\},&|x| < a,\\\\ 0,&|x|\geq a,\end\{array\}\right.$ $\displaystyle b$ 为常数; [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / \begin\{aligned\} \hat f(\xi)&=\frac\{1\}\{\sqrt\{2\pi\}\} \int\_\{-a\}^a \mathrm\{e\}^\{b x\}\mathrm\{e\}^\{-\mathrm\{ i\} x\xi\}\mathrm\{ d\} x =\frac\{1\}\{\sqrt\{2\pi\}\} \frac\{\mathrm\{e\}^\{a(b-\mathrm\{ i\} \xi)\}-\mathrm\{e\}^\{-a(b-\mathrm\{ i\} \xi)\}\}\{b-\mathrm\{ i\} \xi\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (4)、 $\displaystyle f(x)=\mathrm\{e\}^\{-a|x|\} \sin bx$, $\displaystyle a$ 为正常数, $\displaystyle b$ 为常数; [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / \begin\{aligned\} \hat f(\xi)&=\frac\{1\}\{\sqrt\{2\pi\}\} \int\_\{\mathbb\{R\}\} \mathrm\{e\}^\{-a|x|\}\sin bx \mathrm\{e\}^\{-\mathrm\{ i\} x\xi\}\mathrm\{ d\} x\\\\ &\xlongequal\{\tiny\mbox\{对称性\}\} \frac\{2\}\{\sqrt\{2\pi\}\}\int\_0^\infty \mathrm\{e\}^\{-ax\} \sin bx(-\mathrm\{ i\} \sin x\xi)\mathrm\{ d\} x\\\\ &=-\sqrt\{\frac\{2\}\{\pi\}\}\frac\{\mathrm\{ i\} ab\xi\}\{ (a^2+b^2)^2+2(a^2-b^2)\xi^2+\xi^4 \}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 这里最后一步我们要用三角函数的积化和差公式, 然后就是数学分析的分部积分了. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (5)、 $\displaystyle f(x)=\left\\{\begin\{array\}\{llllllllllll\}\mathrm\{e\}^\{\mathrm\{ i\} bx\}, &|x| < L,\\\\ 0,&|x| > L,\end\{array\}\right.$ $\displaystyle b$ 为常数; [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / \begin\{aligned\} \hat f(\xi)&=\frac\{1\}\{\sqrt\{2\pi\}\}\int\_\{-L\}^L \mathrm\{e\}^\{\mathrm\{ i\} bx\}\mathrm\{e\}^\{-\mathrm\{ i\} x\xi\}\mathrm\{ d\} x =\frac\{1\}\{\sqrt\{2\pi\}\}\int\_\{-L\}^L \mathrm\{e\}^\{\mathrm\{ i\} (b-\xi)x\}\mathrm\{ d\} x\\\\ &\xlongequal\{\tiny\mbox\{对称性\}\} \frac\{2\}\{\sqrt\{2\pi\}\} \int\_0^L \cos (b-\xi)x\mathrm\{ d\} x =\sqrt\{\frac\{2\}\{\pi\}\} \frac\{\sin (b-\xi)L\}\{b-\xi\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (6)、 $\displaystyle f(x)=\mathrm\{e\}^\{-ax^2+\mathrm\{ i\} bx+c\}$, $\displaystyle a,b,c$ 为常数; [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle (\mathrm\{e\}^\{-ax^2\})^\wedge=\frac\{1\}\{\sqrt\{2a\}\}\mathrm\{e\}^\{-\frac\{\xi^2\}\{4a\}\}$ 知 \begin\{aligned\} \left\[\mathrm\{e\}^\{-a\left(x-\frac\{\mathrm\{ i\} b\}\{2a\}\right)^2\}\right\]^\wedge =\mathrm\{e\}^\{-\mathrm\{ i\} \frac\{\mathrm\{ i\} b\}\{2a\}\xi\}(\mathrm\{e\}^\{-ax^2\})^\wedge =\frac\{\mathrm\{e\}^\frac\{b\xi\}\{2a\}\}\{\sqrt\{2a\}\}\mathrm\{e\}^\{-\frac\{\xi^2\}\{4a\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 再由 $\displaystyle f(x)=\mathrm\{e\}^\{-a\left(x-\frac\{\mathrm\{ i\} b\}\{2a\}\right)^2+\frac\{4ac-b^2\}\{4a\}\}$ 知 \begin\{aligned\} \hat f(\xi)=\mathrm\{e\}^\frac\{4ac-b^2\}\{4a\}\frac\{\mathrm\{e\}^\frac\{b\xi\}\{2a\}\}\{\sqrt\{2a\}\}\mathrm\{e\}^\{-\frac\{\xi^2\}\{4a\}\} =\frac\{1\}\{\sqrt\{2a\}\}\mathrm\{e\}^\{\frac\{-\xi^2+2b\xi+4ac-b^2\}\{4a\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (7)、 $\displaystyle f(x)=\frac\{1\}\{(a^2+x^2)^k\}$, $\displaystyle a$ 为正常数, $\displaystyle k$ 为正整数; [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / \begin\{aligned\} \hat f(\xi)&=\frac\{1\}\{\sqrt\{2\pi\}\} \int\_\{\mathbb\{R\}\} \frac\{1\}\{(a^2+x^2)^k\} \mathrm\{e\}^\{-\mathrm\{ i\} x\xi\}\mathrm\{ d\} x \stackrel\{x=at\}\{=\}\frac\{1\}\{\sqrt\{2\pi\}\} \int\_\{\mathbb\{R\}\} \frac\{1\}\{a^\{2k\}(1+t^2)^k\} \mathrm\{e\}^\{-\mathrm\{ i\} t\cdot a\xi\} a\mathrm\{ d\} t\\\\ &=\frac\{1\}\{\sqrt\{2\pi\} a^\{2k-1\}\} \int\_\{\mathbb\{R\}\} \frac\{\mathrm\{e\}^\{-\mathrm\{ i\} \cdot a\xi \cdot t\}\}\{(1+t^2)^k\}\mathrm\{ d\} t =\frac\{1\}\{\sqrt\{2\pi\} a^\{2k-1\}\} \int\_\{\mathbb\{R\}\} \frac\{\cos a\xi t-\mathrm\{ i\} \sin a\xi t\}\{(1+t^2)^k\}\mathrm\{ d\} t\\\\ &\xlongequal\{\tiny\mbox\{对称性\}\} \frac\{1\}\{\sqrt\{2\pi\} a^\{2k-1\}\} \int\_\{\mathbb\{R\}\} \frac\{\cos a\xi t\}\{(1+t^2)^k\}\mathrm\{ d\} t =\frac\{1\}\{\sqrt\{2\pi\} a^\{2k-1\}\} \int\_\{\mathbb\{R\}\} \frac\{\mathrm\{e\}^\{\mathrm\{ i\} \cdot a|\xi| \cdot t\}\}\{(1+t^2)^k\}\mathrm\{ d\} t \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 又由 $\displaystyle f(z)=\frac\{\mathrm\{e\}^\{\mathrm\{ i\} a|\xi| z\}\}\{(1+z^2)^k\}$ [想一下: 为啥要弄个绝对值出来? 这样复变函数才会当 $\displaystyle z\to\infty$ 时, 函数极限为 $\displaystyle 0$!] 在上半平面上只有一个 $\displaystyle k$ 阶极点 $\displaystyle \mathrm\{ i\}$, 而由留数定理知 \begin\{aligned\} \hat f(\xi)=\frac\{1\}\{\sqrt\{2\pi\}a^\{2k-1\}\}\cdot 2\pi \mathrm\{ i\} \cdot \frac\{1\}\{(k-1)!\} \left.\frac\{\mathrm\{ d\}^\{k-1\}\}\{\mathrm\{ d\} z^\{k-1\}\}\right|\_\{z=\mathrm\{ i\}\} \left\[\frac\{\mathrm\{e\}^\{\mathrm\{ i\} a|\xi| z\}\}\{(z+\mathrm\{ i\})^k\}\right\]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (8)、 $\displaystyle f(x)=\frac\{x\}\{(a^2+x^2)^k\}$, $\displaystyle a$ 为正常数, $\displaystyle k$ 为正整数; [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由上一小题知 \begin\{aligned\} \hat f(\xi)=\frac\{1\}\{\sqrt\{2\pi\}a^\{2k-1\}\}\cdot 2\pi \cdot \frac\{1\}\{(k-1)!\}\frac\{\mathrm\{ d\}\}\{\mathrm\{ d\} \xi\}\left\\{ \left.\frac\{\mathrm\{ d\}^\{k-1\}\}\{\mathrm\{ d\} z^\{k-1\}\}\right|\_\{z=\mathrm\{ i\}\} \left\[\frac\{\mathrm\{e\}^\{\mathrm\{ i\} a\xi z\}\}\{(z+\mathrm\{ i\})^k\}\right\]\right\\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (9)、 $\displaystyle f(x)=x\mathrm\{e\}^\{-\frac\{1\}\{2\}(x-1)^2\}$; [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / \begin\{aligned\} &\left(\mathrm\{e\}^\{-ax^2\}\right)^\wedge =\frac\{1\}\{\sqrt\{2\pi\}\} \mathrm\{e\}^\{-\frac\{\xi^2\}\{4a\}\} \Rightarrow \left(\mathrm\{e\}^\{-\frac\{1\}\{2\}x^2\}\right)^\wedge =\mathrm\{e\}^\{-\frac\{\xi^2\}\{2a\}\}\\\\ \Rightarrow& \left\[\mathrm\{e\}^\{-\frac\{1\}\{2\}(x-1)^2\}\right\]^\wedge =\mathrm\{e\}^\{-\mathrm\{ i\} \xi\}\mathrm\{e\}^\{-\frac\{\xi^2\}\{2a\}\} \Rightarrow\hat f(\xi)=\mathrm\{ i\}\frac\{\mathrm\{ d\}\}\{\mathrm\{ d\} \xi\}\mathrm\{e\}^\{-\frac\{\xi^2\}\{2a\}-\mathrm\{ i\} \xi\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (10)、 $\displaystyle f(x)=(x\sin x)^2\mathrm\{e\}^\{-|x|\}$; [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由题 1 (6) 知 \begin\{aligned\} \hat f(\xi)=-\frac\{\mathrm\{ d\}^2\}\{\mathrm\{ d\} \xi^2\}\left\[2\sqrt\{\frac\{2\}\{\pi\}\} \frac\{5-3\xi^2\}\{ [1+(\xi-2)^2] (1+\xi^2)[1+(\xi+2)^2]\}\right\]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (11)、 $\displaystyle f(x)=\frac\{x^2\}\{(a^2+x^2)^2\}$, $\displaystyle a > 0$. [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由题 1 (6) 知 \begin\{aligned\} \hat f(\xi)&=\left\[\frac\{a^2+x^2-a^2\}\{(a^2+x^2)^2\}\right\]^\wedge =\left(\frac\{1\}\{a^2+x^2\}\right)^\wedge -a^2\left\[\frac\{1\}\{(a^2+x^2)^2\}\right\]^\wedge\\\\ &=\frac\{1\}\{\sqrt\{2\pi a\}\}2\pi \mathrm\{ i\} \frac\{\mathrm\{e\}^\{-a|\xi|\}\}\{2\mathrm\{ i\}\} -a^2\frac\{1\}\{\sqrt\{2\pi\}a^3\} 2\pi \mathrm\{ i\} \left.\frac\{\mathrm\{ d\} \}\{\mathrm\{ d\} \xi\}\right|\_\{\xi=\mathrm\{ i\}\} \left\[\frac\{\mathrm\{e\}^\{\mathrm\{ i\} a|\xi|z\}\}\{(z+\mathrm\{ i\})^2\}\right\]\\\\ &=\sqrt\{\frac\{\pi\}\{2\}\} \frac\{\mathrm\{e\}^\{-a|\xi|\}(1-a\xi)\}\{2a\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 3、 求下列函数的 Fourier 逆变换: (1)、 $\displaystyle F(\xi)=\mathrm\{e\}^\{-a^2\xi^2t\}$, $\displaystyle a$ 为正常数, $\displaystyle t > 0$ 为参数; [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle A > 0\Rightarrow (\mathrm\{e\}^\{-A x^2\})^\wedge =\frac\{1\}\{\sqrt\{2A\}\}\mathrm\{e\}^\{-\frac\{\xi^2\}\{4A\}\}$ 知 \begin\{aligned\} (\mathrm\{e\}^\{-a^2\xi^2t\})^\wedge (x) =\left(\mathrm\{e\}^\{-a^2t \cdot \xi^2\}\right)^\wedge (x) =\frac\{1\}\{\sqrt\{2\cdot a^2t\}\}\mathrm\{e\}^\{-\frac\{x^2\}\{4a^2t\}\} =\frac\{1\}\{a\sqrt\{2t\}\}\mathrm\{e\}^\{-\frac\{x^2\}\{4a^2t\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 而 \begin\{aligned\} \left(\mathrm\{e\}^\{-a^2\xi^2 t\}\right)^\vee(x) =\left(\mathrm\{e\}^\{-a^2\xi^2t\}\right)^\wedge (-x) =\frac\{1\}\{a\sqrt\{2t\}\}\mathrm\{e\}^\{-\frac\{x^2\}\{4a^2t\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (2)、 $\displaystyle F(\xi)=\mathrm\{e\}^\{-|\xi|t\}$, $\displaystyle t > 0$ 为参数. [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由题 1 (4) 知 \begin\{aligned\} \left(\mathrm\{e\}^\{-a|x|\}\right)^\wedge=\frac\{2a\}\{\sqrt\{2\pi\}(\xi^2+a^2)\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 而 \begin\{aligned\} \left\[\mathrm\{e\}^\{-|\xi|t\}\right\]^\vee(x)=\left\[\mathrm\{e\}^\{-|\xi|t\}\right\]^\wedge(-x) =\frac\{2t\}\{\sqrt\{2\pi\}(x^2+t^2)\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 4、 求下列函数的 Fourier 变换: (1)、 $\displaystyle f(x\_1,\cdots,x\_n)=\mathrm\{e\}^\{-(x\_1^2+\cdots+x\_n^2)+2x\_1\}$; [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / \begin\{aligned\} &\hat f(\xi\_1,\cdots,\xi\_n) =\mathrm\{e\} \left\[\mathrm\{e\}^\{-(x\_1-1)^2\}\right\]^\wedge(\xi\_1) \cdot \left(\mathrm\{e\}^\{-x\_2^2\}\right)^\wedge(\xi\_2) \cdots \left(\mathrm\{e\}^\{-x\_n^2\}\right)^\wedge(\xi\_n)\\\\ =&\mathrm\{e\}\cdot \mathrm\{e\}^\{-\mathrm\{ i\} \xi\_1\}\frac\{1\}\{\sqrt\{2\}\} \mathrm\{e\}^\{-\frac\{\xi\_1^2\}\{4\}\}\cdot \frac\{1\}\{\sqrt\{2\}\} \mathrm\{e\}^\{-\frac\{\xi\_2^2\}\{4\}\}\cdots \frac\{1\}\{\sqrt\{2\}\} \mathrm\{e\}^\{-\frac\{\xi\_n^2\}\{4\}\} =\frac\{1\}\{2^\frac\{n\}\{2\}\} \mathrm\{e\}^\{-\frac\{\xi\_1^2+\cdots+\xi\_n^2\}\{4\}-\mathrm\{ i\} \xi\_1+1\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (2)、 $\displaystyle f(x\_1,\cdots,x\_n)=x\_1\mathrm\{e\}^\{-a(|x\_1|+\cdots+|x\_n|)\}$, $\displaystyle a$ 为正常数. [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由题 1 (4) 知 \begin\{aligned\} \left(\mathrm\{e\}^\{-a|x|\}\right)^\wedge=\frac\{2a\}\{\sqrt\{2\pi\}(\xi^2+a^2)\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 \begin\{aligned\} &\hat f(\xi\_1,\cdots, \xi\_n) =\left(x\_1\mathrm\{e\}^\{-a|x\_1|\}\right)^\wedge(\xi\_1) \cdot \left(\mathrm\{e\}^\{-a|x\_2|\}\right)^\wedge(\xi\_2) \cdots \left(\mathrm\{e\}^\{-a|x\_n|\}\right)^\wedge(\xi\_n)\\\\ =&\mathrm\{ i\} \frac\{\mathrm\{ d\}\}\{\mathrm\{ d\} \xi\_1\} \left\[\frac\{2a\}\{\sqrt\{2\pi\}(\xi\_1^2+a^2)\}\right\] \frac\{2a\}\{\sqrt\{2\pi\}(\xi\_2^2+a^2)\} \cdots\frac\{2a\}\{\sqrt\{2\pi\}(\xi\_n^2+a^2)\}\\\\ =&-\frac\{(2a)^n2\mathrm\{ i\} \xi\_1\}\{(2\pi)^\frac\{n\}\{2\} (\xi\_1^2+a^2)^2(\xi\_2^2+a^2)\cdots(\xi\_n^2+a^2)\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 5、 求下列函数的 Fourier 变换, 并证明相应的积分等式成立: (1)、 $\displaystyle f(x)=\mathrm\{e\}^\{- a|x|\}\ ( a > 0)$, 证明 \begin\{aligned\} \int\_0^\infty \frac\{\cos \xi x\}\{ a^2+\xi^2\}\mathrm\{ d\} \xi=\frac\{\pi\}\{2 a\}\mathrm\{e\}^\{- a|x|\}; \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由题 1 (4) 知 $\displaystyle \left\[\mathrm\{e\}^\{- a |x|\}\right\]^\wedge =\frac\{2 a\}\{\sqrt\{2\pi\}(\xi^2+ a^2)\}$, 而 \begin\{aligned\} \mathrm\{e\}^\{- a |x|\} =&\left\[\frac\{2 a\}\{\sqrt\{2\pi\}(\xi^2+ a^2)\}\right\]^\vee =\frac\{1\}\{\sqrt\{2\pi\}\}\int\_\{\mathbb\{R\}\} \frac\{2 a\}\{\sqrt\{2\pi\}(\xi^2+ a^2)\}\mathrm\{e\}^\{\mathrm\{ i\} \xi x\}\mathrm\{ d\} \xi\\\\ \xlongequal\{\tiny\mbox\{对称性\}\}& \frac\{ a\}\{\pi\}\int\_\{\mathbb\{R\}\}\frac\{\cos \xi x\}\{ a^2+\xi^2\}\mathrm\{ d\} \xi \xlongequal\{\tiny\mbox\{对称性\}\} \frac\{2 a\}\{\pi\}\int\_0^\infty \frac\{\cos \xi x\}\{ a^2+\xi^2\}\mathrm\{ d\} \xi. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (2)、 $\displaystyle f(x)=\mathrm\{e\}^\{-|x|\}\cos x$, 证明 \begin\{aligned\} \int\_0^\infty \frac\{\xi^2+2\}\{\xi^4+4\}\cos\xi x\mathrm\{ d\} \xi=\frac\{\pi\}\{2\}\mathrm\{e\}^\{-|x|\}\cos x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由题 1 (5) 知 \begin\{aligned\} (\mathrm\{e\}^\{-|x|\cos x\})^\wedge =\frac\{1\}\{\sqrt\{2\pi\}\}\left\[\frac\{1\}\{1+(\xi+1)^2\} +\frac\{1\}\{1+(\xi-1)^2\}\right\] =\sqrt\{\frac\{2\}\{\pi\}\}\frac\{\xi^2+2\}\{\xi^4+4\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 \begin\{aligned\} \mathrm\{e\}^\{-|x|\}\cos x =&\left(\sqrt\{\frac\{2\}\{\pi\}\}\frac\{\xi^2+2\}\{\xi^4+4\}\right)^\vee(x) =\frac\{1\}\{\sqrt\{2\pi\}\}\int\_\{\mathbb\{R\}\} \sqrt\{\frac\{2\}\{\pi\}\}\frac\{\xi^2+2\}\{\xi^4+4\}\mathrm\{e\}^\{\mathrm\{ i\} \xi x\}\mathrm\{ d\} \xi\\\\ \xlongequal\{\tiny\mbox\{对称性\}\}&\frac\{1\}\{\pi\}\int\_\{\mathbb\{R\}\} \frac\{\xi^2+2\}\{\xi^4+4\} \cos \xi x\mathrm\{ d\} \xi \xlongequal\{\tiny\mbox\{对称性\}\} \frac\{2\}\{\pi\}\int\_0^\infty \frac\{\xi^2+2\}\{\xi^4+4\}\cos \xi x\mathrm\{ d\} \xi. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (3)、 $\displaystyle f(x)=\left\\{\begin\{array\}\{llllllllllll\}\sin x,&|x|\leq \pi,\\\\ 0,&|x| > \pi,\end\{array\}\right.$ 证明 \begin\{aligned\} \int\_0^\infty \frac\{\sin \xi \pi \sin \xi x\}\{1-\xi^2\}\mathrm\{ d\} \xi =\left\\{\begin\{array\}\{llllllllllll\}\frac\{\pi\}\{2\}\sin x, &|x|\leq \pi,\\\\ 0,&|x| > \pi.\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由题 1 (3) 知 \begin\{aligned\} \hat f(\xi)&=\frac\{\mathrm\{ i\}\}\{\sqrt\{2\pi\}\}\left\[\frac\{\sin (1-\xi)\pi\}\{1-\xi\}-\frac\{\sin (1+\xi)\pi\}\{1+\xi\}\right\]\\\\ &=\frac\{\mathrm\{ i\}\}\{\sqrt\{2\pi\}\}\left(\frac\{1\}\{1-\xi\}+\frac\{1\}\{1+\xi\}\right) \sin \xi \pi =\frac\{\mathrm\{ i\}\}\{\sqrt\{2\pi\}\} \frac\{2\sin \xi \pi\}\{1-\xi^2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 \begin\{aligned\} f(x)=&\left\[\frac\{\mathrm\{ i\}\}\{\sqrt\{2\pi\}\} \frac\{2\sin \xi \pi\}\{1-\xi^2\}\right\]^\vee(x) =\frac\{1\}\{\sqrt\{2\pi\}\} \int\_\{\mathbb\{R\}\} \frac\{\mathrm\{ i\}\}\{\sqrt\{2\pi\}\} \frac\{2\sin \xi \pi\}\{1-\xi^2\} \mathrm\{e\}^\{\mathrm\{ i\} \xi x\}\mathrm\{ d\} x\\\\ \xlongequal\{\tiny\mbox\{对称性\}\}&\frac\{2\}\{\pi\}\int\_0^\infty \frac\{\sin \xi \pi \sin \xi x\}\{1-\xi^2\}\mathrm\{ d\} \xi. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 6、 设 $\displaystyle f(x)$ 在 $\displaystyle \mathbb\{R\}$ 上分段连续、绝对可积, 函数 \begin\{aligned\} g(x)=\int\_\{-\infty\}^x f(\xi)\mathrm\{ d\} \xi \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 在 $\displaystyle \mathbb\{R\}$ 上绝对可积, $\displaystyle \alpha$ 是一个不为零的常数. 证明: (1)、 $\displaystyle [f(\alpha x)]^\wedge(\xi)=\frac\{1\}\{|\alpha|\} [f(x)]^\wedge\left(\frac\{\xi\}\{\alpha\}\right)$; (2)、 $\displaystyle \hat g(\xi)=\frac\{1\}\{\mathrm\{ i\} \xi\} \hat f(\xi)$. [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 若 $\displaystyle \alpha > 0$, 则 \begin\{aligned\} &\left\[f(\alpha x)\right\]^\wedge (x) =\frac\{1\}\{\sqrt\{2\pi\}\}\int\_\{\mathbb\{R\}\} f(\alpha x)\mathrm\{e\}^\{-\mathrm\{ i\} x\xi\}\mathrm\{ d\} x\\\\ \stackrel\{\alpha x=t\}\{=\}&\frac\{1\}\{\sqrt\{2\pi\}\}\int\_\{\mathbb\{R\}\} f(t)\mathrm\{e\}^\{-\mathrm\{ i\} t\frac\{\xi\}\{\alpha\}\} \frac\{\mathrm\{ d\} t\}\{\alpha\} =\frac\{1\}\{\alpha\}\hat f\left(\frac\{\xi\}\{\alpha\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 若 $\displaystyle \alpha < 0$, 则 \begin\{aligned\} &\left\[f(\alpha x)\right\]^\wedge (x) =\frac\{1\}\{\sqrt\{2\pi\}\}\int\_\{\mathbb\{R\}\} f(\alpha x)\mathrm\{e\}^\{-\mathrm\{ i\} x\xi\}\mathrm\{ d\} x\\\\ \stackrel\{\alpha x=t\}\{=\}&\frac\{1\}\{\sqrt\{2\pi\}\}\int\_\{+\infty\}^\{-\infty\} f(t)\mathrm\{e\}^\{-\mathrm\{ i\} t\frac\{\xi\}\{\alpha\}\} \frac\{\mathrm\{ d\} t\}\{\alpha\} =\frac\{1\}\{-\alpha\}\hat f\left(\frac\{\xi\}\{\alpha\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (2)、 $\displaystyle g'(x)=f(x)$ 求 Fourier 变换知 \begin\{aligned\} \mathrm\{ i\} \xi \hat g(\xi)=\hat f(\xi) \Rightarrow \hat g(\xi)=\frac\{1\}\{\mathrm\{ i\} \xi\} \hat f(\xi). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/