傅里叶变换： 与

以下以

$$\mathcal{F}\{f\}(\omega)=\int_{-\infty}^{\infty} f(x)e^{-i\omega x}\,dx$$

为定义（结果是分布意义下的）。

因为

$$\cos x=\frac{e^{ix}+e^{-ix}}{2},\\qquad \sin x=\frac{e^{ix}-e^{-ix}}{2i},$$

且 $\mathcal{F}\{e^{iax}\}=2\pi\,\delta(\omega-a)$，所以

$$\boxed{\mathcal{F}\{\cos x\}(\omega)=\pi\bigl[\delta(\omega-1)+\delta(\omega+1)\bigr]},$$ $$\boxed{\mathcal{F}\{\sin x\}(\omega)=\frac{\pi}{i}\bigl[\delta(\omega-1)-\delta(\omega+1)\bigr]}.$$

不同傅里叶变换约定（是否带 $2\pi$ 或 $1/\sqrt{2\pi}$）只会改变常数因子。

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英文解析

The following is defined as
$$\mathcal{F}\{f\}(\omega)=\int_{-\infty}^{\infty} f(x)e^{-i\omega x}\,dx$$
(in the sense of distributions).

Since
$$\cos x=\frac{e^{ix}+e^{-ix}}{2},\\qquad \sin x=\frac{e^{ix}-e^{-ix}}{2i},$$
and $\mathcal{F}\{e^{iax}\}=2\pi\,\delta(\omega-a)$, it follows that
$$\boxed{\mathcal{F}\{\cos x\}(\omega)=\pi\bigl[\delta(\omega-1)+\delta(\omega+1)\bigr]},$$$$\boxed{\mathcal{F}\{\sin x\}(\omega)=\frac{\pi}{i}\bigl[\delta(\omega-1)-\delta(\omega+1)\bigr]}.$$
Different Fourier transform conventions (whether or not they include $2\pi$ or $1/\sqrt{2\pi}$) only change the constant factor.