请求提供正切函数tan(x)的傅里叶变换表达式

First off, let's clarify a key point: the Fourier transform of $\tan(x)$ cannot be defined using standard Riemann or Lebesgue integrals because $\tan(x)$ has poles at $x = (n + \frac{1}{2})\pi$ for all integers $n$, and the integral $\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} \tan(x)\ e^{-ikx}\ dx$ diverges in the ordinary sense. Instead, we need to work with distributional Fourier transforms (generalized functions) to get a meaningful result.

Step 1: Represent $\tan(x)$ as a Distribution #

We start with the known principal value expansion of $\cot(x)$:
$$\cot(x) = \text{PV}\sum_{n=-\infty}^{\infty} \frac{1}{x - n\pi}$$
Since $\tan(x) = \cot\left(x - \frac{\pi}{2}\right)$, substituting gives us:
$$\tan(x) = \text{PV}\sum_{n=-\infty}^{\infty} \frac{1}{x - (n + \frac{1}{2})\pi}$$
Here, $\text{PV}$ denotes the Cauchy principal value, which is necessary to handle the poles of $\tan(x)$.

Step 2: Compute the Distributional Fourier Transform #

We use the fact that the Fourier transform of $\text{PV}\frac{1}{x - a}$ (under your normalization $\hat{f}(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{{\infty}f(x)e}{-ikx}dx$) is:
$$\mathcal{F}\left{\text{PV}\frac{1}{x - a}\right}(k) = \frac{-i\pi \text{sgn}(k)}{\sqrt{2\pi}} e^{-ika}$$
where $\text{sgn}(k)$ is the sign function ($1$ for $k>0$, $-1$ for $k<0$, $0$ for $k=0$).

Applying this to our expansion of $\tan(x)$:
$$\hat{\tan}(k) = \frac{1}{\sqrt{2\pi}} \sum_{n=-\infty}^{\infty} \mathcal{F}\left{\text{PV}\frac{1}{x - (n + \frac{1}{2})\pi}\right}(k)$$
Substitute the transform result:
$$\hat{\tan}(k) = \frac{-i\pi \text{sgn}(k)}{\sqrt{2\pi}} \sum_{n=-\infty}^{\infty} e^{-ik(n + \frac{1}{2})\pi}$$

Step 3: Evaluate the Infinite Series #

Using the Poisson summation formula, we know:
$$\sum_{n=-\infty}^{\infty} e^{-ikn\pi} = 2\sum_{m=-\infty}^{\infty} \delta(k - 2m)$$
where $\delta$ is the Dirac delta distribution. Multiplying by $e^{-ik\pi/2}$ (from the $(n + \frac{1}{2})\pi$ term) gives:
$$\sum_{n=-\infty}^{\infty} e^{-ik(n + \frac{1}{2})\pi} = e^{-ik\pi/2} \cdot 2\sum_{m=-\infty}^{\infty} \delta(k - 2m)$$
When $k = 2m$ (integer $m$), $e^{-ik\pi/2} = e^{-im\pi} = (-1)^m$. For $k$ not an even integer, the delta terms vanish.

Step 4: Final Simplified Expression #

Putting it all together, we get:
$$\hat{\tan}(k) = -i\sqrt{2\pi} \text{sgn}(k) \sum_{\substack{m=-\infty \ m \neq 0}}^{\infty} (-1)^m \delta(k - 2m)$$
Or, written more compactly:
$$\hat{\tan}(k) = i\sqrt{2\pi} \sum_{m=-\infty}^{\infty} (-1)^{m+1} \text{sgn}(m) \delta(k - 2m)$$

Why Your Mathematica and Manual Attempts Failed #

• Manual Calculation: The integral diverges in the ordinary sense, so standard integration techniques won't work. You need to leverage distribution theory to handle the poles and divergent series.
• Mathematica: By default, Mathematica tries to compute the integral as a regular convergent integral, which doesn't exist. It gets stuck trying to sum an infinite divergent series without recognizing the need for distributional methods. To compute this in Mathematica, you'd need to explicitly use distributional Fourier transform functions or define the principal value integral properly.

内容的提问来源于stack exchange，提问作者user266764