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2017/04/15

1、$\text{p.v.}\,\frac{1}{x}$

因为$(x \ln x - x)‘ = \ln x$， 所以$\int_0^a \ln x \mathrm{\,d}x = \lim_{\epsilon \to 0^+} \int_\epsilon^a \ln x \mathrm{\,d}x = \lim_{\epsilon \to 0^+}(x \ln x - x)\big|_\epsilon^a = a \ln a - a$，即是说对任意的$\varphi \in C_c^\infty(\mathbb{R}^1)$， $\langle \ln|x|, \varphi \rangle$有意义，且连续性也显而易见。所以$\ln|x| \in \mathcal{D}‘(\mathbb{R}^1)$。所以$(\ln|x|)‘ \in \mathcal{D}‘(\mathbb{R}^1)$。将$(\ln|x|)‘$定义为$\text{p.v.}\,\frac{1}{x}$：$$\boxed{\text{p.v.}\,\frac{1}{x} \triangleq (\ln|x|)‘. \quad\text{Then}\quad \text{p.v.}\,\frac{1}{x} \in \mathcal{D}‘(\mathbb{R}^1).}$$

2、$\frac{1}{x \pm i0}$

设$\{f_n\} \subset \mathcal{D}‘(\mathbb{R}^1)$，则$\{f_n‘\} \subset \mathcal{D}‘(\mathbb{R}^1)$。若存在$f \in\mathcal{D}‘(\mathbb{R}^1)$使得$f_n \to f \text{ in }\mathcal{D}‘(\mathbb{R}^1)$，则$\lim_{n \to +\infty} f_n‘$存在且$f_n‘ \to f‘ \text{ in }\mathcal{D}‘(\mathbb{R}^1)$.

当$\epsilon > 0$时，$\ln(x + i\epsilon) = \ln|x + i\epsilon| + i \cdot \arg(x + i\epsilon)$。所以 $\frac{1}{x + i\epsilon} = \big( \ln(x + i\epsilon) \big)‘ \in \mathcal{D}‘(\mathbb{R}_x^1), \,\forall \epsilon > 0$。因为$\lim_{\epsilon \to 0^+} \ln(x + i\epsilon) = \ln|x| - i\pi \cdot (H(x)-1) \in \mathcal{D}‘(\mathbb{R}_x^1)$，所以在$\mathcal{D}‘(\mathbb{R}_x^1)$上$\lim_{\epsilon \to 0^+} \frac{1}{x + i\epsilon}$存在且

\begin{align*}
\lim_{\epsilon \to 0^+} \frac{1}{x + i\epsilon} & = \lim_{\epsilon \to 0^+} \big( \ln(x + i\epsilon) \big)‘
= \big( \lim_{\epsilon \to 0^+} \ln(x + i\epsilon) \big)‘ \\
& = \big( \ln|x| - i\pi \cdot (H(x)-1) \big)‘ \\
& = \text{p.v.}\,\frac{1}{x} - i\pi\delta(x).
\end{align*}

现在将$\frac{1}{x + i0}$定义为$\lim_{\epsilon \to 0^+} \frac{1}{x + i\epsilon}$：$$\boxed{\frac{1}{x + i0} \triangleq \lim_{\epsilon \to 0^+} \frac{1}{x + i\epsilon}. \quad\text{Then}\quad \frac{1}{x + i0} = \text{p.v.}\,\frac{1}{x} - i\pi\delta(x) \in \mathcal{D}‘(\mathbb{R}^1).}$$ 同理，对正负的$i0$，我们有$$\frac{1}{x \pm i0} = \text{p.v.}\,\frac{1}{x} \mp i\pi\delta(x) \in \mathcal{D}‘(\mathbb{R}^1).$$

两个1/x类的广义函数