Define \(u\in\mathcal D'(\mathbb R)\) formally by \[ \langle u,\varphi\rangle = \lim_{\varepsilon\to0^+} \left( \int_{-\infty}^{-\varepsilon}\frac{\varphi(x)}x\,dx + \int_{\varepsilon}^{\infty}\frac{\varphi(x)}x\,dx \right), \qquad \varphi\in C_c^\infty(\mathbb R). \] Show that the limit exists and that \[ u=(\log|x|)' \] in the sense of distributions.
Solution
1. The principal-value limit exists.
Fix \(\varphi\in C_c^\infty(\mathbb R)\). For \(0<\varepsilon<1\), symmetry gives \[ \int_{-1}^{-\varepsilon}\frac{\varphi(0)}x\,dx + \int_{\varepsilon}^{1}\frac{\varphi(0)}x\,dx=0. \] Hence the truncated principal value can be rewritten as \[ \begin{aligned} &\int_{\varepsilon<|x|<1}\frac{\varphi(x)-\varphi(0)}x\,dx + \int_{|x|\ge1}\frac{\varphi(x)}x\,dx. \end{aligned} \] The second integral is absolutely convergent because \(\varphi\) has compact support. Near \(0\), the mean-value theorem gives \[ |\varphi(x)-\varphi(0)|\le C|x|, \] so \[ \left|\frac{\varphi(x)-\varphi(0)}x\right|\le C. \] Thus the first integral converges absolutely as \(\varepsilon\to0^+\). Therefore the principal-value limit exists.
2. The locally integrable function \(\log|x|\) defines a distribution.
Since \[ \int_0^1|\log x|\,dx<\infty, \] we have \(\log|x|\in L^1_{\mathrm{loc}}(\mathbb R)\). Hence it defines a distribution by \[ \langle \log|x|,\varphi\rangle = \int_{\mathbb R}\log|x|\,\varphi(x)\,dx. \]
3. Its distributional derivative equals \(u\).
By definition, \[ \langle (\log|x|)',\varphi\rangle =-\int_{\mathbb R}\log|x|\,\varphi'(x)\,dx. \] Split at \(\pm\varepsilon\) and integrate by parts: \[ \begin{aligned} &-\int_{-\infty}^{-\varepsilon}\log|x|\,\varphi'(x)\,dx -\int_{\varepsilon}^{\infty}\log x\,\varphi'(x)\,dx\\ &\qquad= \int_{-\infty}^{-\varepsilon}\frac{\varphi(x)}x\,dx + \int_{\varepsilon}^{\infty}\frac{\varphi(x)}x\,dx + \log\varepsilon\,[\varphi(\varepsilon)-\varphi(-\varepsilon)]. \end{aligned} \] Because \(\varphi\) is smooth, \[ \varphi(\varepsilon)-\varphi(-\varepsilon)=O(\varepsilon), \] and therefore \[ \log\varepsilon\,[\varphi(\varepsilon)-\varphi(-\varepsilon)]\to0 \] as \(\varepsilon\to0^+\). Taking the limit yields \[ \langle (\log|x|)',\varphi\rangle =\operatorname{p.v.}\int_{\mathbb R}\frac{\varphi(x)}x\,dx =\langle u,\varphi\rangle. \] Thus \[ \boxed{u=(\log|x|)'} \] in \(\mathcal D'(\mathbb R)\).