# principal value as distribution, written as integral over singularity

Let $C_0^\infty(\mathbb{R})$ be the set of smooth functions with compact support on the real line $\mathbb{R}.$ Then, the map

$$\operatorname{p.\!v.}\left(\frac{1}{x}\right)\,: C_0^\infty(\mathbb{R}) \to \mathbb{C}$$

defined via the Cauchy principal value as

$$\operatorname{p.\!v.}\left(\frac{1}{x}\right)(u)=\lim_{\varepsilon\to 0+} \int_{\mathbb{R}\setminus [-\varepsilon;\varepsilon]} \frac{u(x)}{x} \, \mathrm{d}x \quad\text{ for }u\in C_0^\infty(\mathbb{R})$$

Now why is $$\lim_{\varepsilon\to 0+} \int_{\mathbb{R}\setminus [-\varepsilon;\varepsilon]} \frac{u(x)}{x} \, \mathrm{d}x = \int_0^{+\infty} \frac{u(x)-u(-x)}{x}\, \mathrm{d}x$$ why is the integral defined on the left.

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We can write $$I(\varepsilon):=\int_{\Bbb R\setminus [-\varepsilon,\varepsilon]}\frac{u(x)}xdx=\int_{-\infty}^{-\varepsilon}\frac{u(x)}xdx+\int_{\varepsilon}^{\infty}\frac{u(x)}xdx.$$ In the first integral of the RHs, we do the substitution $t=-x$, then $$I(\varepsilon)=-\int_{\varepsilon}^{+\infty}\frac{u(t)}tdt+\int_{\varepsilon}^{\infty}\frac{u(x)}xdx=\int_{\varepsilon}^{+\infty}\frac{u(t)-u(-t)}tdt.$$ Now we can conclude, since, by fundamental theorem of analysis, the integral $\int_0^{+\infty}\frac{u(t)-u(-t)}tdt$ is convergent. Indeed, $$u(t)-u(-t)=\int_{-t}^tu'(s)ds=\left[su'(s)\right]_{-t}^t-\int_{-t}^tsu''(s)ds\\= t(u'(t)+u'(-t))-\int_{-t}^tsu''(s)ds$$ hence, for $0<t\leq 1$ $$\frac{|u(t)-u(-t)|}t\leq 2\sup_{|s|\leq 1}|u'(s)|+2\sup_{|s|\leq 1}|u''(s)|.$$

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can you please give me an more exact argument why $\int_0^{\infty} (u(t)-u(-t))/i \mathrm{d}t$ is convergent, i guess it still has an singularity at 0? – Stefan Jun 25 '12 at 18:12
@Stefan: Your guess is wrong. By Taylor's theorem, $u(t) = u(0) + tu'(s_t)$ for some $s_t \in [0,t]$. Now for $u(t) - u(-t)$ the constant expression $u(0)$ vanishes and the integral exists. – Vobo Jun 25 '12 at 18:33

Because $1/x$ is an odd function. So, decomposing $u(x)$ in its odd and even parts, that is

$$u(x)=\frac{u(x)+u(-x)}{2}+\frac{u(x)-u(-x)}{2}$$

we have

$$\lim_{\varepsilon \to 0} \int_{\lvert x \rvert > \varepsilon} \frac{u(x)}{x}\, dx= \lim_{\varepsilon \to 0} \left(\int_{\lvert x \rvert > \varepsilon} \frac{u(x)+u(-x)}{2x}\, dx + \int_{\lvert x \rvert > \varepsilon} \frac{u(x)-u(-x)}{2x}\, dx\right)$$

and the first integral on the right hand side vanishes because its argument is odd. On the contrary, the second integral has an even argument, so we can rewrite it as follows:

$$\lim_{\varepsilon \to 0}\int_{\lvert x \rvert > \varepsilon} \frac{u(x)-u(-x)}{2x}\, dx = \lim_{\varepsilon \to 0} \int_\varepsilon^\infty \frac{u(x)-u(-x)}{x}\, dx.$$

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