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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.

StefanH
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2 Answers2

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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 $0<t\leq 1$ $$\frac{|u(t)-u(-t)|}t\leq 2\sup_{|s|\leq 1}|u'(s)|+2\sup_{|s|\leq 1}|u''(s)|.$$

Davide Giraudo
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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? – StefanH Jun 25 '12 at 18:12
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    @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
  • Why are you taking $0 < t \leq 1$? And also, why suddenly are you using the absolute value $|u(t) - u(-t)|$? Thanks for your help! – user110320 Dec 16 '17 at 14:05
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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:

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

Giuseppe Negro
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