Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers 2

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:

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

share|improve this answer
add comment

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

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.