# Seeing $1/x$ as a distribution

I have to show that by defining $$\langle u, f\rangle=\lim_{\varepsilon\rightarrow 0}\int_{-\infty}^{-\varepsilon}+\int_{\varepsilon}^{\infty}\frac{f(x)}{x}dx$$

with $f\in\mathcal{D}(\mathbb{R})$, we have $u\in\mathcal{D}'(\mathbb{R})$ and find $u'$.

The tip is to use the mean value theorem. I did the second part with $u'$ but my problem is to show that it is well defined.

So I got $$\int_{-\infty}^{-\varepsilon}\frac{f(x)}{x}dx=\int_{\varepsilon}^{\infty}\frac{f(-x)}{x}dx$$

so

$$\int_{-\infty}^{-\varepsilon}+\int_{\varepsilon}^{\infty}\frac{f(x)}{x}dx=\int_{\varepsilon}^{\infty} \frac{f(x)+f(-x)}{x}dx$$

It would be great if I got $\frac{f(x)-f(-x)}{x}$ because then

$$\frac{f(x)-f(-x)}{x}=2\frac{f(x)-f(-x)}{x-(-x)}=2f'(y)$$

for some $y\in(-x,x)$. But now how can I apply the mean value theorem to show that it is well defined?

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Sign error: $$\int_{-\infty}^{-\varepsilon}\frac{f(x)}{x}dx= -\int_{\varepsilon}^{\infty} \frac{f(-x)}{x}dx$$ Just a remark: another way to approach this construction is to begin with the distribution $\log|x|$ and take its derivative. –  user53153 Jan 15 '13 at 3:03
@rom As the integral domain and sup $|f'|$ are bounded you are done. –  Vobo Jan 15 '13 at 7:01
@5PM with the change of variables $x=-y$ we have $dx=-dy$ so I think that I have the sign correctly. Initially I got the minus but then I couldn't find $u'$. –  rom Jan 15 '13 at 12:08
@Vobo I don't quite see it, how do I make $f'$ to appear. –  rom Jan 15 '13 at 12:10
@rom You also flipped the limits of integration; that changes the sign. –  user53153 Jan 15 '13 at 12:12