Question : Show that $<f_a, \varphi> = \int_{-\infty}^{\infty} \frac{\varphi(x)}{|x|} dx$ for any $\varphi \in D(\mathbb{R})$ for which $\varphi(0)=0$.
I am a little bit confused how to solve this problem. I know that $\varphi(0)=0=\int_{-\infty}^{\infty} \delta(x) \varphi(x) dx$. Is there anyone could give me a simple hint to solve that problem? Do I have to use integration by part?
The distribution $f_a$ was defined for every $a>0$ in Strichartz's book by $$ \langle f_a , \varphi \rangle = \int_{-\infty}^{-a} \frac{\varphi(x)}{|x|}\, dx + \int_a^{+\infty} \frac{\varphi(x)}{|x|}\, dx + \int_{-a}^a \frac{\varphi(x)-\varphi(0)}{|x|}\, dx. $$ If $\varphi(0)=0$, then $$ \langle f_a , \varphi \rangle = \int_{-\infty}^{-a} \frac{\varphi(x)}{|x|}\, dx + \int_a^{+\infty} \frac{\varphi(x)}{|x|}\, dx + \int_{-a}^a \frac{\varphi(x)}{|x|}\, dx. \tag{1} $$ We need to prove that $x \mapsto \frac{\varphi(x)}{|x|}$ is integrable, and then we simply get that $\langle f_a,\varphi \rangle = \int_{\mathbb{R}} \frac{\varphi(x)}{|x|}dx$.
The first and the second terms in (1) are obviously finite, because the support of $\varphi$ is compact. The last term in (1) is finite because $\varphi(0)=0$. Hence there is a constant $L>0$ such that $|\varphi(x)| \leq L |x|$ for $x \in [-a,a]$. As a consequence, the integrand is bounded on the domain of integration, and the improper integral converges.