Let $a>0$ and $$\langle f_a,\psi\rangle=\int_{|x|>a} \frac{\psi(x)}{|x|} dx + \int_{|x|<a} \frac{\psi(x)-\psi(0)}{|x|}dx$$ Prove that $f_a$ does not depend on a.
Proof: To prove that $f_a$ does not depend on a we have to take the derivate of the $f_a$ and set it equal to 0.
$$\langle f_a',\psi \rangle =-\langle f_a,\psi'\rangle=-[\int_{|x|>a} \frac{\psi'(x)}{|x|} dx + \int_{|x|<a} \frac{\psi'(x)-\psi'(0)}{|x|}dx] =-[\int_{|x|>a} \frac{\psi'(x)}{|x|} dx + \int_{|x|<a} \psi'(c)dx] =-[\int_{-\infty}^{-a} \frac{\psi'(x)}{|x|} dx + \int_a^{\infty} \frac{\psi'(x)}{|x|} dx + \int_{|x|<a} \psi'(c)dx] =-[\int_{-\infty}^{-a} \frac{\psi'(x)}{|x|} dx + \int_a^{\infty} \frac{\psi'(x)}{|x|} dx + \int_{|x|<a} \psi'(c)dx] =-[\int_{-\infty}^{-a} \frac{\psi'(x)}{|x|} dx + \int_a^{\infty} \frac{\psi'(x)}{|x|} dx + 2a\psi'(c)$$
How do I proceed?
Let $a>0$. Put
$$T_a= \int_{|x| \geq a} \frac{\psi(x)}{|x|} dx + \int_{|x|<a} \frac{\psi(x)-\psi(0)}{|x|}dx$$ and for $\varepsilon\in ]0,a[$: $$T_a(\varepsilon)=\int_{|x| \geq a} \frac{\psi(x)}{|x|} dx + \int_{\varepsilon<|x|<a} \frac{\psi(x)-\psi(0)}{|x|}dx$$
We have that $T_a(\varepsilon)\to T_a$ if $\varepsilon\to 0$.
Now:
$$T_a(\varepsilon)=\int_{|x| \geq a} \frac{\psi(x)}{|x|} dx +\int_{\varepsilon<|x|<a} \frac{\psi(x)}{|x|}dx-\psi(0) \int_{\varepsilon<|x|<a} \frac{1}{|x|}dx$$
We have:
$$ \int_{\varepsilon<|x|<a} \frac{1}{|x|}dx=2\log(\frac{a}{\varepsilon})$$ Hence
$$T_a(\varepsilon)=\int_{|x| >\varepsilon} \frac{\psi(x)}{|x|} dx-2\psi(0)\log(\frac{a}{\varepsilon})$$
We have also for $b>a$:
$$T_b(\varepsilon)=\int_{|x| >\varepsilon} \frac{\psi(x)}{|x|} dx-2\psi(0)\log(\frac{b}{\varepsilon})$$
Hence: $$T_b(\varepsilon)-T_a(\varepsilon)=-2\psi(0)(\log b - \log a)$$ And now if $\varepsilon \to 0$: $$T_b-T_a=-2\psi(0)(\log b-\log a)$$ (So this is $T_a+2\psi(0)\log a$ that is independent of $a>0$).