Does $\int_0^\infty \frac{\phi(x)-\phi(0)}{x} \text{d}x$ exist if $\phi \in C_0^\infty$?

Question

Good day,

In the lectures on distribution theory we had a distribution $f \in \mathscr{D}'$ (the dual space of $\mathscr{D}:=C_0^\infty$) defined as $$f(\phi):=\langle f,\phi \rangle := \int_0^\infty \frac{\phi(x)-\phi(0)}{x} \text{d}x$$ for all $\phi \in \mathscr{D}$ and it was stated that linearity and continuity can be proved easily...

But do I even know that the integral exists?

We proved that $$\psi(x):= \frac{\phi(x)-\phi(0)}{x} \in C^\infty. $$ Additionally we know that $\phi \in \mathscr{D}$ has compact support, so let $\phi$ be an arbitrary test function with $\text{supp} (\phi) \subset [-M,M], ~M>0$. So I thought I can write

$$\langle f,\phi \rangle := \int_0^\infty \frac{\phi(x)-\phi(0)}{x} \text{d}x"="\int_0^M \frac{\phi(x)-\phi(0)}{x} \text{d}x - \int_M^\infty \frac{\phi(0)}{x} \text{d}x $$ but the right integral is clearly infinite if $0 \in \text{supp}(\phi)$, so I am quite irritated how to prove that the integral exists.

Can someone please help me?

Thanks a lot, Marvin

Answer 1

As written, the integral does not exist if $\phi(0) \ne 0$. Hence, either you have an error in your transcription or the error was present in the lecture.

What could be meant? First, the integral could be $\int_{-\infty}^\infty$ (to be understood in a Cauchy-PV sense), or $\int_0^M$ for some fixed $M> 0$.