I was trying to reason about the possible definition of a distribution from the principal value of $1/x$ to the power $n$ ($n\in\mathbb{N}$).
I have managed to go on up to $1/x^3$, which is correctly defined and linear on $\mathcal{D}(R)$ but now I am curious of what we could obtain by going on.
Applying the possible distribution of a test function we would get
$$\langle P\left(\dfrac{1}{x^n}\right), \phi \rangle = \int_{|x|>\epsilon} \phi^{(n)}(x)\ln|x|\ \text{d}x$$
I started with the logarithm for it's locally integrable and then it defines a distribution via $\phi$ in that way. Assuming that everything is done with diligence, the integration by parts transfers the derivatives on the logarithm, given little by little the $1/x^n$ factor.
First assumption, which could be wrong, but it's useful for the calculations: I assume that $\phi$ is odd and the $n-$th derivative of $\phi$ is even, hence I take
$$\int_{\epsilon}^{+\infty} \phi^{(n)}(x)\ln(x)\ \text{d}x$$
Oh by the way: $\epsilon > 0$.
Now we can integrate by parts $n$ times. The $n$-th derivative of the logarithm is a well known amusing exercise. What I get is
$$2\phi^{(n-1)}(\epsilon)\ln(\epsilon) + 2\sum_{k = 2}^n (-1)^{k-1}(k-1)! \frac{\phi^{(n-k)}}{\epsilon^k}$$
Now my problem is: how to show that this still defines a distribution? And what kind of distribution is? Is there a way to show that all the terms but one (the last one $k = n$) "are" or "go to" zero?
I don't know how to proceed so I ask for help of who here really know where to put the hands. Anyway, the behaviour of that distribution, due to the $\epsilon^{-k}$ factor, reminds me of meromorphic functions. Maybe it has nothing to do with this... Just a flash.
Thank you!