Can we assign meaningful values to $\int_{-a}^bf(x)/x^n\,\mathrm dx$ for $n=2,3,\dots$ The question title pretty much says it all. Let $f(x)\geq 0$ on $x\in[a,b]$ with $f(0)>0$ and $a,b>0$.  In the case for $n=1$ we have the principal value intepretation which assigns a (usually finite) value by taking a symmetric limit around the pole via
$$
I\overset{\text{p.v.}}{=}\lim_{\epsilon\nearrow 0}\left(\int_{-a}^{-\epsilon}+\int_\epsilon^b\right)f(x)\frac{\mathrm dx}{x}.
$$
I understand that the choice of this symmetric limit is in some sense natural because it assigns a principal value of zero to odd functions when the interval of integration is symmetric about the origin.  For $n=2$ our problem is much more severe since there is no way to cancel out divergent parts of the integral with a limit. However, for all $n=2m+1$, $m=1,2,\dots$ we again have a situation where at least some sort of symmetric limit about the pole has promise.

Is there a meaningful way in which we can assign values $\int_{-a}^bf(x)/x^n\,\mathrm dx$ for $n=2,3,\dots$?

I understand that the word meaningful may very well be context dependent. My main interest is the case where $f$ is a probability density on $\operatorname{supp}(X)=[-a,b]$ with nonzero density at the origin. In this context, the integrals described above would assign values to negative moments to the random variable $X\sim f$.

So as it turns out, regularizations of integrals of this form are not always unique. One way to generalize the Cauchy principal value is found in A Generalization of the Cauchy Principal Value, where the author gives the regularization (denoted $\mathcal P$):
$$
\mathcal P\int_a^b f(x)\frac{\mathrm dx}{(x-u)^{n+1}}=\lim_{\epsilon\nearrow 0}\left(\int_a^{u-\epsilon} f(x)\frac{\mathrm dx}{(x-u)^{n+1}}+\int_{u+\epsilon}^b f(x)\frac{\mathrm dx}{(x-u)^{n+1}}-H_n(u,\epsilon)\right),
$$
for some $a<u<b$ and $n\in\Bbb N_0$ where
$$
H_n(u,\epsilon)=%
\begin{cases}
0, &n=0\\
\sum_{\ell=0}^{n-1}\frac{f^{(\ell)}(u)}{\ell!}\frac{1-(-1)^{n-\ell}}{(n-\ell) \epsilon^{n-\ell}}, &n\in\Bbb N.
\end{cases}
$$
For $n=0$ we get the Cauchy principal value and $n=1$ gives the Hadamard finit part.
