= Solution
There is an important convention issue in the printed statement. The ordinary <Cauchy principal value> exists only for $n=1$. Closing the Fourier contour in the lower half-plane, with a small indentation above the real-axis pole, gives
$$
\boxed{\operatorname{PV}\int_{\mathbb R}\frac{e^{-ix}}x\,dx=-i\pi}.
$$
Equivalently, the odd cosine contribution cancels and the <Dirichlet integral> $\int_0^\infty\sin x/x\,dx=\pi/2$ supplies the imaginary part.
For $n\ge2$, the paired integrand near zero is $2\cos x/x^n$ if $n$ is even and $-2i\sin x/x^n$ if $n$ is odd. Its leading even power is nonintegrable: $x^{-n}$ in the first case and $x^{1-n}$ in the second. Thus \b[the ordinary principal value diverges for every $n>1$].
If the intended convention is the <Hadamard finite-part integral>, define it consistently by distributional differentiation,
$$
\operatorname{Pf}\frac1{x^n}
=\frac{(-1)^{n-1}}{(n-1)!}\frac{d^{n-1}}{dx^{n-1}}\operatorname{PV}\frac1x.
$$
Integration by parts in this distributional pairing, or regularized Fourier transformation, then gives
$$
\boxed{\operatorname{Pf}\int_{\mathbb R}\frac{e^{-ix}}{x^n}\,dx
=\frac{\pi(-i)^n}{(n-1)!}}.
$$
The distinction is essential: subtracting the divergent local Taylor terms supplies a finite part, but symmetric exclusion alone does not.
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