General Fourier inversion formula (Gil-Pelaez)

Question

Gil-Pelaez (1951) proves the Fourier inversion formula \begin{align*} F(x) &= \frac{1}{2} + \frac{1}{2\pi} \int_0^\infty \frac{e^{itx}\phi(-t)-e^{-itx}\phi(t)}{it}dt \\ &= \frac{1}{2} - \frac{1}{\pi} \int_0^\infty \Im\left(\frac{e^{-itx}\phi(t)}{t}\right)dt, \end{align*} where $F$ is the cdf of a random variable and $\phi$ the characteristic function, $\phi(t)=\int_\mathbb{R}e^{itx}dF(x)$.

Is it possible to relate the integral (or its principal value) $$\int_0^\infty \Im\left(\frac{e^{-itx}\phi(t)}{t^n}\right)dt$$ for $n\in\mathbb{N}$ to the cdf $F$?

Gil-Pelaez' original proof wouldn't work because it would then involve integrals like $\int_0^\infty \frac{\sin(x)}{x^n}dx$ and $\int_0^\infty \frac{\cos(x)}{x^n}dx$ which don't necessarily exist but perhaps somebody has seen a different proof that is adaptable for powers in the denominator?

Answer 1

Whenever the distribution with characteristic function $\phi$ has a finite mean $a$, we have $\phi(t)=1+iat+o(t)$ (as $t\downarrow0$). So, for any real $x\ne a$, the integrand in your integral is $\sim (a-x)t^{1-n}$ and hence for any $n\ge2$ the integral diverges to $\pm\infty$ in a right neighborhood of $0$. So, your integral does not exist for any $n\ge2$ and any real $x\ne a$, even as a principal value.

However, in this paper or its arXiv version, one can find many formulas of the same flavor as the Gil--Pelaez one, with $t^p$ for however large $p$ in the denominator of the integrand.