For my first post on Math Stackexchange, I ask your help on a specific issue regarding the Gil-Pelaez formula. I have tried various versions of the formula to get the result right but I still cannot spot where my error is.
The Gil-Pelaez formula (1951) for computing the cumulative density function of random variable $ X $, noted $ F(X) $, can be defined as follows:
$ F_{X}(x) = \dfrac{1}{2} - \dfrac{1}{\pi}\int_{0}^{+\infty}\Im\left(\dfrac{e^{- ixs}\varphi_{X}(s)}{s}\right)ds $ (1)
or
$ F_{X}(x) = \dfrac{1}{2} + \dfrac{1}{2\pi}\int_{0}^{+\infty}\dfrac{e^{ixs}\varphi_{X}(-s) - e^{-ixs}\varphi_{X}(s)}{is}ds $ (2)
, where $ \varphi_{X}(s) $ is the characteristic function of $ X $. Assume that all moments of $ X $ are known and $ \varphi_{X}(s) $ can be written in series expansion form, we have:
$ \varphi_{X}(s) = \sum_{j=0}^{\infty} \dfrac{(is)^{j}}{j!}\mathbb{E}[X^{j}] $
Then, we can write:
$ \int_{0}^{+\infty}\Im\left(\dfrac{e^{-ixs}}{s} \varphi_{X}(s)\right)ds = \int_{0}^{+\infty}\Im\left(\sum_{j=0}^{\infty} \dfrac{(is)^{j}\mathbb{E}[X^{j}] }{j!}\dfrac{e^{-ixs}}{s}\right)ds = \int_{0}^{+\infty}\Im\left(\sum_{j=0}^{\infty} \dfrac{i^{j}\mathbb{E}[X^{j}] }{j!} s^{j-1}e^{-ixs}\right)ds = \Im\left(\sum_{j=0}^{\infty} \dfrac{i^{j}\mathbb{E}[X^{j}] }{j!}\int_{0}^{+\infty} s^{j-1} e^{-ixs} ds\right) $
One recognizes the Mellin transform of $ e^{-ixs} $, which for integer $ p, p > 0 $, is equal to $ \mathcal{M}\left[e^{-ixs}\right](p) = (ix)^{-p}\Gamma(p) $. Thus, we have:
$ \int_{0}^{+\infty}\Im\left(\dfrac{e^{-ixs}}{s} \varphi_{X}(s)ds\right) = \Im\left(\sum_{j=0}^{\infty} \dfrac{i^{j}\mathbb{E}[X^{j}] }{j!} (ix)^{-j}\Gamma(j) \right) = \Im\left(\sum_{j=0}^{\infty} \dfrac{i^{j}\mathbb{E}[X^{j}] }{i^{j}j!}x^{-j}\Gamma(j) \right) = \Im\left(\sum_{j=0}^{\infty} \dfrac{\mathbb{E}[X^{j}] }{j!}x^{-j}\Gamma(j) \right) = 0 $
The result seems obviously wrong to me but I don't know where my calculations go wrong. I have tried decomposing the $ e^{-ixs} $ into cosine and sine components, or use formula (2) instead. We have in that case, using the fact that $ (-1)^{j} = (i^{2})^{j} = i^{2j} $:
$ \int_{0}^{+\infty}\dfrac{e^{ixs}}{is} \varphi_{X}(-s)ds = \int_{0}^{+\infty} \left(\sum_{j=0}^{\infty} (-1)^{j}\dfrac{e^{ixs}}{is}\dfrac{(is)^{j}}{j!}\mathbb{E}[X^{j}] ds \right) = \int_{0}^{+\infty} \left(\sum_{j=0}^{\infty} (-1)^{j}s^{j-1}e^{ixs}\dfrac{i^{j-1}}{j!}\mathbb{E}[X^{j}] ds \right) = \int_{0}^{+\infty} \left(\sum_{j=0}^{\infty} i^{3j-1}s^{j-1}e^{ixs}\dfrac{\mathbb{E}[X^{j}]}{j!} ds \right) = \sum_{j=0}^{\infty}\left(\dfrac{i^{3j-1}\mathbb{E}[X^{j}]}{j!} \int_{0}^{+\infty} s^{j-1}e^{ixs}ds \right) = \sum_{j=0}^{\infty}\left(\dfrac{i^{3j-1}\mathbb{E}[X^{j}]}{j!} (-ix)^{-j}\Gamma(j) \right) = \sum_{j=0}^{\infty}\left(\dfrac{i^{3j-1}\mathbb{E}[X^{j}]}{j!} (-i)^{-j}(x)^{-j}\Gamma(j) \right) = \sum_{j=0}^{\infty}\left(\dfrac{i^{3j-1}\mathbb{E}[X^{j}]}{j!} (-1) ^{-j}i^{-j}(x)^{-j}\Gamma(j) \right) = \sum_{j=0}^{\infty}\left(\dfrac{i^{3j-1}\mathbb{E}[X^{j}]}{j!}i^{-3j}(x)^{-j}\Gamma(j) \right) = \sum_{j=0}^{\infty}i^{-1}\mathbb{E}[X^{j}](x)^{-j} $
And:
$ \int_{0}^{+\infty}\dfrac{e^{-ixs}}{is} \varphi_{X}(s)ds = \int_{0}^{+\infty} \left(\sum_{j=0}^{\infty} \dfrac{e^{-ixs}}{is}\dfrac{(is)^{j}}{j!}\mathbb{E}[X^{j}] ds \right) = \int_{0}^{+\infty} \left(\sum_{j=0}^{\infty} s^{j-1}e^{-ixs}\dfrac{i^{j-1}}{j!}\mathbb{E}[X^{j}] ds \right) = \sum_{j=0}^{\infty}\left(\dfrac{i^{j-1}\mathbb{E}[X^{j}]}{j!} \int_{0}^{+\infty} s^{j-1}e^{-ixs}ds \right) = \sum_{j=0}^{\infty}\left(\dfrac{i^{j-1}\mathbb{E}[X^{j}]}{j!}(ix)^{-j}\Gamma(j) \right) = \sum_{j=0}^{\infty}i^{-1}\mathbb{E}[X^{j}](x)^{-j} $
We thus see that:
$ \int_{0}^{+\infty}\dfrac{e^{ixs}\varphi_{X}(-s) - e^{-ixs}\varphi_{X}(s)}{is}ds = 0 $
This is again equal to 0.
Any hint or help would be greatly appreciated ! Thanks in advance !
I am not sure. Can we interchange the integral with sum over $j$ here? $\int_{0}^{+\infty}\Im\left(\sum_{j=0}^{\infty} \dfrac{i^{j}\mathbb{E}[X^{j}] }{j!} s^{j-1}e^{-ixs}\right)ds = \Im\left(\sum_{j=0}^{\infty} \dfrac{i^{j}\mathbb{E}[X^{j}] }{j!}\int_{0}^{+\infty} s^{j-1} e^{-ixs} ds\right) $
For example, if $x=0$, the integral in the RHS seems to be infinite, while the LHS could still be finite.