I am trying to compute the series representation of the standard normal distribution CDF from its characteristic function.
Given a random variable $ X $ following a standard normal distribution with mean $ 0 $ and standard deviation $ 1 $, we know its characteristic function is $ \varphi(s) = e^{-s^{2}/2} $, with series representation:
$ \varphi(s) = \sum_{j=0}^{\infty} (-1)^{j}\dfrac{s^{2j}}{2^{j}j!} $
We also know the Taylor series representation of the standard normal distribution CDF:
$ F(x) = \dfrac{1}{2} + \dfrac{1}{\sqrt{2\pi}}\sum_{j=0}^{\infty} \dfrac{(-1)^{j}x^{2j+1}}{2^{j}j!(2j+1)} $
Using Gil-Pelaez inversion formula, we have:
$ F(x) = \dfrac{1}{2} - \dfrac{1}{\pi}\int_{0}^{+\infty}\Im\left(\dfrac{e^{- ixs}\varphi(s)}{s} \right)ds \\ = \dfrac{1}{2} - \dfrac{1}{\pi}\int_{0}^{+\infty}\Im\left(\dfrac{e^{- ixs}}{s}e^{-s^{2}/2} ds\right) \\ = \dfrac{1}{2} + \dfrac{1}{\pi}\int_{0}^{+\infty} \dfrac{\sin(xs)}{s}e^{-s^{2}/2}ds \\ = \dfrac{1}{2} + \dfrac{1}{\pi}\int_{0}^{+\infty} \sin(xs) \sum_{j=0} ^{\infty} \dfrac{(-1)^{j}}{2^{j}j!}s^{2j-1} ds \\ \\ = \dfrac{1}{2} + \dfrac{1}{\pi}\sum_{j=0}^{\infty} \dfrac{(-1)^{j}}{2^{j}j!} \int_{0}^{+ \infty} s^{2j-1}\sin(xs) ds $
For the equality to hold, we should have:
$ \int_{0}^{+ \infty} s^{2j-1}\sin(xs) ds = \sqrt{\dfrac{\pi}{2}}\dfrac{x^{2j+1}}{2j+1} $
Developing the above expression, we get:
$ \sqrt{\dfrac{\pi}{2}}\dfrac{x^{2j+1}}{2j+1} \\ = \sqrt{\dfrac{\pi}{2}}\dfrac{x^{2j+1}(2j)!}{(2j+1)!} \\ = \left[\sqrt{\pi}\dfrac{(2j)!2^{2j}}{2^{2j}}\right]\times \dfrac{\sqrt{2}}{2} \times \dfrac{x^{2j+1}}{(2j+1)!} \\ = \Gamma\left(\dfrac{2j+1}{2}\right) \times 2^{2j-1/2} \times \dfrac{\Gamma(j)x^{2j+1}}{\Gamma(2j+1)} $
One recognizes the Gamma function, which appears in the Mellin transform of the sine function. However, the above result is in now way equal to the Mellin transform of $ \sin(xs) $.
I am stuck from this point and I don't know how to get to the Taylor series representation of $ F(x) $ from there. Does anyone have any idea ? Thanks ! :)