Difference between characteristic function and F-transform

I'm struggling to understand the difference between this two functions.

I have this condition: $$P_j:=\mathbb{Q}(S_T>K):=\frac{1}{2}+\frac{1}{\pi}\int_{0}^{+\infty}Re[\frac{e^{iuK}f_j(u,x,v)}{iu}]\textrm{d}u$$ with $$Re$$ the real part of integrand function, $$i:=\sqrt{-1}$$ and $$f_j$$ the characteristic function of $$x:=\textrm{ln}(S)$$ with density $$P_j$$. Well. From this…

…I know that characteristic function $$\varphi$$ is the F-transform. So, if the functions are equal, why here (page 31, Proposition 3.1, Formule 3.14) we are being told that there is the necessity to see the form of $$f_j:=\phi_j$$ (that, with a "guess and verify" approach, it is assumed that be the same as $$e^{C_J(\tau,u)+D_j(\tau,u)+iux}$$)? Don't we already know the form of the characteristic function $$\phi_j$$, namely that of Fourier transform?

I really thank anyone who wants to help me!

• Something is missing from this post. Your question begins with "if the functions are equal." What functions are you referring to? The remainder of your question isn't comprehensible in English, so we need additional help figuring out what you're trying to ask. – whuber Jun 18 at 20:55
• @whuber Thanks for answer. Functions to which I's referring are the F-transform and the characteristic function. Like you can see here at pag. 14 (politesi.polimi.it/bitstream/10589/72489/1/2012_12_Cozzi.pdf) the author says that "The idea was to evaluate from the formula (1.6) the probability distribution of the ﬁnal stock price and from this evaluate the characteristic function, namely the Fourier transform of the probability distribution. C. and M. suggested a very eﬃcient method to compute vanilla prices using the F- transform for the valuation of this characteristic function. – Marco Pittella Jun 18 at 21:04