In (Gil-Pelaez, 1951), the following is stated
$F(x)= \frac{1}{2}+\frac{1}{2\pi}\int_0^\infty\frac{e^{itx}\phi(-t)-e^{-itx}\phi(t)}{it}dt.$
I am trying to show that it can be equivalently stated as
$F(x)= \frac{1}{2}+\frac{1}{\pi}\int_0^\infty\Re \left[\frac{e^{-itx}\phi(t)}{it}\right]dt$.
To that end, I have tried using the following relation:
$\Re[f(t)]=\frac{f(t)+\bar{f(t)}}{2}$,
where $\bar{f(t)}$ is the complex conjugate of $f(t)$. This yields
$\Re \left[\frac{e^{-itx}\phi(t)}{it}\right] = \frac{1}{2}\left[\frac{e^{-itx}\phi(t)}{it}+\frac{ie^{itx}\phi(-t)}{t}\right]$.
In order to complete the argument, I therefore need the relation below to hold
$\frac{1}{2}\frac{e^{itx}\phi(-t)-e^{-itx}\phi(t)}{it} = \frac{1}{2}\left[\frac{e^{-itx}\phi(t)}{it}+\frac{ie^{itx}\phi(-t)}{t}\right]$,
which only seems to hold if I multiply either the right hand side or left hand side by -1.
My question is therefore, what have I missed in the above?
I found the answer after stepping away from the problem for a bit:
I had managed to switch up $F(x)$ and $1-F(x)$ in earlier derivations...
So it actually holds that
$1-F(x) = \frac{1}{2}-\frac{1}{2\pi}\int_0^\infty \frac{e^{itx}\phi(-t)-e^{-itx}\phi(t)}{it}dt$
is equivalent to the stated equation.