Stating the Gil-Pelaez Theorem in terms of a real integral

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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?

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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.