Probability density function with the help of the Laplace (Fourier) transform I am reading a paper that derives a closed form expression of the following probability using properties from Fourier transform, 
$$ \mathbb{P} \biggl( F \geq T \ (I+W) \biggl)$$
Assumptions:
1- Assume that the $F$ has a finite first moment and admits a square integrable density $f(y)$.
2- $I$ and $W$ are independent and either one admits a density which is square integrable. 
3- $I+W$ admits a density $g(t)$.
4- $T$ is a constant
The proof is presented next, 
Proof
$$\begin{align}
\mathbb{P} \biggl( F \geq T(I+W) \biggl)&= \int_{0}^{\infty} f(y) \  \mathbb{P} \biggl(I+W<y/T \biggl)\\
&= \int_{0}^{+\infty} f(y) \ \int_{-\infty}^{+\infty} g(t) \ 1(0<t<y/T) dt \\
&\stackrel{?}{=} \int_{0}^{+\infty} f(y)\int_{-\infty}^{+\infty} \mathcal{L}_{I}(-2 i \pi s)\mathcal{L}_{W}(-2i\pi s)\times \frac{e^{2i\pi ys/T }-1}{2i\pi s} ds dy\\
&\stackrel{?}{=} \int_{0}^{+\infty} f(y)\int_{-\infty}^{+\infty} \mathcal{L}_{I}(-2 i \pi s T)\mathcal{L}_{W}(-2i\pi sT)\times \frac{e^{2i\pi ys }-1}{2i\pi s} ds dy\\
&\stackrel{?}{=}\int_{-\infty}^{+\infty} \mathcal{L}_{I}(2 i \pi s T)\mathcal{L}_{W}(2i\pi sT)\times \frac{\mathcal{L_F}(-2i\pi s )-1}{2i\pi s} ds 
\end{align}$$
where the notation $\mathcal{L}_{I}(.),\mathcal{L}_{W}(.), \mathcal{L}_{F}(.)$ represent the Laplace transform of $I,W$ and $F$.
Questions
My questions are regarding the last three steps marked with (?). 
I understand that there is a convolution step for the probability density function $I+W$ which becomes in the frequency domain multiplication between two transforms, however I don't understand how the authors arrive to the argument of the Laplace that is $-2 i \pi s$.
The authors claim that this can be done using Plancherel-Parseval theroem. I have added the theorem below 
$$\int_{\mathbb{R}} f(t)\ g(t) \ dt = \int_{\mathbb{R}}\hat{f}(s)\ \overline{\hat{g}(s)} \ ds $$
where for all functions $f,g$ from $\mathbb{R}$ to $\mathbb{R}$, the Fourier of Transform at $s\in \mathbb{R}$ when it exists is given by  $$\hat{f}(s)= \int_{\mathbb{R}}e^{-2i\pi t s} f(t) \ dt $$
$$\hat{g}(s)= \int_{\mathbb{R}}e^{-2i\pi t s} g(t) \ dt $$
and $ \ \overline{\hat{g}(s)}$ denotes the complex conjugate of $\hat{g}(s)$.
Many thanks
 A: First, it seems to me that the densities of $I,W$ are only supported
on $\left[0,\infty\right)$, otherwise I don't see how the derivation
can be true. Hence, I will assume this in the following.
Then, let us denote the densities of $I,W$ by $g_{I},g_{W}$, respectively.
This implies
$$
\overline{\widehat{g_{I}}\left(\xi\right)}=\overline{\int_{-\infty}^{\infty}g_{I}\left(x\right)\cdot e^{-2\pi ix\xi}\,{\rm d}x}=\int_{0}^{\infty}g_{I}\left(x\right)\cdot e^{2\pi ix\xi}\,{\rm d}x=\mathcal{L}_{I}\left(-2\pi i\xi\right).
$$
An analogous formula holds for $\mathcal{L}_{W}$.
Now note that
\begin{eqnarray*}
\left(\mathcal{F}1_{0<t<y/T}\right)\left(\xi\right) & = & \int_{0}^{y/T}e^{-2\pi it\xi}\,{\rm d}t\\
 & = & \frac{e^{-2\pi it\xi}}{-2\pi i\xi}\bigg|_{t=0}^{y/T}\\
 & = & \frac{e^{-2\pi iy\xi/T}-1}{-2\pi i\xi}
\end{eqnarray*}
Finally, $g=g_{I}\ast g_{W}$ (because $I,W$ are independent), so
that $\mathcal{F}g=\widehat{g_{I}}\cdot\widehat{g_{W}}$ (by the convolution theorem).
As $g_{I},g_{W}$
are square integrable, so are the Fourier transforms $\widehat{g_{I}},\widehat{g_{W}}$
(by Plancherel's theorem). But densities are always integrable, so
that $\widehat{g_{I}},\widehat{g_{W}}\in C_{0}\left(\mathbb{R}\right)$.
Hence $\widehat{g_{I}}\cdot\widehat{g_{W}}\in L^{1}\left(\mathbb{R}^{d}\right)\cap L^{\infty}\left(\mathbb{R}^{d}\right)\subset L^{p}\left(\mathbb{R}^{d}\right)$
for all $p\in\left[1,\infty\right]$. Furthermore, $1_{0<t<y/T} \in L^2$, so that the same is true of its Fourier transform.
Now Plancherel (Parseval if you want) implies
\begin{eqnarray*}
 &  & \int_{-\infty}^{\infty}\mathcal{L}_{I}\left(-2\pi is\right)\cdot\mathcal{L}_{W}\left(-2\pi is\right)\cdot\frac{e^{-2\pi iys/T}-1}{-2\pi is}\,{\rm d}s\\
 & = & \int_{-\infty}^{\infty}\overline{\widehat{g_{I}}\left(s\right)\cdot\widehat{g_{W}}\left(s\right)}\cdot\left(\mathcal{F}1_{0<t<y/T}\right)\left(s\right)\,{\rm d}s\\
 & = & \int_{-\infty}^{\infty}\overline{\widehat{g}\left(s\right)}\cdot\widehat{1_{0<t<y/T}}\left(s\right)\,{\rm d}s\\
 & \overset{\text{Plancherel}}{=} & \int_{-\infty}^{\infty}\overline{g\left(t\right)}\cdot1_{0<t<y/T}\,{\rm d}t\\
 & = & \int_{-\infty}^{\infty}g\left(t\right)\cdot1_{0<t<y/T}\,{\rm d}t.
\end{eqnarray*}
In the first integral, we can now make the substitution $\omega=s/T$
(with ${\rm d}s=T\cdot{\rm d}\omega$), which yields
\begin{eqnarray*}
 &  & \int_{-\infty}^{\infty}g\left(t\right)\cdot1_{0<t<y/T}\,{\rm d}t\\
 & = & \int_{-\infty}^{\infty}\mathcal{L}_{I}\left(-2\pi is\right)\cdot\mathcal{L}_{W}\left(-2\pi is\right)\cdot\frac{e^{-2\pi iys/T}-1}{-2\pi is}\,{\rm d}s\\
 & = & \int_{-\infty}^{\infty}\mathcal{L}_{I}\left(-2\pi i\omega T\right)\mathcal{L}_{W}\left(-2\pi i\omega T\right)\cdot\frac{e^{-2\pi i\omega y}-1}{-2\pi i\omega T}\, T\cdot{\rm d}\omega\\
 & = & \int_{-\infty}^{\infty}\mathcal{L}_{I}\left(-2\pi i\omega T\right)\mathcal{L}_{W}\left(-2\pi i\omega T\right)\cdot\frac{e^{-2\pi i\omega y}-1}{-2\pi i\omega}\,{\rm d}\omega.
\end{eqnarray*}
If we finally integrate this against $f\left(y\right)$ and use Fubini's
theorem, we arrive at
\begin{eqnarray*}
 &  & \int_{0}^{\infty}f\left(y\right)\cdot\int_{-\infty}^{\infty}g\left(t\right)\cdot1_{0<t<y/T}\,{\rm d}t\\
 & = & \int_{0}^{\infty}f\left(y\right)\cdot\int_{-\infty}^{\infty}\mathcal{L}_{I}\left(-2\pi i\omega T\right)\mathcal{L}_{W}\left(-2\pi i\omega T\right)\cdot\frac{e^{-2\pi i\omega y}-1}{-2\pi i\omega}\,{\rm d}\omega\,{\rm d}y\\
 & = & \int_{-\infty}^{\infty}\mathcal{L}_{I}\left(-2\pi i\omega T\right)\mathcal{L}_{W}\left(-2\pi i\omega T\right)\int_{0}^{\infty}f\left(y\right)\cdot\frac{e^{-2\pi i\omega y}-1}{-2\pi i\omega}\,{\rm d}y\,{\rm d}\omega\\
 & \overset{\left(\ast\right)}{=} & \int_{-\infty}^{\infty}\mathcal{L}_{I}\left(-2\pi i\omega T\right)\mathcal{L}_{W}\left(-2\pi i\omega T\right)\left[\frac{1}{-2\pi i\omega}\underbrace{\int_{0}^{\infty}f\left(y\right)\cdot e^{-2\pi i\omega y}\,{\rm d}y}_{=\mathcal{L}_{F}\left(2\pi i\omega\right)}+\frac{1}{2\pi i\omega}\right]\,{\rm d}\omega\\
 & = & \int_{-\infty}^{\infty}\mathcal{L}_{I}\left(-2\pi i\omega T\right)\mathcal{L}_{W}\left(-2\pi i\omega T\right)\left[\frac{\mathcal{L}_{F}\left(2\pi i\omega\right)-1}{-2\pi i\omega}\right]\,{\rm d}\omega\\
 & \overset{\gamma=-\omega}{=} & \int_{-\infty}^{\infty}\mathcal{L}_{I}\left(2\pi i\gamma T\right)\mathcal{L}_{W}\left(2\pi i\gamma T\right)\left[\frac{\mathcal{L}_{F}\left(-2\pi i\gamma\right)-1}{2\pi i\gamma}\right]\,{\rm d}\gamma.
\end{eqnarray*}
At $\left(\ast\right)$, we used the fact that $f$ is a density on
$\left(0,\infty\right)$, so that
$$
\int_{0}^{\infty}\frac{f\left(y\right)}{2\pi i\omega}\,{\rm d}y=\frac{1}{2\pi i\omega}.
$$
Hence, the final result is true. I can not completely reproduce the
exact intermediate steps that you posted (I don't completely get the step at the first "?"). But the arguments used in my derivation are essentially the same that are used to justify the last two "?" and the first "?" is similar to what I did above with Plancherel's theorem. I hope this helps :)
