On using Fourier transforms to solve the root of a convolution In continuation of Lower bounds of laplace transform of characteristic functions.

My question is:
Can anyone point out where i'm going wrong in the derivation below. 

It's been a while since I've used transforms and i know I'm missing something as the product of the derivation does not agree with direct calculation.
Thanks in advance.

We wish to solve $K = f(x;a)$ where $K$ is some complex number and
\begin{equation}
f(x;a) = \int_{-\infty}^\infty h(x-\omega;a)g(\omega) \text{d}\omega
\end{equation}
with $g$ some probability distribution, $a,x$ are real numbers and 
\begin{equation}
h(x;a) = \frac{1}{a+ i x},\quad a>0
\end{equation}
We let $\bar{\varphi}(t)= \hat{g}(t)$ and calculate the Fourier transform of $h$
\begin{equation}
\hat{h}(t) = 2\pi e^{-at}\Theta(t),
\end{equation}
where $\Theta(t)$ is the Heaviside step function. Thus $\mathcal{F}[K](t) = \mathcal{F}[f](t) $ gives
\begin{equation}
2\pi K\delta(t) =2\pi e^{-at}\Theta(t)\bar{\varphi}(t) \implies \frac{K}{\bar{\varphi}(t)}\delta(t)=e^{-at}\Theta(t)
\end{equation}
Taking the IFT we get
\begin{equation}
\frac{K}{\bar{\varphi}(0)} = \int_0^\infty e^{-(a-ix)t}\text{d}t
\end{equation}
Which gives $\frac{1}{a-ix} = \frac{K}{\bar{\varphi(0)}}$ thus 
$$a+ix = \frac{\varphi(0)}{\bar{K}}$$
However this does no agree with direct calculation.
Take the Cauchy distribution with median $0$ $g(\omega;\sigma) = \frac{\sigma}{\pi(\sigma^2+\omega^2)}$ and its analytic continuation $G(\omega;\sigma) =\frac{1}{\pi(\sigma-i \omega)}$
$$f(x;a) =i\int_{-\infty}^\infty \frac{g(z)}{z - i(a +ix)}dz = \pi G(ia-x)$$
by the Schwarz Integral formula.
Which implies $f(x;a) =\frac{1}{x+ia + \sigma}$.
Thus if $f(x;a) = K$, then
$$ x+ia = \frac{1}{K} - \sigma$$
and from the section above, this is not equal to $x+ia = \frac{\varphi(0)}{\bar{K}} = \frac{1}{\bar{K}}$

Then you pass to 
$$
2\pi K\delta(t) =2\pi e^{-at}\Theta(t)\bar{\varphi}(t)
$$
it means that you are solving equation 
$$
K\equiv \int_{-\infty}^\infty h(x-\omega;a)g(\omega) \text{d}\omega
$$
and it's not that is required. The Fourier transform is not applicable in this way
since the lhs is not given, $K$ being just a value of rhs at some point $x$.
