Existence of inverse Fourier transform through validity of calculations

Question

Let \begin{equation} f(x)=\left\{\begin{array}{cl}e^{-x}\,e^{-\gamma \min(x,k)}&,x\ge 0\\ 0&,x<0\end{array}\quad,\gamma>0.\right. \end{equation} $f$ clearly lies in $L^1$, so it's Fourier transform $\hat{f}$ is well defined. I calculated it to be \begin{equation} (\mathcal{F}f)(y)=\hat{f}(y)=e^{-k(1+\gamma+iy)}\left(\frac{1}{1+iy}-\frac{1}{1+\gamma+iy}\right)+\frac{1}{1+iy} \end{equation} For the inverse transform to exist, $\hat{f}$ also needs to be $L^1$. I don't really know how to show this though ($\rvert\hat{f}\lvert$ is a rather complicated term (see comments)), so I just went ahead and tried to calculate it. Since the transform has a rather nice form, looking through Fourier transform tables gets me \begin{equation} \mathcal{F}^{-1}(\hat{f})(x)=e^{-x}\Big(\Theta(x-k)\big(e^{-k\gamma}-e^{-\gamma x}\big) + \Theta(x)e^{-\gamma x}\Big), \end{equation} where $\Theta$ is the Heaviside step function.
This is exactly the same as $f$, just using step functions.
Is this a legitimate approach to show the existence of the inverse transform? If not, why does it work out to be correct nonetheless?

Answer 1

As it was pointed out in this post, I can make use of the Plancherel Theorem.
The function $f$ lies in $(L^1 \bigcap L^2) $, since it's in $L^1$ and $f(x)\le 1\ \forall x$. So it's Fourier transform also lies in $L^2$ and the inversion formula holds.