Characteristic Function and Inverse Fourier Transform By studying the limiting distribution of a particular random variable $X$ I found that the asymptotic characteristic function of the variable is 
$$
\varphi_{X}\left(\omega\right)= e^{i\,\omega\,Y_0}\,\left[\frac{2\,\gamma}{2\,\gamma-\omega^2}\,\left(e^{-\frac{1}{2}\,t\,\omega^2}-e^{-t\,\gamma}\right)\right]\quad (1)
$$
where $Y_0>0$, $t>0$ and $\gamma>0$ are given parameters. My purpose is to find the distribution function of the random variable $X$, but I cannot recover the density since it seems that the inverse Fourier transform of (1) is very difficult to be found (I'v tried unsuccessfully with standard symbolic softwares). Any idea? Note that the function in $(1)$ is well-behaved at $\omega^2=2\,\gamma$. 
Some additional thoughts: I suspect that $(1)$ is the characteristic function of a pure jump process at time $t$, a sort of compound poisson process but more complicated (it is pretty evident that $(1)$ it is not the characteristic function of a compound poisson, unless a complicated characteristic function it is chosen for the jump amplitude). This suspect is confirmed by some numerical simulation.    
 A: There is some problem with the fact that $\varphi_{X}\left(\omega\right)$ is not 1 for $\omega=1$. Let this aside, we should calculate the inverse Fourier transform
$$ P(x) = \mathcal{F}^{-1}\left\{ e^{i\,\omega\,Y_0}\,\left[\frac{2\,\gamma}{2\,\gamma-\omega^2}\,\left(e^{-\frac{1}{2}\,t\,\omega^2}-e^{-t\,\gamma}\right)\right]\right\}.$$
The tricky part is the evaluation of
$$I(x)= \mathcal{F}^{-1}\left\{\frac{2 \gamma}{2\gamma -\omega^2}  e^{-\frac{1}{2}\,t\,\omega^2}\right\} .$$
We proceed by employing the convolution theorem. For that we need the results
$$ \mathcal{F}^{-1}\left\{e^{-\frac{1}{2}\,t\,\omega^2}\right\} = \frac{1}{2 \pi t^{1/2}} e^{-\frac1{2t} x^2 } $$
and
$$ \mathcal{F}^{-1}\left\{\frac{2 \gamma}{2 \gamma -\omega^2}\right\} =
\frac{\gamma^{1/2}}{2 \pi^{1/2}} \sin(\sqrt{2 \gamma} x) \mathop{\rm sgn}(x).$$
We obtain the inverse Fourier transform
$$I(x) =\frac{\gamma^{1/2}}{2 \pi t^{1/2}} \int\!dy \,e^{-\frac{1}{2t}(x-y)^2}\sin(\sqrt{2 \gamma} y) \mathop{\rm sgn}(y)\\=
\mathop{\rm Im}\left\{\sqrt{2\pi t} e^{i \sqrt{2 \gamma} x - \gamma t} \mathop{\rm erf}\left(\frac{x}{\sqrt{2 \gamma}}  + i \sqrt{\gamma t} \right)  \right\}. $$
The additional factor $e^{i\omega Y_0}$ simply shifts the argument $x$. In conclusion, we have the result
$$P(x) = I(x- Y_0) -e^{-\gamma t} \frac{\gamma^{1/2}}{2 \pi^{1/2}} \sin(\sqrt{2 \gamma} x-Y_0) \mathop{\rm sgn}(x-Y_0). $$
A: Assume the inverse transform of $\varphi_{X}(\omega)$ is $f_X(x)$. 


*

*Assume inside the bracket as $Y(w)$. Then $\varphi_{X}(\omega)=e^{i\omega Y_0}Y(w)$. Hence, $f_X(x)=y(x-Y_0)$, where $y(x)=\mathcal{F}^{-1}\{Y(\omega)\}$.

*Looking at inside the bracket, you can distinguish two terms, i.e., $Y(\omega)=Y_1(\omega)+Y_2(\omega)$, where $Y_1(\omega)=\frac{2\,\gamma}{2\,\gamma-\omega^2}\,e^{-\frac{1}{2}\,t\,\omega^2}$, and $Y_2(\omega)=-\frac{2\,\gamma}{2\,\gamma-\omega^2}e^{-t\,\gamma}$. Now you need to calculate $\mathcal{F}^{-1}\{Y_1\}$ and $\mathcal{F}^{-1}\{Y_2\}$.


From here you may use partial fraction expansion or any other technique to calculate the inverse Fourier transform which seems to be much easier to get
 $$y_2(x)=-\frac{\sqrt{-2\gamma}\left(u(x)+u(-x)e^{2\sqrt{-2\gamma}x}\right)}{2e^{\sqrt{-2\gamma}x}}e^{-t\gamma}$$ and the problem reduces to $\mathcal{F}^{-1}\{Y_1(\omega)\}$ only, which seems not so difficult. Particularly, since $Y_1(\omega$) is the product of two parts, the inverse Fourier transform is equal to their convolution: $$y_1(x)=e^{t\gamma}y_2(x)*g(x)$$
where $$g(x)=\mathcal{F}^{-1}\{e^{-\frac{1}{2}\,t\,\omega^2}\}=\sqrt{\frac{1}{2\pi  t}}e^{-\frac{1}{2t}x^2}$$
