How to evaluate this inverse fourier transform integral I am trying to do the following integral to evaluate an inverse fourier transform 
$$I = \frac{1}{2\pi}\int _{-\infty }^{\infty }k^{-i \xi } \left(c^{i \xi } - c^{1-i \xi}\right) \exp \left[- i  m \xi  -  m \xi^2 - i \xi z \right] d \xi,$$
where $c$,$k$ and $m$ are some positive constants and $z$ is the new variable and $c > k$.
We have
$$I = \frac{1}{2\pi}\int _{-\infty }^{\infty }(\frac{c}{k})^{i \xi }\exp \left[- i  m \xi  -  m \xi^2 - i \xi z \right] d \xi - \frac{c}{2\pi}\int _{-\infty }^{\infty }(ck)^{-i \xi } \exp \left[- i  m \xi  -  m \xi^2 - i \xi z \right] d \xi$$
I have been looking through formulae online, but i have not had much luck.
Any help will be greatly appreciated.
Thank you.
 A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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$\ds{I =
{1 \over 2\pi}\int_{-\infty }^{\infty }k^{-\ic\xi}
\pars{c^{\ic\xi} - c^{1 - \ic\xi}}\exp\pars{-\ic m\xi - m\xi^{2} - \ic\xi z}\,
\dd\xi\,;\qquad c, k, m > 0}$

Lets $\ds{\quad c \equiv \expo{a}\,,\quad k \equiv \expo{b}\,,\quad
m \equiv \expo{s}\,,\quad a,b,s \in \mathbb{R}}$:
\begin{align}
\color{#f00}{I} & =
{1 \over 2\pi}\int_{-\infty }^{\infty }
\bracks{\expo{\ic\pars{a - b}\xi} - \expo{a - \pars{a + b}\ic\xi}\,}
\exp\pars{-m\braces{\xi^{2} + \ic\bracks{1 + {z \over m}}\xi}}\,\dd\xi
\\[3mm] & =
\,\mathrm{f}\pars{b - a + z \over m} -
\expo{a}\,\mathrm{f}\pars{b + a + z \over m} =
\,\mathrm{f}\pars{\ln\pars{k/c} + z \over m} -
c\,\mathrm{f}\pars{\ln\pars{kc} + z \over m}
\end{align}

where
\begin{align}
\,\mathrm{f}\pars{x} & \equiv
{1 \over 2\pi}\int_{-\infty}^{\infty}
\exp\pars{-m\bracks{\xi^{2} + \ic x\xi}}\,\dd \xi =
{1 \over 2\pi}\int_{-\infty}^{\infty}
\exp\pars{-m\bracks{\xi + \half\,\ic x}^{2} - {1 \over 4}\,mx^{2}}\,\dd \xi
\\[3mm] & =
{1 \over 2\pi}\expo{-mx^{2}/4}\int_{-\infty}^{\infty}\expo{-m\xi^{2}}\,\dd\xi =
{1 \over 2\pi}\expo{-mx^{2}/4}\,\root{\pi \over m} =
{\pi^{-1/2} \over 2\root{m}}\expo{-mx^{2}/4}
\end{align}

\begin{align}
\color{#f00}{I} & = \color{#f00}{%
{\pi^{-1/2} \over 2\root{m}}\bracks{%
\exp\pars{-\,{\bracks{\ln\pars{k/c} + z}^{2} \over 4m}} -
c\exp\pars{-\,{\bracks{\ln\pars{kc} + z}^{2} \over 4m}}}}
\end{align}
