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.
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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}