I would very much like to compute the Fourier transform of:
\begin{equation} f(t)=e^{A(t)}\tag{1} \end{equation}
\begin{equation} A(t):=-\frac{1}{2}\sum_{n=0}^{N}(\alpha_{n}e^{i{\omega}nt}-Q_{n})^{2}\tag{2} \end{equation}
Where $\alpha_{n}$, $\omega$, and $Q_{n}$ are constants for given $n$.
Or equivalently, of course:
\begin{equation} f(t)=\prod_{n=0}^{N}\exp[-\frac{1}{2}(\alpha_{n}e^{i{\omega}nt}-Q_{n})^{2}]\tag{3} \end{equation}
The Fourier transform should be given by:
\begin{equation} {F(\Omega)=}\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{i{\Omega}t}\prod_{n=0}^{N}\exp[-\frac{1}{2}(\alpha_{n}e^{i{\omega}nt}-Q_{n})^{2}]dt\tag{4} \end{equation}
I'm not aware of any technique which can perform this integration but granted, I'm in physics and lack knowledge of the level of sophisticated mathematics that may have very useful methods of solving this. I was hoping someone could point me in the direction of some of those methods, help me solve this directly, or tell me explicitly that this doesn't have a solution.
Thanks in any case.
I think I may have just about figured this out, but I would certainly appreciate if anyone wanted to check my work:
We have: \begin{equation} f(t):=\exp[-\frac{1}{2}\sum_{n=0}^{N}(\alpha_{n}e^{i{\omega}nt}-Q_{n})^{2}] \end{equation}
This is the same as:
\begin{equation} f(t)=-\sum_{k=0}^{\infty}\frac{1}{2k!}\sum_{n=0}^{N}[\alpha_{n}e^{i{\omega}nt}-Q_{n}]^{2k} \end{equation}
Then the Fourier transform of this is:
\begin{equation} {\tilde{F}(\Omega)}=-\frac{1}{2\sqrt{2\pi}}\sum_{k=0}^{\infty}\sum_{n=0}^{N}\frac{1}{k!}\int_{-\infty}^{\infty}e^{i{\Omega}t}[\alpha_{n}e^{i{\omega}nt}-Q_{n}]^{2k}dt \end{equation}
Where I've been a slightly bad mathematician and assumed that I can reverse the order of integration and summation because my integral is well-behaved at $\infty$ without rigorously proving that.
Focusing just on the integral for a moment: \begin{equation} \int_{-\infty}^{\infty}e^{i{\Omega}t}[\alpha_{n}e^{i{\omega}nt}-Q_{n}]^{2k}dt\\=\int_{-\infty}^{\infty}[\alpha_{n}e^{i{\omega}nt}e^{\frac{i{\Omega}t}{2k}}-Q_{n}e^{\frac{i{\Omega}t}{2k}}]^{2k}dt\\=\int_{-\infty}^{\infty}[\alpha_{n}e^{i\frac{(n\omega-\Omega)t}{2k}}-Q_{n}e^{i\frac{{\Omega}t}{2k}}]^{2k} dt \end{equation}
Using the binomial theorem, this becomes:
\begin{equation} -\sum_{j=0}^{2k}{2k\choose j}\int_{-\infty}^{\infty}(\alpha_{n}e^{i\frac{(n\omega-\Omega)t}{2k}})^{2k-j}(Q_{n}e^{i\frac{{\Omega}t}{2k}})^{j})dt\\=-\sum_{j=0}^{2k}{2k\choose j}\int_{-\infty}^{\infty}\alpha_{n}^{2k-j}Q_{n}^{j}\exp[{i(n\omega+\Omega-\frac{n{\omega}j}{2k})t}]dt \end{equation}
Noticing that $\alpha_{n}^{2k-j}Q_{n}^{j}$ is just a constant for given $(n,j,k)$, it can be pulled out of the integral, giving: \begin{equation} -\sum_{j=0}^{2k}{2k\choose j}\alpha_{n}^{2k-j}Q_{n}^{j}\int_{-\infty}^{\infty}\exp[{i(n\omega+\Omega-\frac{n{\omega}j}{2k})t}]dt\\=-\sum_{j=0}^{2k}{2k\choose j}\alpha_{n}^{2k-j}Q_{n}^{j}\delta(\Omega-(\frac{n{\omega}j}{2k}-n\omega)) \end{equation}
To clean this up, let $\lambda(n,\omega,k)\equiv(\frac{n{\omega}j}{2k}-n\omega)$ and $\mathcal{E}(n,k,j)\equiv\alpha_{n}^{2k-j}Q_{n}^{j}$. Then the whole Fourier transform is: \begin{equation} {\tilde{F}(\Omega)}=\frac{1}{2\sqrt{2\pi}}\sum_{k=0}^{\infty}\sum_{n=0}^{N}\sum_{j=0}^{2k}{2k\choose j}\frac{\mathcal{E}}{k!}\delta(\Omega-\lambda) \end{equation}
I don't know if this can be simplified any further. If anyone sees a further simplification or development on this, I would appreciate if you pointed it out.