So I've been attempting to solve this definite integral in my spare time for a couple of weeks, now, and I think I've used my entire bag of tricks.
$$\int_{-1}^{1} e^\frac{1}{x^2-1}\cos{ax}\,dx$$
I think this function should be differentiable within the range of $(-1,1)$.
I already found the following two topics which have helped greatly:
Evaluate integral $\int_{-1}^{1} x^2 \exp(\frac{1}{x^2-1}) dx$
Evaluate definite integral $\int_{-1}^1 \exp(1/(x^2-1)) \, dx$
However, the addition of the $\cos{ax}$ term is confounding.
It is an even function, so
$$\int_{-1}^{1} e^\frac{1}{x^2-1}\cos{ax}\,dx = 2 \int_{0}^{1} e^{\frac{1}{x^2-1}}\cos{ax}\,dx.$$
I have also managed to express it as something like looks like a Fourier transform, but I got stuck at the end:
$$\int_{-1}^{1} e^\frac{1}{x^2-1}\cos{ax}\,dx.$$
Following Euler's Formula, $e^{i\theta} = \cos\theta + i\sin\theta$,
$$\cos\theta = e^{i\theta} - i\sin\theta$$
$$\int_{-1}^{1} e^\frac{1}{x^2-1}\cos{ax}\,dx$$
$$=\int_{-1}^{1} e^\frac{1}{x^2-1}\left(e^{iax} - i\sin{ax}\right)\,dx$$
$$=\int_{-1}^{1} e^\frac{1}{x^2-1} e^{iax}\,dx - i\int_{-1}^{1} e^{\frac{1}{x^2-1}} \sin{ax}\,dx.$$
Since the $\sin{ax}$ part of the equation is odd, it goes to zero (it also goes to zero since it's imaginary, and our original function is real):
$$=\int_{-1}^{1} e^\frac{1}{x^2-1} e^{iax}\,dx.$$
By substituting $x\rightarrow t$ and $a\rightarrow \omega$, we get something that is essentially a Fourier transform,
$$=\int_{-1}^{1} e^\frac{1}{t^2-1} e^{i\omega t}\,dt,$$
or
$$=\int_{-\infty}^{\infty} f(t)\, e^{i\omega t}\,dt,$$
where
$$f(t) = \begin{cases} e^\frac{1}{t^2-1} & -1 < t < 1 \\ 0 & \text{otherwise} \end{cases}.$$
While I can't seem to solve the example of $\mathcal{F}[f(t)]$, it occurs to me that I may be able to solve the example of $\mathcal{F}[\ln{f(t)}]$:
$$\mathcal{F}[\ln{f(t)}] = \mathcal{F}\left[\frac{1}{t^2-1}\right],$$
which Wolfram Alpha gives as,
$$=-\sqrt{\frac{\pi}{2}}\operatorname{sgn}\omega\sin\omega.$$
However, that doesn't seem to help. I can get this far:
$$\frac{d}{dt}\ln f(t) = \frac{f'(t)}{f(t)},$$
and
$$\mathcal{F}\left[\frac{d}{dt}f(t)\right] = i\omega\mathcal{F}[f(t)],$$
therefore
$$\mathcal{F}\left[\frac{d}{dt}\ln f(t)\right] = i\omega\mathcal{F}[\ln f(t)] = \mathcal{F}\left[\frac{f'(t)}{f(t)}\right],$$
however, I don't think that gets me any closer to solving for $\mathcal{F}[f(t)].$
I also looked at this short discussion about determining $\mathcal{F}\left[e^{f(t)}\right]$ in terms of $\mathcal{F}[f(t)]$, but I didn't really get anywhere.
Anyone have any good ideas?
