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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:

  1. Evaluate integral $\int_{-1}^{1} x^2 \exp(\frac{1}{x^2-1}) dx$

  2. 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?

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    $\begingroup$ in case you are interested in asymptotics: arxiv.org/pdf/1508.04376v1.pdf $\endgroup$ – tired Sep 9 '16 at 14:32
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    $\begingroup$ for what stand $a$ here in this case? $\endgroup$ – Dr. Sonnhard Graubner Sep 9 '16 at 15:07
  • $\begingroup$ @Dr.SonnhardGraubner: $a$, here, is just an arbitrary scalar/constant. $\endgroup$ – B. Young Sep 9 '16 at 16:44
  • $\begingroup$ @tired: This is interesting, thanks. $\endgroup$ – B. Young Sep 9 '16 at 16:49

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