# A tricky integral

1. Jan 23, 2014

### Jip

Hi,
I'm working the following paper ""Plane waves viewed from an accelerated frame, K Srinivasan, L Sriramkumar, T Padmanabhan - Physical Review D, 1997"

and there's this integral:
$$\int_{-\infty}^{+\infty} e^{- i \Omega t} \cos \left( \beta - e^{a(\phi/\Omega-t)}\right) dt$$
whose result seems to be
$$= \frac{e^{- i \phi}}{2 a}\Gamma\left(\frac{i \Omega}{a}\right) \left( e^{\Omega/4\Omega_0} e^{i \beta}+ e^{-\Omega/4\Omega_0} e^{-i \beta}\right)$$
where $$\Omega_0 = a/2 \pi$$

Following the paper I changed variable $$z= e^{a(\phi/\Omega -t)}$$. The integral is then proportional to
$$\int_{0}^{\infty} z^{\frac{i \, \, \Omega}{a} -1}\left(e^{i (\beta -z)}+e^{-i (\beta -z)}\right) dz$$

This is looking a bit a like a Gamma function. The paper then says "analytically continuing to Im z". This is not clear to me. Shall I integrate along some path in the complex plane? Which one? I tried along the first quadrant of C (between R+ and Im+), avoiding the pole in z=0, but its not clear to me how to control the contributions of the paths of very small radius of very large.

I could use some help! :)
Thanks

Last edited: Jan 23, 2014
2. Nov 21, 2015

### </Henry>

Ok, seems that someone has to write something.. I have some idea, that will lead to the exact result, but I can't write you the whole process. In these works all you need is ability & fantasy.

This is what you have to do:

1) Call for simplicity $$\Delta = \frac{\phi}{\Omega}$$

2) Make the substitution $$y = \beta - e^{a(\Delta - t)}$$ so you will get $$t = \frac{\Delta a - \ln(\beta - y)}{a}$$ and your extrema will run from minus infinity to beta.

3) After you re arranged a bit, shift with $$y = \beta - p$$ with new extrema from $0$ to Infinity.

4) At this point you'll have to compute the integral $$\int_{0}^{\infty} \frac{\cos(\beta-p)}{p} e^{q\ln(p)}$$ where you wrote $$q = \frac{i\Omega}{a}$$

5) Done that and you'll get the result.