# Saddle Points in Complex Plane of trig function

I am trying to analytically Fourier transform a set of functions that have the form

$f(k) e^{-\rho~\psi(k)}$

where the general $f(k)$ is some linear combination of trig functions without poles, $\psi(k) = - a \sin(k)^{2} + i~c~ (\cos(k)^{2}+\gamma^{2}\sin(k)^{2})^{1/2}$ and $\rho \gg 1$. The parameters $a \gt 0,c$ are all real and order one, while $0 \lt \gamma \lt 1$ and real.

The issue is that I want to find the fourier transform/ fourier component that also has this large scale $\rho$ in it, i.e the factor in the integrand for the FT of $e^{i k x} = e^{\rho~i k b}$. The form of the integral then becomes

$\int^{\pi/2}_{-\pi/2} f(k) ~e^{\rho~g(k)} dk$,

where $g(k) = - a \sin(k)^{2} + i~c~ (\cos(k)^{2}+\gamma^{2}\sin(k)^{2})^{1/2} + i b k$

My first idea was to use integral asymtopics, method of steepest descent since $\rho \gg 1$ or approximate about the saddle points but I get stuck right away at finding the saddle points. The problem I find is dealing with the root in the third term when finding the saddle points or the imaginary part (to find the contours of constant phase). For example, differentiating gives

$g '(k) = i b -a \sin(2k) + \dfrac{i c (\gamma^{2}-1)\sin(2k)}{2(\cos(k)^{2}+\gamma^{2}\sin(k)^{2})^{1/2}} = 0$

which when squared leads to a 6th order polynomial, once all the trig functions have been written in terms of $\tan(k)$. Explicitly equating real and imaginary parts also leads to difficult equations.

Also, the function is well approximated by replacing $-a \sin(k)^{2} \to -a k^{2}$ which might help but then you have to deal with a polynomial term with the trig terms.

Does this look possible to evaluate analytically in this way? Or are there other techniques that can be used? Mathematica gives the solution in terms of its root functions, so doesn't seem that helpful.

• Typically, in Fourier integrals, you end up with something of the form $$\int_a^b dk \, F(k) \, e^{\rho g(k)} \, e^{i k x}$$ In this case, we are taking the FT of the function $F(k) e^{\rho g(k)}$. This function is independent of $x$, the argument of the FT, as it must be. (Otherwise, the integral doesn't represent a transform.) In this case, the typical problem asks for the asymptotic behavior of the integral as $\rho \to \infty$. Then then $e^{i k x}$ piece may be considered as "slowly varying" relative to the other exponential and lumped in with $F$. This simplifies the problem. – Ron Gordon Nov 15 '14 at 15:16
• Yes, this is simpler. However, I would like to find the Fourier component that is of similar size to the large parameter $x$. This wasn't clear at all in my question! I've edited to try and make this clearer. (also I've used the notation with $x$ as my large parameter, this is clearly not the best idea, I will change to the notation you used above) – JohnD Nov 15 '14 at 15:55