Integration of the function exp(-i*r(x)*cos(x))

[bianyuanren2]

Hi,i am trying to integrate the function exp(-i*r(x)*cos(x)) from -pi to pi,where
r(x)=sqrt(A^2*cos(x)^2+B^2*sin(x)^2),A and B are constant.the function
exp(-i*r(x)*cos(x)) is similar with the function exp(-i*r*cos(x)) which integrate from -pi to pi is 2*pi*BesselJ(0,r).In the function exp(-i*r*cos(x)) , r is constant and usually means radius, but in the function exp(-i*r(x)*cos(x)), r(x) is a variable and define from ellipse.so the constant A can be considered as the major axis of the ellipse ; B the constant can be considered as the short axis of the ellipse.
Can anybody help me? Thank you very much.

 

[gtoguy97]

The integration of exp(-i*r(x)*cos(x)) from -pi to pi can be done using the following steps:1. Express r(x) in polar coordinates: r(x) = sqrt(A^2*cos(x)^2+B^2*sin(x)^2) = sqrt(A^2 + B^2 - (A^2-B^2)*cos(2x)) = sqrt((A^2 + B^2)*(1-(A^2-B^2)*cos(2x)/(A^2 + B^2)) = C*sqrt(1-k*cos(2x)) where C = sqrt(A^2 + B^2) and k = (A^2-B^2)/(A^2 + B^2) 2. Rewrite the integral in terms of polar coordinates: integral exp(-i*r(x)*cos(x))dx from -pi to pi = integral exp(-i*C*sqrt(1-k*cos(2x))*cos(x))dx from -pi to pi = integral exp(-i*C*sqrt(1-k*cos(2x)))d(2x) from -2pi to 2pi = 2*integral exp(-i*C*sqrt(1-k*cos(u)))du from -pi to pi 3. Use the substitution u = 2x: integral exp(-i*C*sqrt(1-k*cos(u)))du from -pi to pi = integral exp(-i*C*sqrt(1-k*cos(2x)))d(2x) from -pi to pi = 1/2*integral exp(-i*C*sqrt(1-k*cos(2x)))dx from -pi to pi 4. Use the formula for the integral of exp(i*z*cos(x)): integral exp(-i*C*sqrt