K = Integral from zero to infinity of x^(-a)dx/[1+2 x cos(theta) + x^2]

where -1 < a < 1. We can compute the integral by considering the integrand as a function defined on the complex plane. Usually this is trivial, in this case, however, we must choose a definition of the function z^(-a). Any point z in the complex plane can be written in polar coordinates as z = r exp(i theta) with 0<= theta < 2 pi. We can define z^(-a) as r^(-a) exp(-i a theta). This definition makes z^(-a) on the positive real axis identical to the value of the real function defined there, but it forces one to put the branch cut infinitesimally below the real axis, as the polar angle drops back to zero there. The function then has a branch point singularity at z = 0, the integrand is meromorphic when we cut away the branch cut that starts at z = 0. We can then compute the integral by considering the following contour integral.

We integrate f(z) = z^(-a)dx/[1+2 z cos(alpha) + z^2] from epsilon to R on the real axis, we then move along a counterclockwise circle with radius R ...

Solution Summary

A detailed solution for x^(-a)dx/[1+2 x cos(theta) + x^2] is given.

$2.19

See Also This Related BrainMass Solution

Solutions to Several First Order ODEs

Solve 1st order DE, please include explaination of process, if possible.

...1/(2R) Integral from -R to R of f(x) cos(n pi/R x)dx = 1/R Integral from 0 to R of f(x) cos(n pi/R x)dx. ... J_n(x) = 1/(2 pi i^n) Integral from 0 to 2 pi of exp ...

...Integral from 0 to infinity of dx/(1+x^2) = pi/2. ... a + b cos^2(theta) = a + b/2 [cos(2 theta) + 1] = a + b/2 + b/2 cos(2 theta). Let's put: a' = a + b/2. ...

... 9 evaluate are dx = 3Sec 2 (θ )dθ and x 2 + 1 = 3Sec(θ ... 3Sec 2 (θ ) 1 dy 1 dθ = dx = (10) 3Sec(θ ... therefore use the lookup of the standard integral for Sec ...

...Integral over theta from theta = 0 to 2 pi of P(z, theta) d theta = P(z) ----. ... Appendix 2: The proof that dx dy = z dz d theta. ...x = z cos(theta). ...