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.

...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. ...

...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). ...

... In what follows we denote int_a^bf(x)dx the integral of ... The steps that follow (namely the triple integral for z ... of the center of mass (¯, y , z ) are x¯¯ 1 1...

... We have used the notation x = dx . ...1 taking L, the so-called Lagrangian, to be the diﬀerence ... that of the two variables, one is the integration variable and ...

...Theta goes from zero to pi, so cos(theta) goes from 1 to minus 1. You can interchange the limits by ...Integral over x from -1 to 1 of (1 - x^2)dx = 4/3. ...