Branch Cuts and Integral Evaluation

[metgt4]

Homework Statement

The function f(z) = (1-z²)^1/2 of complex variable z is defined to be real and positive in the range -1 < z < 1. Using cuts running along the real axis for 1 < x < infinity and -infinity < x < -1, show how f(z) is made single valued and evaluate it on the upper and lower sides of both cuts.

Use these results and a suitable contour in the complex plane to evaluate the integral of:

(dx)/(x(x² - 1)^1/2)

from 1 to infinity

The Attempt at a Solution

I've gotten the first part of the problem, but I'm not sure how to apply that to the integral. I've attached my work so that you know I've actually done something, but I'm confused as to how I use that work to help with solving that integral.

 

[mighty2000]

Thank you for your post. It seems like you have made good progress in understanding the behavior of the function f(z) = (1-z2)1/2 in the given range. To use this understanding to evaluate the integral, we can use the concept of contour integration.

First, we can rewrite the given integral as:

∫(dx)/(x(x2 - 1)1/2) = ∫(dx)/(x2 - 1) - ∫(dx)/(x(x2 - 1)1/2)

We can see that the second integral can be expressed in terms of the first one by using the substitution u = (x2 - 1)1/2. This gives us:

∫(dx)/(x(x2 - 1)1/2) = ∫(dx)/(x2 - 1) - ∫(du)/u

Using the result from the first part of the problem, we know that the integral ∫(dx)/(x2 - 1) is single-valued in the range 1 < x < infinity and -infinity < x < -1, and it can be evaluated using the upper and lower sides of the cuts.

Now, to evaluate the second integral, we can choose a suitable contour in the complex plane that includes the cuts and the points 1 and infinity. One possible choice is a semi-circular contour in the upper half plane with radius R, centered at the origin. This contour includes the points 1 and infinity and avoids the cuts.

Using the residue theorem, we can evaluate the integral along this contour as:

∫(du)/u = 2πi Res(f(z), z = 0)

where Res(f(z), z = 0) is the residue of the function f(z) at the point z = 0. We can calculate this residue using the Laurent series expansion of f(z) around z = 0, which is given by:

f(z) = (1-z2)1/2 = 1 - z2/2 + z4/8 - z6/16 + ...

Therefore, the residue at z = 0 is given by:

Res(f(z), z = 0) = -1/2

Substituting this into the integral, we get:

∫(du)/u = 2πi (-1/2) = -πi

Finally