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Residue of a branch cut

  • Thread starter daoshay
  • Start date
14
0
1. Homework Statement
I'm finding the residues of the branch cut of [tex] \int^\infty_0 \frac{dx}{x^{1/4}(x^2+1)}dx [/tex]


2. Homework Equations



3. The Attempt at a Solution
I am trying to find the residue of i

I am not sure how to handle lim z->i of [tex] \frac{1}{z^\frac{1}{4}(z+i)} [/tex]

Any nudges would be vastly appreciated.
 
14
0
I've got
[tex]
\frac{1}{e^\frac{i\pi}{8}2i}
[/tex]

as one residue and
[tex]
\frac{1}{e^\frac{3i\pi}{8}(-2i)}
[/tex]

As the other. Any encouragement to continue or abandon ship? I'll be going to my professors last office hours with this stuff, so I'll find out one way or another.

Thanks...
 
1,786
53
That is very good. So I assume you're using a pac-man contour with the branch-cut on the positive real axis. Assuming you can show the integral goes to zero on the large circular part and the small indentation around the origin, then can you see how the integral over the remainder of the contour is:

[itex]\left(1-\frac{1}{i}\right)\int_0^{\infty} \frac{1}{x^{1/4}(x^2+1)}dx=2\pi i(r_1+r_2)[/itex]

Also, it's not correct to say residue of a branch cut. Only isolated singular points have residues.
 
Last edited:
1,827
7
Also note that you can just integrate of the half circle in the upper half plane and come back along the negative real axis. That saves you work as you then only have to evaluate the residue at z = i. Also, you can then choose the branch cut along any curve in the lower half plane.
 
1,786
53
A third way: use a branch of [itex]\sqrt[4]{x}[/itex] different than what was used above to compute the integral.
 
1,827
7
Fourth way: Suppose you use the complete circle but don't like to put the branch cut along the positive real axis. That looks impossible, but you can then chop the contour into two parts and use different branches of x^1/4 for the two parts, so that the function on one part of the contour is obtained by analytic continuation from the other part.
The branch cut for part 2 can intersect part 1 of the contour and vice versa.


If you do this then you see that the computations are almost identical to what you would do when working with the original contour. The only real difference is that you now don't have to bother moving from epsilon to R just above the real axis and returning from R to epsilon just below the real axis. Both paths are on the real axis, but on the return path you are integrating a different branch of the integrand.
 
1,786
53
Fourth way: Suppose you use the complete circle but don't like to put the branch cut along the positive real axis. That looks impossible, but you can then chop the contour into two parts and use different branches of x^1/4 for the two parts, so that the function on one part of the contour is obtained by analytic continuation from the other part.
The branch cut for part 2 can intersect part 1 of the contour and vice versa.
I don't see that at all but I'm quite sure you are right about it. :)
 
1,827
7
I don't see that at all but I'm quite sure you are right about it. :)
I wasn't too clear about how to split the contour into two parts. Take some alpha not equal to pi/2 or 3/2 pi/2. Then we define:

Contour 1: Line from epsilon to R, from R a counterclockwise circle segment to R exp(i alpha), from there a line to epsilon exp(i alpha), from there a clockwise circle segment back to epsilon

Contour 2: Line from epsilon exp(i alpha) to R exp(i alpha), from there a counterclockwise circle segment to R, from there a line to epsilon, from there a clockwise circle segment back to epsilon exp(i alpha).

The function we integrate along contour 1 has a branch cut that intersects contour 2. This is not a problem, but we then can't integrate the same function over contour 2. However, we can analytically continue that function (with its domain restricted a bit) into the region bounded by contour 2. That function will then necessarily have to have a branch cut that will intersect contour 1. The two functions will then agree in the region inbetween that branch cut and the branch cut of the original function.

ALthought this looks more complicated, in reality is simpler as the choice of the branch cuts is not restricted a lot. The integrals along the
r exp(i alpha) lines cancel and it is then easy to see that all you are really doing is integrating a truly multivalued function without branch cuts along the original contour. So, this leads to:

Fifth way: Work with multivalued functions defined on Riemann surfaces.
 

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