# Math Help - integral, closed path

1. ## integral, closed path

Evaluate $\frac{1}{2 \pi i } \int_{\gamma} \frac{dz}{z(z^2-1)}$, where $\gamma$ is the closed path indicated in the picture.

Hint: Proceed directly with partial fractions.

Attempt:

Using partial fractions, we get $\frac{.5}{z+1}+\frac{.5}{z-1}-\frac{1}{z}$. Now, I do not see where to go from here. How does using partial fractions help in this problem? Thank you.

2. Originally Posted by pascal4542
Evaluate $\frac{1}{2 \pi i } \int_{\gamma} \frac{dz}{z(z^2-1)}$, where $\gamma$ is the closed path indicated in the picture.

Hint: Proceed directly with partial fractions.

Attempt:

Using partial fractions, we get $\frac{.5}{z+1}+\frac{.5}{z-1}-\frac{1}{z}$. Now, I do not see where to go from here. How does using partial fractions help in this problem? Thank you.
My geometric intuition is not that good but it seems as if your curve only goes around 0 one time, so your integral would be $2\pi i$

3. You know, I don't think my first answer is correct because a closer look makes me doubt that the curve does not go around at least one time over -1 and 1. It's probably better if you wait and see if someone else has a better eye for these kind of things.

4. The path goes around z= 0 and z= 1 once and around z= -1 twice.

5. Originally Posted by pascal4542
Evaluate $\frac{1}{2 \pi i } \int_{\gamma} \frac{dz}{z(z^2-1)}$, where $\gamma$ is the closed path indicated in the picture.

Hint: Proceed directly with partial fractions.

Attempt:

Using partial fractions, we get $\frac{.5}{z+1}+\frac{.5}{z-1}-\frac{1}{z}$. Now, I do not see where to go from here. How does using partial fractions help in this problem? Thank you.
Hi guys. Looks to me it's twice around $-1$ and once around $1$ as the loops around the origin cancel or no? Is it not then:

$\frac{1}{2\pi i}\mathop\oint\limits_{\gamma}\frac{1}{z(z^2-1)}dz=2\mathop\text{Res}_{z=-1}+\mathop\text{Res}_{z=1}=1+1/2=3/2$

but isn't that still quite unacceptable? Don't you guys have any fun in these classes? How about demonstrating that? Well I don't know exactly, that's the fun part. I guess blow up the figure, superimpose it on some grid, estimate a hundred points or so, run some least-square fit on the figure piecewise, then solve the integral numerically or if it's a jpeg file then the file itself would have the data points making up the curve. That would make it a lot easier. The exact details though, I'd leave up to all the smart students in class.

6. I could be wrong you know Pascal. Let's see then: I can draw the contour free-hand in Mathematica (see below) and "capture" the points by just doing a cut-and-paste then assigning the paste to some variable name which I called mycontourpts. I now have all the points (crudely) making up the contour. I can then extract those points and calculate interpolation functions for the x and y values and even calculate their derivatives. I then set $z(t)=x(t)+iy(t)$ and then calculate numerically the following integral:

$\frac{1}{2\pi i}\mathop\oint\limits_{\text{mycontourpts}}\frac{x '(t)+i y'(t)}{z(z^2-1)}\biggr|_{z=x(t)+iy(t)}$

This is the Mathematica code I used (minus the capture of mycontourpts):

Code:
lns = Cases[Normal[First[mycontourpts]],
Line[pts_] -> pts, {0, Infinity}];
myvals = First[lns];
myxval = (#1[[1]] & ) /@ myvals;
myyval = (#1[[2]] & ) /@ myvals;
myx = ListInterpolation[myxval]
myxd[t_] = D[myx[t], t]
myy = ListInterpolation[myyval]
myyd[t_] = D[myy[t], t]
i1 = NIntegrate[(myxd[t] + I*myyd[t])/
(z*(z^2 - 1)) /.
z -> myx[t] + I*myy[t], {t, 1, 989},
WorkingPrecision -> 25]
N[i1/(2*Pi*I)]
The result is $1.49633 - 0.0043122 I$

It's kinda' close to 3/2 with a little bit of imaginary part that could be due to the crudeness of the method. Really, I would have liked to get it closer or else I made a mistake. I'm thinkin' another B at best here.