Complex Countour Integral

1. Oct 8, 2009

ryanj123

1. The problem statement, all variables and given/known data

Let C be the contou starting at z=-1 going around the circle lzl=1 and ending back at z=-1 (counterclockwise) determine. Int. (Log(z)/z) dz.

2. Relevant equations

z=e^(Log(z))

3. The attempt at a solution

So I've been stuck on this problem for quite some time.

Im not sure if the integral is suppose to be broken up into:

Int f(z(t))*z'(t) dt.

If so, I'm unsure what to call my f(z) and my z(t).
I tried calculating:

z(t)=e^it
z'(t)=ie^it
Log(z(t))=Log(e^it)=it

Then

(0,pi)Int((it/e^it)*ie^it) dt
= (0,pi)Int (-t) dt
= -pi.

I'm sure this isn't correct.
Any help would be appreciated.

2. Oct 8, 2009

Dick

The integral of t from 0 to pi isn't pi. Is it?

3. Oct 8, 2009

ryanj123

Ha. I got ahead of myself and didn't even compute the integral.

(0,pi)Int(-t) dt
= -t^2/2 evaluated at (0,pi)
= -pi^2/2

Is this on the right track? Should the bounds be changed? Or is my integral entirely incorrect?

4. Oct 8, 2009

Dick

It looks right to me. Any reason to doubt it?

5. Oct 9, 2009

latentcorpse

why doesn't this work using the Cauchy Integral formula?

i set f(z)=log z and a=0 but then f(a)=f(0)=-infinity.

that's how it doesn't work but what's the reason for this?
is it because f isn't holomorphic on the unit disc? (and it's not holomorphic because it "blows up" at z=0)

6. Oct 9, 2009

Dick

I don't know what that's supposed to be about. You want to integrate Log(z)/z, not Log(z) and the limits are 1 to -1, not 0 to 'a' whatever that is. If you are asking why you can't do it as an antiderivative, you can. As long as you're careful about not crossing branch cuts. Evaluate (Log(z))^2/2 between 1 and -1.

7. Oct 9, 2009

latentcorpse

i havent done anything with the Cauchy integral formula in a while....can you talk me through the procedure here:

we have $f(a)=\oint \frac{f(z)}{z-a}dz$

so why can't we set f(z)=log z and a=0 then the RHS is just $\oint \frac{\log{z}}{z} dz$ and it will be equal to the LHS $f(a)=f(0)=\log{0}=-\infty$
i think the reason this doesn't work is because the f i have used in this case isn't holomorphic on the region we are integrating over i.e. at z=0 it has a pole (is that correct???)

so what should i use for my f(z)?

thanks.

8. Oct 9, 2009

Dick

Well, yeah. Log isn't holomorphic in the disk. It's worse than a pole, it has a branch cut. The given contour isn't even closed. You can't really integrate around anything. You have to find another way to solve it. There are two alternative methods in this thread.

9. Oct 9, 2009

latentcorpse

ok. so you can't use the formula because log isn't holomorphic. but the contour is closed surely? its a circle of radius 1 from z=-1 to z=-1 isn't it?

10. Oct 9, 2009

Office_Shredder

Staff Emeritus
Yes, but there's a branch cut, so somewhere on that circle log won't even be defined, or at least is discontinuous

11. Oct 9, 2009

Dick

Hmm. I was paying more attention to the limits of integration in the attempt. If the first z=-1 isn't a typo then ryanj123 had better rework the limits.

12. Oct 9, 2009

latentcorpse

would this be possible using a semicircular contour from z=1 to z=-1 then along the real axis to z=1 again but with a small semicircular indent to avoid the origin and then use the Cauchy Residue Theorem?

13. Oct 9, 2009

Office_Shredder

Staff Emeritus
You don't know that the integral around that little indent around the origin goes to zero as it becomes small

14. Oct 9, 2009

latentcorpse

why not? because log goes to infinity at 0?