Why is integral of 1/z over unit circle not zero?

  • Thread starter Poopsilon
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  • #1
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Ok I can do the integral and see that it is equal to 2∏i, but thinking about it in terms of 'adding up' all the points along the curve I feel like every every point gets canceled out by its antipode, e.g. 1/i and -1/i.
 

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  • #2
HallsofIvy
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Integration is NOT simply "adding up" points.
 
  • #3
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Well maybe you could point me to the appropriate resources where I might reconcile my understanding of the Riemann integral and why it does not behave how I would expect it to in the aforementioned case.
 
  • #4
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Poopsilon's question can be phrased more precisely as follows (and I would be interested in hearing the answer to this): the integral of an odd real function over an open interval containing zero is zero, because the integral from -a to 0 cancels the integral from 0 to a. Well, 1/z is odd in the sense that 1/(-z) = -1/z. So wouldn't you expect the integral of 1/z over the unit circle to be zero for analogous reasons?
 
  • #5
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In fact upon performing the appropriate pullback I obtain the integral i∫dt from 0 to 2∏. So in this case I now do seem to be simply adding up the points along the curve. I'm not sure what this means..

@lugita: yes that seems like a good way to phrase it.
 
  • #6
Office_Shredder
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When you're integrating over the circle you're taking a path integral. So it's not just the value of the function that's important, but the direction you're moving in. In the one dimensional case your integral is always going in the same direction (towards positive values) so it's just the value of the function that's important. In this case, when you look at 1/z vs 1/(-z), they're negative, but they're also going in opposite directions, which has the effect of cancelling the negative sign.
 
  • #7
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Ok I can do the integral and see that it is equal to 2∏i, but thinking about it in terms of 'adding up' all the points along the curve I feel like every every point gets canceled out by its antipode, e.g. 1/i and -1/i.

It's because there's a hole in the domain. 1/z blows up at z = 0.

http://en.wikipedia.org/wiki/Cauchy_integral_theorem

Now a question that always bothered me is, how does the function being integrated around a circle "know" there's a hole in the middle of the circle?

That's what the OP is asking.
 
  • #8
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Office_Shredder could you expand on what you said, maybe with special attention to our 1/z case, as from my POV it's this very fact that direction matters which I'm using in my argument that intuitively the integral should be zero, and not 2∏i.
 
  • #9
Fredrik
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Office_Shredder could you expand on what you said, maybe with special attention to our 1/z case, as from my POV it's this very fact that direction matters which I'm using in my argument that intuitively the integral should be zero, and not 2∏i.
One way of explaining his point is that in a Riemann sum ##\sum_k \frac{1}{z_k} \Delta z_k##, it's not just the 1/zk that's equal in magnitude and opposite in direction on the opposite side of the circle. ##\Delta z_k## is also equal in magnitude and opposite in direction. So if the contribution to the integral from one short segment of the circle is 1/z Δz, then the contribution from the "opposite" segment is (-1/z)(-Δz)=1/z Δz, i.e. exactly the same.
 
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  • #10
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But Fredrik, why doesn't this argument hold for f(z)=z? And furthermore, I was under the impression that our direction of integration is not changing, it's either counter-clockwise or clockwise, and thus positive or negative accordingly.
 
  • #11
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When you're at an angle of pi/4 and you're going around counterclockwise, the small change in z is pointing up and to the left ( the direction you're traveling in) and when you're at an angle of 5pi/4 the change in z is going to be down and to the right.

The fact that the integral of z around the unit circle is zero is not because z is an odd function. All the argument says is that a function being odd is not sufficient to make it zero when going around a circle
 
  • #12
Fredrik
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But Fredrik, why doesn't this argument hold for f(z)=z?
I would say that it does. The contribution from opposite sides are the same in this case too, for precisely the same reason.

And furthermore, I was under the impression that our direction of integration is not changing, it's either counter-clockwise or clockwise, and thus positive or negative accordingly.
It's precisely because it's either clockwise the whole way through or counter-clockwise the whole way through that that Δz has "opposite" values on opposite sides.
 
  • #13
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But that is only perceived change in direction, you don't multiply by an extra negative just because you're down at 5pi/4.

But then why wouldn't the integral of z over the unit circle also be non zero?
 
  • #14
Office_Shredder
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We haven't argued that it MUST be non-zero. All we've said is the function being odd is not a sufficient condition for the integral being zero.

It's like asking why the integral from -pi to pi of cos(x) is zero even though cos(x) isn't odd... sometimes integrals are just zero because they are

But that is only perceived change in direction, you don't multiply by an extra negative just because you're down at 5pi/4.

The delta(z)s are complex numbers. The norm is whatever step size you're taking, and the direction in which these complex numbers are pointing (once you've taken many slices) is the direction of motion along the unit circle. Let's think about delta z for zk=1 and zk=-1. If our step size is 1/100, at zk=1 we get that delta z is 1/100*i. When zk = -1, delta z is -1/100*i . So those two pieces contribute to our Riemann sum:
[tex]\frac{1}{100*i}*\frac{1}{1} + \frac{-1}{100*i}\frac{1}{-1} = \frac{2}{100*i}[/tex]
and the two pieces do NOT cancel
 
  • #15
Fredrik
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But that is only perceived change in direction, you don't multiply by an extra negative just because you're down at 5pi/4.
How is it "only perceived"? The Δz clearly has the same magnitude but opposite sign on the opposite side. That's not a matter of perception.

But then why wouldn't the integral of z over the unit circle also be non zero?
Why would it be? Office_Shredder and I have only used this observation about contributions from opposite segments to explain why the opposite sign of 1/z on opposite sides doesn't automatically make the integral zero.

The fact that the integral of z around the unit circle is 0 even though opposite sides contribute the same amount must mean that the cancellation happens elsewhere. Maybe the contribution from a segment is cancelled by a segment an angle π/2 away (I didn't think about it, I just made up a number), maybe it's not cancelled by any segment. All we know (unless we think it through), is that the contributions from all the segments add up to zero.
 
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  • #16
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Looking at the graph of cos(x) just from 0 to pi and then using the fact that it's an even function, it's quite obvious why its integral from -pi to pi is zero. I can't personally accept the fact that sometimes integrals are just zero because they are. Clearly the reason z is zero and 1/z is not is that 1/z has a singularity inside the unit circle. The reason for this is I feel for some deep reason involving the proof of Cauchy's Integral Theorem and the proof of Pullbacks for integrals.

Fredrik, could you explain to me exactly what you mean by Δz.

Edit: I guess maybe odd functions were not the right way to frame this, I'm more interested in the discrepancy between the integrals of z and 1/z.

Edit #2: Ok I see what office_shredder is saying about Δz, but I'm still curious about why this phenomenon doesn't cause the integral of z to be non zero.
 
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  • #17
Office_Shredder
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"This phenomenon" is that the portions in the Riemann sum contributed by [itex] 1/z_k \delta z_k[/itex] and [itex]1/(-z_k) \delta(-z_k)[/itex] are equal.

Consider the Riemann sum of cos(x). [itex] cos(x_k)\delta x = cos(-x_k) \delta x[/itex] so the contributions from x and -x are equal. But the integral is still zero because of more global properties of cos(x).
 
  • #18
Fredrik
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Clearly the reason z is zero and 1/z is not is that 1/z has a singularity inside the unit circle. The reason for this is I feel for some deep reason involving the proof of Cauchy's Integral Theorem and the proof of Pullbacks for integrals.
The theorem we need is the one that says that if B and C are two contours with the same endpoints, and B can be continuously deformed to C without going outside an open set on which f is analytic, then ##\int_B\, f(z)dz=\int_C\, f(z)dz##. That is a pretty deep theorem, but it's surprisingly easy to prove (see e.g. Saff & Snider). It follows almost immediately that if C is a closed contour, and f is analytic on an open set that contains C, then ##\int_C\, f(z)dz=0##.
 
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  • #19
lavinia
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In fact upon performing the appropriate pullback I obtain the integral i∫dt from 0 to 2∏. So in this case I now do seem to be simply adding up the points along the curve. I'm not sure what this means..

@lugita: yes that seems like a good way to phrase it.

along the circle,

dz = e[itex]^{i\theta}[/itex]dr + ire[itex]^{i\theta}[/itex]d[itex]\theta[/itex]

but dr = 0 since the circle has constant radius.

So dz/z = ire[itex]^{i\theta}[/itex]d[itex]\theta[/itex]/re[itex]^{i\theta}[/itex] = id[itex]\theta[/itex]

So dz/z is just the differential of the angle function multiplied by i
 
  • #20
Fredrik
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Fredrik, could you explain to me exactly what you mean by Δz.
If ##\{z_0,\dots,z_n\}## are points on the circle, then
$$\int f(z) dz\approx\sum_{k=1}^n f(z_k)(z_k-z_{k-1})=\sum_{k=1}^n f(z_k)\Delta z_k.$$

Edit: I should also have said that we're assuming that for all k with 1≤k≤n, zk-1 is an "earlier" point on the circle than zk, in the sense that we encounter zk-1 before zk if we go around the circle in the counter-clockwise direction starting at 1.
 
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  • #21
Bacle2
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But I think we can see that the net contributions of antipodes do not cancel out
(we can ignore the contribution of r, since it cancels out). While it is true that:

1/z= 1/e =e-iθ =cos(-θ)+isin(-θ)=cosθ-isinθ

and, for the antipode ei(θ+∏)= -cosθ+isinθ ,

The contributions of θ=0 must cancel out that of both θ=∏. But the contribution of ∏

must also cancel out that of 2∏.
 
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  • #22
Office_Shredder
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But I think we can see that the net contributions of antipodes do not cancel out
(we can ignore the contribution of r, since it cancels out). While it is true that:

1/z= 1/e =e-iθ =cos(-θ)+isin(-θ)=cosθ-isinθ

and, for the antipode ei(θ+∏)= -cosθ+isinθ ,

The contributions of θ=0 must cancel out that of both θ=∏. But the contribution of ∏

must also cancel out that of 2∏.

This is incorrect. When integrating from 0 to 2pi the angles 0, pi and 2pi are the only angles for which you get this happening. And assuming your Riemann sum is a left or right hand sum the angles 0 and 2pi won't both show up
 
  • #23
Bacle2
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This is incorrect. When integrating from 0 to 2pi the angles 0, pi and 2pi are the only angles for which you get this happening. And assuming your Riemann sum is a left or right hand sum the angles 0 and 2pi won't both show up

Of course; I was referring to the talk about odd real-valued functions canceling out,
that some where talking about ,and it is an informal argument. I should have been more
clear on that.

A rigorous argument would have required , as you said, Riemann sums, choices of
elements in partitions, etc.
 

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