Does an antiderivative of e^z/z^3 exist in the punctured complex plane?

[platinumtucan]

Hi, so my question is the subject line. In the multiply connected domain |z|>0, does the function f(z) = e^z/z^3 have an antiderivative?

I'm learning from Brown and Churchill, and they have a theroem on pg. 142 that leads me to believe it does. I don't remember what my prof said about this though :)

If f(z) is continuous on a domain D and if anyone of the following statements is true, then so are the others:

c)The integrals of f(z) around closed contour lying entirely in D all have value zero
a)f(z) has an antiderivative F(z) throughout D

Now, we have a taylor series for $f(z) = \displaystyle\sum_{i=0}^\infty \dfrac{z^(n-3)}{n!}, 0<|z|<\infty$, but we don't even need this. The only singularity of f is not even a point in the domain of definition of our function.

If it has an antiderivative there, does it have an antiderivative throughout the whole complex plane?

Please someone help me out :)

 

[snipez90]

Yes, that sounds like Morera's theorem (not that the name really matters). This was a somewhat subtle point in complex integration, I think. You can construct an antiderivative locally (say in a disk U) with very basic requirements, namely continuity of your function f and the requirement that the integral of f over the boundary of any closed rectangle in U is 0. But if your function is holomorphic, then by Goursat (which guarantees that the integral over the closed rectangle is 0), we have an antiderivative. This is a standard way to prove Morera's theorem for closed rectangles.

Thus, e^z/z^3 has an antiderivative on a domain away from 0. On the whole plane, I think the answer should be negative, because for instance you could integrate around the unit circle and get pi*i (this should be right, and judging from where you are in the course, you could verify this via Cauchy's integral formula).

 

[mathwonk]

If it has an antiderivatoive then the integral around any closed curve would be zero. so integrate it around the unit circle. if you do not get zero, the answer is no.

since e^z = 1 + z + z^2/2 + z^3/6 +z^4(...), dividing by z^3 gives a term of 1/2z. and this term has a non zero integral around that curve. the other terms can be antidifferentiated by the power rule. so the otherv etrms give integral zero.

thus it is exactly as snipez90 says. no.

 

[jackmell]

I believe it has an antiderivative A(z) everywhere in the punctured plane except that antiderivative is multi-valued and the integral between two points z0 and z1 along a path C, which traverses an analytical path over this multivalued antiderivative can be calculated by the difference between the value of this "extended" antiderivative at the endpoints of the path:

[tex]\int_{z_0}^{z_1} \frac{e^{z}}{z^3}dz=A(z)\biggr|_{z_0}^{z_1}[/tex]

where it is understood A(z) remains differentiable along the entire path and may traverse multiple single-valued determinations of the function.

And in particular, for some closed contour say |z|=1 we could write:

[tex]\mathop\oint\limits_{|z|=1} \frac{e^{z}}{z^3}dz=A(z)\biggr|_{1}^{1}=\pi i[/tex]

Likewise:

[tex]\mathop\oint\limits_{|z|=1} \frac{1}{z}dz=\log(1)-\log(1)=2\pi i[/tex]

And that 2pi i results from taking the difference between log(1) at the end of the differentiable path, minus the value of log(1) at the start of the path. And that difference is 2pi i. Just plot it and you'll see what I mean.

I really think we should make some changes to the way this topic is taught in Complex Analysis.

 

[Office_Shredder]

A multi-valued function isn't a function, so that's a tough pill to swallow.

 

[jackmell]

A multi-valued function isn't a function, so that's a tough pill to swallow.

But it is wonderfully analytic, beautiful in fact, in all it's detail, nicely smooth and differentiable and thus satisfying the criteria for the Fundamental Theorem of Calculus for line integrals although I believe we need to amend that theorem in the following way to more explicitly address analytic extensions of multifunctions:

The Fundamendal Theorem of Calculus for Line Integrals over Multifunctions:

Suppose that [itex] \gamma: [a,b]\to C[/itex] is a smooth curve and [itex] M[/itex] is a multifunction defined and multianalytic (multivalued but analytic on any local determination of the function) in an open set [tex] G\backslash\{p_n\}[/itex] where the set [tex] \{p_n\}[/itex] is a set of points where the function is not analytic such as poles. Assume [itex] \gamma[/itex] tracks over analytic extensions of [itex] M'[/itex] and [itex] M[/itex]. Then:

[tex] \int_{\gamma} M'(z)dz= M(\gamma(b))-M(\gamma(a))[/tex]

and using this theorem, it's immediately clear what the value of the following integral is upon inspection of the antiderivative where the subscript zero means [itex] \gamma[/itex] tracks an analytic extension of the root function and the antiderivative is interpreted as it's analytic extension:

[tex]\mathop\oint\limits_{|\gamma(t)| = r > 1}\frac{z_{0}+z_{1}z+z_{2}z^{2}}{\left(\sqrt{z^{2}-1}\right)_{0}}dz=\left(z_{1}+\frac{z z_{2}}{2}\right)\sqrt{z^{2}-1}+1/2\left(2z_{0}+z_{2}\right)\log[2(z+\sqrt{z^{2}-1})]\biggr|_{\gamma(a)}^{\gamma(b)} [/tex]

and therefore for the original problem we can write (via two parts with the Ei function being multi-valued):

[tex]\mathop\oint\limits_{|z|=2} \frac{e^z}{z^3}dz=M(z)\biggr|_2^2=\frac{-e^z (1+z)+z^2 \text{ExpIntegralEi}[z]}{2 z^2}\biggr|_2^2=\pi i[/tex]

and therefore I challenge the argument that this function does not have an (extended) antiderivative throughout the punctured plane.

 

[mathwonk]

To me you are confusing the two concepts, line integral, and antiderivative. Your multivalued antiderivative is just the line integral. An antiderivative is by definition single valued. I.e. if you have a multivalued function you need some way to tell which value it has in a given situation. That is determined by choosing a path, which means you are really just considering path integrals. Or you can introduce a covering space (Riemann surface) on which the antiderivative is defined and single valued, while being multivalued down on the base space. But again to evaluate it you need a path to lift into the covering space. So your discussion is to me just a matter of language. Of course if you change the meaning of words, then the (words used in the) answer to a question changes.

 

[jackmell]

Hi Mathwonk. Afraid you might get a little annoyed with me. Congrats on the medal too!

The line integral I'm addressing is over a function which is the antiderivative of the integrand. This antiderivative happens to be multivalued. You stated the antiderivative is be definition single-valued. Why can't they be multivalued as long as a single-valued "determination" is choosen in a specific analysis where a single value is required. I'm doing that above: the path trajects along a single surface (single-valued) of the underlying Riemann surface (of the multi-valued function) during the entire length of the contour. Hence, as I see it, the analysis is relying on a single-valued component of the multifunction antiderivative. Personally, I think this makes beautiful sense and I think that perspective would help others understand what's going on in a integral involving Riemann surfaces.

I've observed over the past several years, students having much difficulty understanding multi-functions, branch-cuts, determinations, and worst of all, contour integration over multifunctions. They are stuck in "flatland" and have not been taught the beauty of the underlying (multivalued) Riemann surfaces. This suggestion I make to explicitly "adjust" the fundamental theorem of calculus to address more explicitly, multifunctions in the form of what I would like to call "multivalued antiderivatives" I believe, along with other things, would help the student better understand this beautiful concept in Complex Analysis. That and some nice plots too. :)

 

[Office_Shredder]

The problem is that M is explicitly multi-valued, and then you write down something like [tex] M(\gamma(a))[/tex]. That has no meaning on its own because M is multi-valued. The only reason you knew which value to pick in your examples is because you knew what was supposed to work for the line integral

 

[jackmell]

The problem is that M is explicitly multi-valued, and then you write down something like [tex] M(\gamma(a))[/tex]. That has no meaning on its own because M is multi-valued. The only reason you knew which value to pick in your examples is because you knew what was supposed to work for the line integral

If I have:

[tex]\int_a^b f(z)dz[/tex]

which is single-valued, and the antiderivative is multivalued as in the example for this thread:

[tex]\mathop\oint\limits_{|z|=2} \frac{e^z}{z^3}dz[/tex]

I was under the assumption that it did not matter which "sheet" of the multivalued antiderivative I start the integration over. As long as I maintain a differentiable path to the end point, the difference:

[tex]M(\gamma(t))-M(\gamma(0))[/tex]

will be the same. Are you telling me Office_Shredder, that is not the case?

However in the case:

[tex]
\mathop\oint\limits_{|\gamma(t)| = r > 1}\frac{z_{0}+z_{1}z+z_{2}z^{2}}{\left(\sqrt{z^{2}-1}\right)_{0}}dz=\left(z_{1}+\frac{z z_{2}}{2}\right)\sqrt{z^{2}-1}+1/2\left(2z_{0}+z_{2}\right)\log[2(z+\sqrt{z^{2}-1})]\biggr|_{\gamma(a)}^{\gamma(b)}
[/tex]

since I'm integrating over a multifunction, I will necessarily have two values for the integral and I'm starting to see there is no way to know which "sheet" of the multivalued antiderivative I would need to begin the integration in order to yield one or the other answer. Dang it.