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Thread Subject:
Essential singularities and residue theorem and contour integral?

Subject: Essential singularities and residue theorem and contour integral?

From: Mike

Date: 28 Aug, 2006 22:47:59

Message: 1 of 14

Hi there,

Does the residue theorem of contour integral still work when there are
essential singularities inside the closed contour?

For example, exp(1/z) inside the closed contour |z|=1? Matlab,
Mathematica, Maple failed in computing the residue of "exp(1/z)" for
me. My quick guess of the residues for exp(1/z) at z=0 is either 1 or
0. Am I right?

Thanks

-M

Subject: Essential singularities and residue theorem and contour integral?

From: israel@math.ubc.ca (Robert Israel)

Date: 29 Aug, 2006 05:59:08

Message: 2 of 14

In article <1156830479.567529.93780@i42g2000cwa.googlegroups.com>,
Mike <housing2006@gmail.com> wrote:

>Does the residue theorem of contour integral still work when there are
>essential singularities inside the closed contour?

Do you know the statement of the theorem? Does it say anything about
whether the singularities are essential?

>For example, exp(1/z) inside the closed contour |z|=1? Matlab,
>Mathematica, Maple failed in computing the residue of "exp(1/z)" for
>me. My quick guess of the residues for exp(1/z) at z=0 is either 1 or
>0. Am I right?

Hint: substitute t=1/z in the series expansion for exp(t).

Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

Subject: Essential singularities and residue theorem and contour integral?

From: Mike

Date: 28 Aug, 2006 23:01:09

Message: 3 of 14


Mike wrote:
> Hi there,
>
> Does the residue theorem of contour integral still work when there are
> essential singularities inside the closed contour?
>
> For example, exp(1/z) inside the closed contour |z|=1? Matlab,
> Mathematica, Maple failed in computing the residue of "exp(1/z)" for
> me. My quick guess of the residues for exp(1/z) at z=0 is either 1 or
> 0. Am I right?
>
> Thanks
>
> -M

It is very strange that I did a numerial contour integral for exp(1/z)
on the closed contour |z|=1?

In Matlab it is:

quad(@(t) exp(1./exp(1i*t)), -pi, pi)

The result is 2*pi,

But I remember contour integral should be:

2*pi*i*Res[exp(1/z), @z=0],

by the residue theorem, assuming it still works here,

but then according to the results,

Res[exp(1/z), @z=0] should be "-i", which is very strange...

Isn't it?

Subject: Essential singularities and residue theorem and contour integral?

From: Mike

Date: 28 Aug, 2006 23:16:02

Message: 4 of 14


Mike wrote:
> Mike wrote:
> > Hi there,
> >
> > Does the residue theorem of contour integral still work when there are
> > essential singularities inside the closed contour?
> >
> > For example, exp(1/z) inside the closed contour |z|=1? Matlab,
> > Mathematica, Maple failed in computing the residue of "exp(1/z)" for
> > me. My quick guess of the residues for exp(1/z) at z=0 is either 1 or
> > 0. Am I right?
> >
> > Thanks
> >
> > -M
>
> It is very strange that I did a numerial contour integral for exp(1/z)
> on the closed contour |z|=1?
>
> In Matlab it is:
>
> quad(@(t) exp(1./exp(1i*t)), -pi, pi)
>
> The result is 2*pi,
>
> But I remember contour integral should be:
>
> 2*pi*i*Res[exp(1/z), @z=0],
>
> by the residue theorem, assuming it still works here,
>
> but then according to the results,
>
> Res[exp(1/z), @z=0] should be "-i", which is very strange...
>
> Isn't it?

Now I did some further experiments:

I tried to use numerical integral to experiment my knowledge of contour
integral with branches:

I was doing the contour integral along the closed circle |z|=1 and the
integrand was z^(p), and I chose different p to trial:

1) p=0:
K>> quad(@(t) (exp(1i*t)).^(0), -pi, pi)

ans =

    6.2832

This is strange: the integrand z^(0)=1, which does not has a
singularity inside |z|=1 at all, thus the residue = 0. The contour
integral should be 0... but it obviously should be 2*pi. That's very
strange for me.

2) p=-1:
K>> quad(@(t) (exp(1i*t)).^(-1), -pi, pi)

ans =

  4.0143e-009

This is also strange, the integrand z^(-1) has a residue = 1, the
contour integral should be 2*pi*i. But the numerical integration result
shows it is 0. Strange!

3) p=2:

K>> quad(@(t) (exp(1i*t)).^(2), -pi, pi)

ans =

  5.4264e-010

The numerical result is 0, this is in accordance with my expectation.

4) p=-2.3: Now it has branches...

K>> quad(@(t) (exp(1i*t)).^(-2.3), -pi, pi)

ans =

   0.7035 - 0.0000i

The integrand is z^(-2.3), I don't know how to compute its residue, so
I don't know if 0.7035 should be the answer I expected or not...

---------------------

I've also done the following tests:

K>> quad(@(t) (exp(1i*t)).^(0.5), -pi, pi)

ans =

   4.0000 - 0.0000i

K>> quad(@(t) (exp(1i*t)).^(1.5), -pi, pi)

ans =

  -1.3333 - 0.0000i

K>> quad(@(t) (exp(1i*t)).^(2.5), -pi, pi)

ans =

   0.8000 + 0.0000i

The results look really strange to me. I am guessing that the branch is
in the play... but I don't know how does the branch affect my
results...

Any thoughts?

Thanx

Subject: Essential singularities and residue theorem and contour integral?

From: Mike

Date: 28 Aug, 2006 23:20:32

Message: 5 of 14


Robert Israel wrote:
> In article <1156830479.567529.93780@i42g2000cwa.googlegroups.com>,
> Mike <housing2006@gmail.com> wrote:
>
> >Does the residue theorem of contour integral still work when there are
> >essential singularities inside the closed contour?
>
> Do you know the statement of the theorem? Does it say anything about
> whether the singularities are essential?
>

No the theorem did not restrict the type of the singularities...

> >For example, exp(1/z) inside the closed contour |z|=1? Matlab,
> >Mathematica, Maple failed in computing the residue of "exp(1/z)" for
> >me. My quick guess of the residues for exp(1/z) at z=0 is either 1 or
> >0. Am I right?
>
> Hint: substitute t=1/z in the series expansion for exp(t).
>

This gives the residue for exp(1/z) at z=0 to be 1, but I really don't
know if it is right or not. There is no way I can check this result up.
A numerical integral gives different results than I had expected... I
am also aware of the fact that the numerical results from a CAM
software may not serve as a complete correct check-up...

Subject: Essential singularities and residue theorem and contour integral?

From: israel@math.ubc.ca (Robert Israel)

Date: 29 Aug, 2006 07:05:20

Message: 6 of 14

In article <1156832432.298433.42880@i3g2000cwc.googlegroups.com>,
Mike <housing2006@gmail.com> wrote:
>
>Robert Israel wrote:
>> In article <1156830479.567529.93780@i42g2000cwa.googlegroups.com>,
>> Mike <housing2006@gmail.com> wrote:
>>
>> >Does the residue theorem of contour integral still work when there are
>> >essential singularities inside the closed contour?
>>
>> Do you know the statement of the theorem? Does it say anything about
>> whether the singularities are essential?
>>
>
>No the theorem did not restrict the type of the singularities...

Well, there's your answer then.

>> >For example, exp(1/z) inside the closed contour |z|=1? Matlab,
>> >Mathematica, Maple failed in computing the residue of "exp(1/z)" for
>> >me. My quick guess of the residues for exp(1/z) at z=0 is either 1 or
>> >0. Am I right?
>>
>> Hint: substitute t=1/z in the series expansion for exp(t).
>>
>
>This gives the residue for exp(1/z) at z=0 to be 1, but I really don't
>know if it is right or not.

Yes, that's right.

> There is no way I can check this result up.
>A numerical integral gives different results than I had expected... I
>am also aware of the fact that the numerical results from a CAM
>software may not serve as a complete correct check-up...

In Maple:

> f:= z -> exp(1/z):
  res:= 1/(2*Pi)*Int(f(exp(I*t))*exp(I*t), t=0..2*Pi):
  evalf(res);

   .9999999995-.7333100980e-17*I

Close enough?

Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

Subject: Essential singularities and residue theorem and contour integral?

From: israel@math.ubc.ca (Robert Israel)

Date: 29 Aug, 2006 07:08:55

Message: 7 of 14

In article <1156831269.598789.111730@m79g2000cwm.googlegroups.com>,
Mike <housing2006@gmail.com> wrote:
>
>Mike wrote:
>> Hi there,
>>
>> Does the residue theorem of contour integral still work when there are
>> essential singularities inside the closed contour?
>>
>> For example, exp(1/z) inside the closed contour |z|=1? Matlab,
>> Mathematica, Maple failed in computing the residue of "exp(1/z)" for
>> me. My quick guess of the residues for exp(1/z) at z=0 is either 1 or
>> 0. Am I right?
>>
>> Thanks
>>
>> -M
>
>It is very strange that I did a numerial contour integral for exp(1/z)
>on the closed contour |z|=1?
>
>In Matlab it is:
>
>quad(@(t) exp(1./exp(1i*t)), -pi, pi)
>
>The result is 2*pi,
>
>But I remember contour integral should be:
>
>2*pi*i*Res[exp(1/z), @z=0],
>
>by the residue theorem, assuming it still works here,
>
>but then according to the results,
>
>Res[exp(1/z), @z=0] should be "-i", which is very strange...
>

Remember you're integrating f(z) dz around the curve.
If z = exp(i t), then dz = i exp(i t) dt. You seem to have
left out the factor i exp(i t) from the dz.

Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

Subject: Essential singularities and residue theorem and contour integral?

From: Julian V. Noble

Date: 29 Aug, 2006 10:00:39

Message: 8 of 14

Mike wrote:
> Hi there,
>
> Does the residue theorem of contour integral still work when there are
> essential singularities inside the closed contour?

Yes. For example, the integral of exp(1/z) around the
unit circle is 2\pi i .


> For example, exp(1/z) inside the closed contour |z|=1? Matlab,
> Mathematica, Maple failed in computing the residue of "exp(1/z)" for
> me. My quick guess of the residues for exp(1/z) at z=0 is either 1 or
> 0. Am I right?
>
> Thanks
>
> -M
>


--
Julian V. Noble
Professor Emeritus of Physics
University of Virginia

Subject: Essential singularities and residue theorem and contour integral?

From: Mike

Date: 29 Aug, 2006 13:56:16

Message: 9 of 14


Robert Israel wrote:
> In article <1156832432.298433.42880@i3g2000cwc.googlegroups.com>,
> Mike <housing2006@gmail.com> wrote:
> >
> >Robert Israel wrote:
> >> In article <1156830479.567529.93780@i42g2000cwa.googlegroups.com>,
> >> Mike <housing2006@gmail.com> wrote:
> >>
> >> >Does the residue theorem of contour integral still work when there are
> >> >essential singularities inside the closed contour?
> >>
> >> Do you know the statement of the theorem? Does it say anything about
> >> whether the singularities are essential?
> >>
> >
> >No the theorem did not restrict the type of the singularities...
>
> Well, there's your answer then.
>

What's your answer?


> .9999999995-.7333100980e-17*I
>
> Close enough?
>

Agreed.

Subject: Essential singularities and residue theorem and contour integral?

From: Mike

Date: 29 Aug, 2006 14:02:52

Message: 10 of 14


Robert Israel wrote:
> In article <1156831269.598789.111730@m79g2000cwm.googlegroups.com>,
> Mike <housing2006@gmail.com> wrote:
> >
> >Mike wrote:
> >> Hi there,
> >>
> >> Does the residue theorem of contour integral still work when there are
> >> essential singularities inside the closed contour?
> >>
> >> For example, exp(1/z) inside the closed contour |z|=1? Matlab,
> >> Mathematica, Maple failed in computing the residue of "exp(1/z)" for
> >> me. My quick guess of the residues for exp(1/z) at z=0 is either 1 or
> >> 0. Am I right?
> >>
> >> Thanks
> >>
> >> -M
> >
> >It is very strange that I did a numerial contour integral for exp(1/z)
> >on the closed contour |z|=1?
> >
> >In Matlab it is:
> >
> >quad(@(t) exp(1./exp(1i*t)), -pi, pi)
> >
> >The result is 2*pi,
> >
> >But I remember contour integral should be:
> >
> >2*pi*i*Res[exp(1/z), @z=0],
> >
> >by the residue theorem, assuming it still works here,
> >
> >but then according to the results,
> >
> >Res[exp(1/z), @z=0] should be "-i", which is very strange...
> >
>
> Remember you're integrating f(z) dz around the curve.
> If z = exp(i t), then dz = i exp(i t) dt. You seem to have
> left out the factor i exp(i t) from the dz.
>
> Robert Israel israel@math.ubc.ca
> Department of Mathematics http://www.math.ubc.ca/~israel
> University of British Columbia Vancouver, BC, Canada

Yes you are correct. I made a mistake here.

Thanks Robert and thanks everybody.

Now my question are solely on the branches:

1) z^(2.5)

K>> quad(@(t) (exp(1i*t)).^(2.5)*1i.*exp(1i*t)/2/pi/1i, -pi, pi)

ans =

  -0.0909 - 0.0000i

2) z^(2.1)

K>> quad(@(t) (exp(1i*t)).^(2.1)*1i.*exp(1i*t)/2/pi/1i, -pi, pi)

ans =

  -0.0317 - 0.0000i

3) z^(-2.1)

K>> quad(@(t) (exp(1i*t)).^(-2.1)*1i.*exp(1i*t)/2/pi/1i, -pi, pi)

ans =

   -0.0894

I don't see a way to find residue at z=0 for z^2.1 and z^(-2.1), but
where do the above "pseudo residues" come from?

Anybody gives me some enlightening?

Thanks a lot!

Subject: Essential singularities and residue theorem and contour integral?

From: The World Wide Wade

Date: 29 Aug, 2006 14:08:33

Message: 11 of 14

In article
<1156830479.567529.93780@i42g2000cwa.googlegroups.com>,
 "Mike" <housing2006@gmail.com> wrote:

> Hi there,
>
> Does the residue theorem of contour integral still work when there are
> essential singularities inside the closed contour?

Why not look up the residue theorem?

> For example, exp(1/z) inside the closed contour |z|=1? Matlab,
> Mathematica, Maple failed in computing the residue of "exp(1/z)" for
> me. My quick guess of the residues for exp(1/z) at z=0 is either 1 or
> 0. Am I right?

The residue of exp(1/z) at 0 is by definition the coefficient of
1/z in the Laurent expansion of exp(1/z) about 0. We have
exp(1/z) = 1 + 1/z + (1/z)^2/2! + ..., so the residue is 1.

Subject: Essential singularities and residue theorem and contour integral?

From: Robert Israel

Date: 29 Aug, 2006 16:00:13

Message: 12 of 14

Mike wrote:

> Now my question are solely on the branches:
>
> 1) z^(2.5)
>
> K>> quad(@(t) (exp(1i*t)).^(2.5)*1i.*exp(1i*t)/2/pi/1i, -pi, pi)
>
> ans =
>
> -0.0909 - 0.0000i
...
> I don't see a way to find residue at z=0 for z^2.1 and z^(-2.1), but
> where do the above "pseudo residues" come from?

There is no such thing as a residue at a branch point. In general, the

integral of the function on a contour around the branch point will
depend on where you cross the branch cut (as well as on which
branch is used). In particular, for f(z) = z^p where p is not an
integer, if you integrate the principal branch on the circle C of
radius r centred at 0,
1/(2 pi i) int_C f(z) dz
        = r^(p+1)/(2 pi) int_{-pi}^pi exp(i t (p+1)) dt
        = r^(p+1)/(2 pi (p+1) i) (exp(pi (p+1) i) - exp(-pi (p+1) i))
        = r^(p+1)/(pi (p+1)) sin(pi (p+1))

In example #1, with r = 1 and p = 5/2,
sin(7 pi/2)/(7 pi/2) = -.09094568174 approximately.

Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

Subject: Essential singularities and residue theorem and contour integral?

From: Mike

Date: 29 Aug, 2006 22:04:13

Message: 13 of 14


Robert Israel wrote:
> Mike wrote:
>
> > Now my question are solely on the branches:
> >
> > 1) z^(2.5)
> >
> > K>> quad(@(t) (exp(1i*t)).^(2.5)*1i.*exp(1i*t)/2/pi/1i, -pi, pi)
> >
> > ans =
> >
> > -0.0909 - 0.0000i
> ...
> > I don't see a way to find residue at z=0 for z^2.1 and z^(-2.1), but
> > where do the above "pseudo residues" come from?
>
> There is no such thing as a residue at a branch point. In general, the
>
> integral of the function on a contour around the branch point will
> depend on where you cross the branch cut (as well as on which
> branch is used). In particular, for f(z) = z^p where p is not an
> integer, if you integrate the principal branch on the circle C of
> radius r centred at 0,
> 1/(2 pi i) int_C f(z) dz
> = r^(p+1)/(2 pi) int_{-pi}^pi exp(i t (p+1)) dt
> = r^(p+1)/(2 pi (p+1) i) (exp(pi (p+1) i) - exp(-pi (p+1) i))
> = r^(p+1)/(pi (p+1)) sin(pi (p+1))
>
> In example #1, with r = 1 and p = 5/2,
> sin(7 pi/2)/(7 pi/2) = -.09094568174 approximately.
>
> Robert Israel israel@math.ubc.ca
> Department of Mathematics http://www.math.ubc.ca/~israel
> University of British Columbia Vancouver, BC, Canada

Thanks a lot Robert. Yes, your argument is very convincing. I've
learned a lot from it.

However it is surprising to me that you said "there is no residue for
branch point", and for negative powers, such as z^(-2.1) etc, the point
z=0 is deemed as a branch point but not a singularity?

Before you told me this, I thought there are both a branch point and a
singularity at z=0 for negative powers for z^p.

Now you are saying there is no singularity there?

Similar for log(z), you deem z=0 as a branch point or singularity?

I've done the numerical contour integration along |z|=1 for log(z)
also:

K>> quad(@(t) (1i*t)*1i.*exp(1i*t)/2/pi/1i, -pi, pi)

ans =

  -1.0000 - 0.0000i

Any confirmation about the correctness?

---------------------------------------------------

If the integrand is complicated with branch points and several other
singularities inside the contour, and then I cannot work out the angle
integral after we change the z into r*exp(i*t), t from -pi to pi... and
also you said there is no residue for branch point so residue theorem
does not apply, under such situations how can I calculate the contour
integral?

Subject: Essential singularities and residue theorem and contour integral?

From: israel@math.ubc.ca (Robert Israel)

Date: 30 Aug, 2006 05:29:17

Message: 14 of 14

In article <1156914253.224777.120900@e3g2000cwe.googlegroups.com>,
Mike <housing2006@gmail.com> wrote:

>However it is surprising to me that you said "there is no residue for
>branch point", and for negative powers, such as z^(-2.1) etc,
>the point
>z=0 is deemed as a branch point but not a singularity?

It is not an _isolated_ singularity, because of the branch cut.

>Similar for log(z), you deem z=0 as a branch point or singularity?

Branch point.

>I've done the numerical contour integration along |z|=1 for log(z)
>also:
>
>K>> quad(@(t) (1i*t)*1i.*exp(1i*t)/2/pi/1i, -pi, pi)
>
>ans =
>
> -1.0000 - 0.0000i
>
>Any confirmation about the correctness?

Yes, it's correct. Now try it along |z| = 2.

>---------------------------------------------------
>
>If the integrand is complicated with branch points and several other
>singularities inside the contour, and then I cannot work out the angle
>integral after we change the z into r*exp(i*t), t from -pi to
>pi... and
>also you said there is no residue for branch point so residue theorem
>does not apply, under such situations how can I calculate the contour
>integral?

A standard trick is to integrate along a "keyhole" contour
that avoids crossing the branch cut.

Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

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