Complex Contour Integral: Does it Equal 0?

[Lizzie11]

Hi,

Does this complex contour integral equal 0?

[Int] (z^2)/(sin z) dz along the closed contour e^2(pi)(i)(t) 0<t<1

It should equal zero cause its analytic in the domain around te curve and the zero in the numerator is of higher order than the zero in the denominator at the point z=0.

thanks

Lizzie11

 

[Hurkyl]

It sound plausable...

Maybe if you expressed yourself more rigorously, you could be more confident in your result?

 

[saltydog]

Hi,

Does this complex contour integral equal 0?

[Int] (z^2)/(sin z) dz along the closed contour e^2(pi)(i)(t) 0<t<1

It should equal zero cause its analytic in the domain around te curve and the zero in the numerator is of higher order than the zero in the denominator at the point z=0.

thanks

Lizzie11

Care to verify this by actually solving:

\oint_C \frac{z^2}{Sin[z]}dz

with:

C=z(t)=e^{(2\pi t)i}

Or no?

Edit: Isn't this NOT a closed path if 0<t<1?
Edit2: If this is so then I should remove the circle off my integral right?

 

[saltydog]

Yep, I'm pretty convinced it's not closed if 0<t<1 and am just about ready to remove my circle. Note also if that's the case then Cauchy Integral Theorem doesn't apply in which case we can't conclude the integral is zero. However, I'm not very good with Complex Analysis and could be wrong. Can someone more knowledgeable than I confirm or reject my suspicions?

Thanks,
Salty

 

[saltydog]

Alright, one more: Maybe we can consider the Cauchy Principle Value, that is, what is the limit of the line integral as t goes to 0 and 1. Same dif maybe

 

[joeboo]

saltydog:

If you were to evaluate the integrals ( explicitly, not using Cauchy or any other such shortcuts), how would you differentiate between the integral over the 0 &lt; t &lt; 1 contour integral, and the 0 \leq t \leq 1 contour?

Additionally, and for the above ( hinted at ) reason, I don't believe a map from an open interval ( (0,1) \rightarrow \mathbb{C} ) is considered to be a path, at least not in complex analysis. In other words, paths need endpoints

 

[saltydog]

saltydog:

If you were to evaluate the integrals ( explicitly, not using Cauchy or any other such shortcuts), how would you differentiate between the integral over the 0 &lt; t &lt; 1 contour integral, and the 0 \leq t \leq 1 contour?

Additionally, and for the above ( hinted at ) reason, I don't believe a map from an open interval ( (0,1) \rightarrow \mathbb{C} ) is considered to be a path, at least not in complex analysis. In other words, paths need endpoints

Thanks Joeboo. I suspected such. Another words: you made a typo Lizzie and I want my money back.

 

[saltydog]

Can someone help me evaluate the integral without Cauchy?

This is what I have so far:

I=\oint_C \frac{z^2}{Sin[z]}dz

with:

C=z(t)=e^{(2\pi t)i}

and:

0\leq t\leq 1

Thus:

\frac{dz}{dt}=2\pi i e^{(2\pi t)i}

Making the substution, I get:

I=2\pi i\int_0^1 \frac{e^{(6\pi i)t}}{Sin[e^{(2\pi i)t}]}dt

I dont' know how to go further.

 

[Hurkyl]

If you really want to do it that way... I figure the best approach is to break it into its real and complex parts, so that you get an integral whose integrand is f(t) + i g(t).

 

[saltydog]

If you really want to do it that way... I figure the best approach is to break it into its real and complex parts, so that you get an integral whose integrand is f(t) + i g(t).

Hello Hurkly,

Been spending some time with Complex Analysis. I don't see anyway of splitting the integrand into separate functions f(t) and g(t). The best I can do in that route is:

\frac{Cos[2t]+iSin[2t]}{Sin[Cos(t)+iSin(t)]}(-2Sin[t]+iCos[t])

That's a mess. Can't even separate it when I use (x+yi). In that case, the integrand becomes:

\frac{(x^2-y^2)+2xyi}{Sin[x+yi]}

Surely, Cauchy's Integral Theorem saves the day with this one, but suppose I wanted to integrate the same function over a non-closed contour, say:

\int_C \frac{z^2}{Sin[z]}dz

with:

C=z(t)=t+it^2

and:

0\leq t\leq 2

This is just a short segement of a parabola in the complex plane. Any ideas how to solve this one?

 

[Hurkyl]

Don't forget the addition of angles trig identity!

And don't forget cosh iz = cos z, sinh iz = i sin z.
(e^u = cosh u + sinh u)

Or, you could convert sin to exponentials, then separate that into its real and imaginary parts.



As for other paths, one thing you can do is to find a different path between the same two endpoints that is easier to integrate over. (e.g. a horizontal line + vertical line) Then, if you can also find the integral around the closed contour formed by the two paths (pay attention to orientation!), you can find your original integral.

 

[saltydog]

Well, I don't get it Hurkly. Let me start from the beginning and perhaps you can show me where my problem is:

\int_C \frac{z^2}{Sin[z]}dz

Letting:

C=z(t)=Cos[t]+iSin[t]

dz=(-Sin[t]+iCos[t])dt

we have for the integrand:


\frac{(Cos[t]+iSin[t])^2}{Sin[Cos(t)]Cosh[Sin(t)]+iCos[Cos(t)]Sinh[Sin(t)]}(-Sin(t)+iCos(t))dt

I can simplify this:

\frac{-Sin[3t]+iCos[3t]}{Sin(u)Cosh(v)+iCos(u)Sinh(v)}

with:

u(t)=Cos(t)
v(t)=Sin(t)

That's still:

\frac{f(t)+ig(t)}{h(t)+ik(t)}

Which I can't separate into real and imaginary parts. Jesus I hope it's not a simple algebraic move I'm missing.

 

[saltydog]

Great . . . just multiply top and bottom by the conjugate of the denominator:

\frac{(f+ig)(h-ik)}{(h+ik)(h-ik)}

\frac{r(x)+iv(x)}{h^2-k^2}

\frac{r(x)}{h^2-k^2}+\frac{iv(x)}{h^2-k^2}

You know, I don't know why I bother with differential equations if I can't even solve the easier stuf . . .

 

[saltydog]

Using the above relations with:

u(t)=Cos[t]

v(t)=Sin[t]

K(t)=Sin[u(t)]Cosh[v(t)]

L(t)=Cos[u(t)]Sinh[v(t)]

M(t)=(Sin[u(t)]Cosh[v(t)])^2

N(t)=(Cos[u(t)]Sinh[v(t)])^2

I obtain:

\int_C\frac{z^2}{Sin[z]}dz=\int_0^{2\pi} \frac{L(t)Cos(3t)-K(t)Sin(3t)}{M(t)+N(t)}dt+i\int_0^{2\pi} \frac{L(t)Sin(3t)+K(t)Cos(3t)}{M(t)+N(t)}dt

Letting the real part be denoted by Re and the Imaginary by I am we have finally:

\int_C\frac{z^2}{Sin[z]}dz=Re+iIm

The plots below are the integrands for Re and I am for 0\leq t\leq 2 \pi[/tex]<br /> <br /> Note the symmetry. Had I obtained functions which were for example positive throughout the range, then I would have concluded that either I&#039;m wrong or Cauchy. In an effort to vindicate us both, I numerically integrated these. Both the real and imaginary integrals were zero (to 15 places) in the specified range. Thus, we have in fact in some ways reduced a complex-contour integral to a geometric representation: The area under two curves and simultaneously vindicated (alright numerically, whatever) Cauchy&#039;s Integral Theorem.

 

[Hurkyl]

I don't think you need to numerically integrate -- you just need to make some substitution relating to the symmetry you found. With the right one, you'll be able to show I = -I, proving I = 0.

 

[saltydog]

I don't think you need to numerically integrate -- you just need to make some substitution relating to the symmetry you found. With the right one, you'll be able to show I = -I, proving I = 0.

Thanks Hurkly. I see it but haven't as yet figured it out. Will work on it more as doing so would be much better than the numerical approach. Apparently, the real and imaginary portions shall always be zero and no doubt showing so must arise from the proof of the general case.

 

[saltydog]

Alright, this is how I did it: First I needed to show:

\int_0^{2\pi} Sin[t]dt=\int_0^\pi Sin[t]dt+\int_\pi^{2\pi} Sin[t]dt=0

This is proven by making the substitution x=2\pi-t in the second integral. Since the integrand above for the real portion is similar to a sine wave, and noting the plot, I'll first break up the integral:

Re=\int_0^\pi \frac{L(t)Cos(3t)-K(t)Sin(3t)}{M(t)+N(t)}dt+\int_\pi^{2\pi} \frac{L(t)Cos(3t)-K(t)Sin(3t)}{M(t)+N(t)}dt

Making the substitution x=2\pi-t in the second integral, we have:

L(2\pi-x)Cos[3(2\pi-x)]=-L(x)Cos(3x)
(messy to show but true)

K(2\pi-x)Sin[3(2\pi-x)]=-K(x)Sin[3x]

Thus the numerator becomes:

-(L(x)Cos(3x)-K(x)Sin(3x))

The denominators are:

Sin[u(2\pi-x)]Cosh[v(2\pi-x)]=Sin[u(x)]Cosh[v(x)]

cos[u(2\pi-x)]Sinh[v(2\pi-x)]=-Cos[u(x)]Sinh[v(x)]

Since these two terms are squared, we then have:

Re=\int_0^\pi \frac{L(t)Cos(3t)-K(t)Sin(3t)}{M(t)+N(t)}dt-\int_0^\pi \frac{L(x)Cos(3x)-K(x)Sin(3x)}{M(x)+N(x)}dx=0

The imaginary part, I'll say what I've read so many times before, "will be left for the reader".

Thanks Hurkly. If I was the teacher, we'd be doin' this one and they'd hate me.

 

[Lizzie11]

Thank you for the replies, but this is no where near as complicated.

y=e^(2(pi)(i)(t)) 0=<t=<1 is indeed a closed contour, since e^(2(pi)(i)) = e^0.

My question was am I right in assuming that the function is analytic (equivalently differentiable) on the domain inside the contour, if it is then the integral is 0.

Thank you again.

 

[cepheid]

My question was am I right in assuming that the function is analytic (equivalently differentiable) on the domain inside the contour, if it is then the integral is 0.

Thank you again.

Analytic?

What is the sine of zero?

 

[saltydog]

Thank you for the replies, but this is no where near as complicated.

Thank you again.

Hello Lizzie. You know, sometimes it's good to things the hard way . . . like cutting grass with a push mower as opposed to a riding mower.

 

[cepheid]

Wait a minute. We haven't had a resolution yet...I was trying to point out that the function is undefined at the origin, so how could you possibly invoke the Cauchy Integral Theorem? So I'm curious...how does one go about solving this integral?

 

[shmoe]

It's a removable singularity.

 

[cepheid]

It's a removable singularity.

Oh...I see. So did you figure that out by showing that the limit as z ---> 0 exists (is finite), using L'Hopital's? That's all I can think of. And so the residue is zero...and so...we're done. But please let me know if I'm off on that one.

 

[shmoe]

sin(z) has a zero of order 1 at z=0 (see it's power series), so z^2/sin(z)->0 as z->0.

 

[Lizzie11]

Right, that is what I thought, since its a removable singularity, the residue at zero is zero, then the integral of that function is zero.

thank you everybody