contour integration problem


by mmzaj
Tags: contour, integration
mmzaj
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#1
Dec29-08, 06:22 PM
P: 99
i'm having trouble trying to evaluate the following contour integration :


[tex]\oint _{C}\frac{ds}{s^{n+1}}[/tex]

where the contour C encloses the right half plane

i'm having crazy results !! i need an explanation of the work .
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Mute
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#2
Dec29-08, 06:59 PM
HW Helper
P: 1,391
What methods are you allowed to use? This is a trivial integral if you're allowed to state the half-residue theorem. If you need to prove the half residue theorem, though, it will be slightly more complicated (but not by much, I expect).

The half-residue theorem states that if your contour goes around a pole in a semicircular arc, then the contribution of the pole to the integral is half the residue of a full circular contour about that pole.

For example, say you have a pole at the origin, and your contour is a large semi circle of radius R that has a straight line piece of the contour along the real axis, with a small semicircular piece of radius epsilon in the contour that goes around the pole at the origin instead of through it. Then although strictly speaking there are no poles inside the contour, the effect of the pole as you take epsilon goes to zero is to contribute half of the residue about that pole to your actual contour.

The 'half-residue' theorem isn't quite as general as it could be: in fact, if you have any fraction of a circle that goes around a pole, its contribution to the integral will be the same fraction of the residue from that pole. e.g., if you have a pole at the origin and a quarter circular contour whose straight line pieces run along the real and imaginary axes, the pole at the origin will contribute a quarter of the residue about it to the integral over the contour.

Uh... does that make sense?
mmzaj
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#3
Dec29-08, 07:17 PM
P: 99
thanks for the quick response . actually the problem popped up in my research on control theory . and i'm really not into complex analysis . i tried parameterizing s , and computing the integration on an arc going from [tex]\frac{-\pi}{2}[/tex] to [tex]\frac{\pi}{2}[/tex] with a radius R [tex]\rightarrow\infty[/tex], and from [tex]i\infty[/tex] to [tex]-i\infty[/tex] . but i had crazy results , they give zero number singularities of my transfer function in the RHP , which isn't the case !!! i'll try and put my work here , but first let me try to understand your post .

jostpuur
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#4
Dec29-08, 10:37 PM
P: 1,983

contour integration problem


At least I don't understand what the original contour is supposed to be, from the first post.

Whatever the contour is supposed to be, the function [tex]s\mapsto \frac{1}{s^{n+1}}[/tex] can be integrated by the fundamental theorem of calculus. I would not recommend getting into the residue stuff.
mmzaj
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#5
Dec30-08, 03:29 PM
P: 99
the contour encloses the right half plane . so it's an arc from [tex]\frac{-\pi}{2}[/tex] to [tex]\frac{\pi}{2}[/tex] with a radius [tex]R\rightarrow\infty[/tex] , and a straight line from iR to -iR .
HallsofIvy
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#6
Dec30-08, 04:16 PM
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There is NO "contour that encloses the right half plane". Do you mean "integrate along the imaginary axis"? If so then you could integrate from -Ri along the arc with radius R, from [itex]-pi/2[/itex] to [itex]pi/2[/itex], as mmzaj says, to Ri and then from iR down to -iR, finally, letting R go to [itex]\infty[/itex].
mmzaj
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#7
Dec30-08, 06:23 PM
P: 99
the integration along the contour is ZERO . but that's not possible !!
let me explain . my transfer function [tex] G(s) [/tex] has a number of singularities ( zeros and poles ) in the RHP . according to Cauchy's argument principle the integration
[tex]\oint _{C}\frac{G^{'}(s)}{G(s)}ds=2\pi i(N-M) [/tex]

N : the number of zeros inside the contour
M : the number of poles inside the contour
where the contour C encloses all the singularities in the right half plane . because in control theory we are interested in the singularities of the transfer function in the RHP . now , under some conditions
[tex]\frac {G^{'}(s)}{G(S)} [/tex]
can be expanded as :

[tex]\frac {G^{'}(s)}{G(S)}= \sum^{\infty}_{n=0} \frac {A_{n}}{s^{n+1}}[/tex]

where [tex] A_{n} [/tex] is some coefficient .

now if ,
[tex]\oint _{C}\frac{ds}{s^{n+1}}=0[/tex]
that indicates that the function has no singularities in the RHP , which isn't a general case , and isn't a condition on the series expansion
[tex]\frac {G^{'}(s)}{G(S)}= \sum^{\infty}_{n=0} \frac {A_{n}}{s^{n+1}}[/tex]

so , here is my problem !!
jostpuur
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#8
Dec31-08, 04:03 AM
P: 1,983
Quote Quote by mmzaj View Post
the contour encloses the right half plane . so it's an arc from [tex]\frac{-\pi}{2}[/tex] to [tex]\frac{\pi}{2}[/tex] with a radius [tex]R\rightarrow\infty[/tex] , and a straight line from iR to -iR .
If [itex]n\geq 0$[/itex], then the straight line from iR to -iR goes directly through the singularity at [itex]z=0[/itex], and the integral doesn't exists.
Mute
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#9
Dec31-08, 12:49 PM
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Quote Quote by jostpuur View Post
If [itex]n\geq 0$[/itex], then the straight line from iR to -iR goes directly through the singularity at [itex]z=0[/itex], and the integral doesn't exists.
Which is why you form a small semi-circle of radius [itex]\epsilon[/itex] that goes around the singularity and take the limit as [itex]\epsilon \rightarrow 0[/itex], which requires the use of the half residue theorem. That is, if your contour goes through a pole, the pole contributes 1/2 the value of its residue to the integral.

So, the integral

[tex]\oint _{C}\frac{ds}{s^{n+1}}[/tex]

is non-zero only for n = 0, as for n < 0 the integrand is analytic so the integral around the contour is zero, and for n > 0 the residue of the integrand is zero.
jostpuur
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#10
Dec31-08, 01:34 PM
P: 1,983
Quote Quote by Mute View Post
Which is why you form a small semi-circle of radius [itex]\epsilon[/itex] that goes around the singularity and take the limit as [itex]\epsilon \rightarrow 0[/itex], which requires the use of the half residue theorem.
I don't believe this is helpful. It will matter whether the pole is passed from left or from right, and how do you know which one is the correct choice now? You cannot know it, unless you know what was originally wanted with the integral. mmzaj is trying to integrate over an invalid contour, but we shouldn't fix it arbitrarily.

The half residue theorem is very unlikely going to become useful, because the integrated function is so simple that it can be integrated with the fundamental theorem of calculus.
HallsofIvy
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#11
Dec31-08, 01:37 PM
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Either way is correct, you just have to be careful of the sign.
Mute
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#12
Dec31-08, 03:22 PM
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Quote Quote by jostpuur View Post
I don't believe this is helpful. It will matter whether the pole is passed from left or from right, and how do you know which one is the correct choice now? You cannot know it, unless you know what was originally wanted with the integral. mmzaj is trying to integrate over an invalid contour, but we shouldn't fix it arbitrarily.

The half residue theorem is very unlikely going to become useful, because the integrated function is so simple that it can be integrated with the fundamental theorem of calculus.
If you take the semicircle of radius [itex]\epsilon[/itex] to be such that the overall contour does not enclose the pole (i.e., to the right of the pole, in this problem), the pole contributes half of its residue to the integral. If you take the semicircle of radius [itex]\epsilon[/itex] around the pole the other direction (to the left of the pole in this problem), then in the limit [itex]\epsilon \rightarrow 0[/itex] it contributes minus its resiude to the integral as the semicircle contour is now traversed in the opposite sense as the previous case, but then you also need to take into account the fact that now the overall contour encloses the pole, which contributes the full residue, and the two contributions sum to give half the residue again. You do need to know which direction the overall contour is going, but that should be stated in the problem already.

(Strictly speaking, if you were to do the integral by calculating its value along each piece of the contour, to get the result you would by the half-residue theorem you're actually doing a principal value integral along the imaginary axis. One might take issue with this but I see no reason for it to be a problem as long as everything's consistent.).
squidsoft
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#13
May9-09, 06:56 PM
P: 54
Quote Quote by mmzaj View Post
the integration along the contour is ZERO . but that's not possible !!
let me explain . my transfer function [tex] G(s) [/tex] has a number of singularities ( zeros and poles ) in the RHP . according to Cauchy's argument principle the integration
[tex]\oint _{C}\frac{G^{'}(s)}{G(s)}ds=2\pi i(N-M) [/tex]

N : the number of zeros inside the contour
M : the number of poles inside the contour
where the contour C encloses all the singularities in the right half plane . because in control theory we are interested in the singularities of the transfer function in the RHP . now , under some conditions
[tex]\frac {G^{'}(s)}{G(S)} [/tex]
can be expanded as :

[tex]\frac {G^{'}(s)}{G(S)}= \sum^{\infty}_{n=0} \frac {A_{n}}{s^{n+1}}[/tex]

where [tex] A_{n} [/tex] is some coefficient .

now if ,
[tex]\oint _{C}\frac{ds}{s^{n+1}}=0[/tex]
that indicates that the function has no singularities in the RHP , which isn't a general case , and isn't a condition on the series expansion
[tex]\frac {G^{'}(s)}{G(S)}= \sum^{\infty}_{n=0} \frac {A_{n}}{s^{n+1}}[/tex]

so , here is my problem !!
I feel the statement above is ill-posed. If it were my test question, this is how I would answer it:

[tex]\oint _{C}\frac{G'(s)}{G(s)}ds=2\pi i(N-M)[/tex]

is valid only if the contour encloses all the zeros and poles and does not pass through any of them.

Now, the statement ``the contour encloses all zeros and poles in the right half-plane'' does not necessarily mean the contour is traversing the imaginary axis. The contour can be larger and just enclosing all the zeros and poles in the right half-plane. However when the statement: ``under certain conditions:''
[tex]\frac{G'(s)}{G(s)}=\sum_{n=0}^{\infty}\frac{a_n}{z^{n+1}}[/tex]

is made, that power series is centered at the origin with a radius of convergence equal to the distance to the nearest zero or pole of [tex]G(z)[/tex] since the expression [tex]\frac{G'(z)}{G(z)}[/tex] is non-analytic at both the zeros and poles of [tex]G(z)[/tex]. Therefore this series cannot be substituted into the the expression:

[tex]\oint_C\frac{G'(z)}{G(z)}dz[/tex]

since the contour [tex]C[/tex] necessarily encloses a region over which the power series fails to represent the quotient [tex]\frac{G'(z)}{G(z)}[/tex].


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