evaluate integral

[alejandrito20]

i need to evaluate [math](A'e^{nA})'[/math],where [math]A=kte |\phi|[/math],with [math]\phi[/math] a angular coordinate between [math](-\pi,\pi)[/math] upon a line integration over a closed path along the coordinate angular [math]\phi[/math].......¿why the result is zero?

 

....' is the derivate on [math]\phi[/math]

Edited April 4, 2010 by alejandrito20

[Amr Morsi]

Do you know the equation of the path? Or, it is simply r*phi?

If so, are you sure that the bounded integral is zero?

[the tree]

Argh, so many letters. I'm guessing the primes are for first derivatives, but with respect to what? And is e|phi| meant to read e^|phi|? Knowing what path would certainly be helpful as well.

[alejandrito20]

Do you know the equation of the path?

 

No

 

If so, are you sure that the bounded integral is zero

YES

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Knowing what path would certainly be helpful as well.

 

in the article ( arxiv.org/abs/0907.1321v1) says: "Note also that the left hand side of the equation (10) vanishes upon a line integration over a closed path along the compact internal space."

[the tree]

i need to evaluate [math](A'e^{nA})'[/math],where [math]A=kte |\phi|[/math],with [math]\phi[/math]

 

in the article ( arxiv.org/abs/0907.1321v1) says: "Note also that the left hand side of the equation (10) vanishes upon a line integration over a closed path along the compact internal space."

But the LHS of equation 10 in that article is [imath]\nabla \cdot (W^\zeta \nabla W)[/imath].

[timo]

But the LHS of equation 10 in that article is [imath]\nabla \cdot (W^\zeta \nabla W)[/imath].

In which case "gradient field" is the term you want to look up. The vanishing of the integral over any closed path is a property of a gradient field.

[Amr Morsi]

If W is only a function of phi, then it will turn to something like the formula written by alejandrito but divided by r^2*(sin(theta))^2...........

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I think that "compact inetrnal space" defines the path.

[the tree]

But the LHS of equation 10 in that article is [imath]\nabla \cdot (W^\zeta \nabla W)[/imath].

 

In which case "gradient field" is the term you want to look up. The vanishing of the integral over any closed path is a property of a gradient field.

I thought that'd apply in this, although I wasn't certain so I didn't want to say it outright. I'm still befuddled as to how it's connected to the expression presented in the OP.

[alejandrito20]

In which case "gradient field" is the term you want to look up. The vanishing of the integral over any closed path is a property of a gradient field.

 

but, i read that :

 

If A = grad F,The line integral over a gradient field is equal to the difference between the values of the potential function at the end-points ...the integral round a closed curve has the value zero...

 

i think that [math]\oint \nabla W \cdot dr=0[/math]

 

but [math]\oint \nabla \cdot (W^\zeta \nabla W) dr[/math], with z a number, ¿is zero too???????

[Amr Morsi]

The former integral you introduced in the first post gets vanished. Just substitute with phi1=phi+pi and integrate from phi1=0 to phi1=2pi. The derivative will cancel out with the integration.

 

Or, you may use the definition you posted before in another thread:

[math]

\frac{d^2|\phi|}{d\phi^2}=2(\delta(\phi)-\delta(\pi-\phi))

[/math]