i need to evaluate the follows integral:

 

[math]\oint dy (A' e^{A})' [/math]

 

where ' is derivation respect [math]y[/math] and [math]A(y)=A(y+2\pi)[/math] and too [math]A(y)'=A(y+2\pi)'[/math]

 

 

 

where [math]y \in (-\pi,\pi)[/math] ia a angular coordinate , and [math]A' [/math] is discontinuous in [math]-\pi,0,\pi[/math]

 

 

 

the result is zero, but i think that is NOT CORRECT to say:

 

[math]\oint dy (A' e^{A})' = A' (\pi)e^{A(\pi)}- A' (-\pi)e^{A(-\pi)} =0 [/math], since [math]A' [/math] is discontinuous in -pi and pi

Edited September 22, 201114 yr by alejandrito20

I think you meant that A' is discontinuous in -pi and pi, not A. Why not simply integrate from 1 to (1+2pi), and not have your problem?

 

EDIT: Oh, I overlooked something: If A' is discontinuous: How can (A' exp(A))' exist?

Edited September 22, 201114 yr by timo

I think you meant that A' is discontinuous in -pi and pi, not A. Why not simply integrate from 1 to (1+2pi), and not have your problem?

 

yes , is discontinuous in A'

 

wy not simply integrate from 1 to (1+2pi), and not have your problem?

 

your says [math]\int_1^{1+2\pi} dy (A' exp(A))' [/math]?????

 

 

 

 

EDIT: Oh, I overlooked something: If A' is discontinuous: How can (A' exp(A))' exist?

(A' exp(A))' there is Not exist in -pi,0,pi

Edited September 22, 201114 yr by alejandrito20

i need to evaluate the follows integral:

 

[math]\oint dy (A' e^{A})' [/math]

 

where ' is derivation respect [math]y[/math] and [math]A(y)=A(y+2\pi)[/math] and too [math]A(y)'=A(y+2\pi)'[/math]

 

 

 

where [math]y \in (-\pi,\pi)[/math] ia a angular coordinate , and [math]A' [/math] is discontinuous in [math]-\pi,0,\pi[/math]

 

 

 

the result is zero, but i think that is NOT CORRECT to say:

 

[math]\oint dy (A' e^{A})' = A' (\pi)e^{A(\pi)}- A' (-\pi)e^{A(-\pi)} =0 [/math], since [math]A' [/math] is discontinuous in -pi and pi

 

 

The integral of the derivative of a periodic function over a period will always be zero. But as Timo observes, you have imposed additional conditions that make the function non-differentiable, in contrtadiction to your problem statement.

 

You have something messed up in the problem statement.

The integral of the derivative of a periodic function over a period will always be zero. But as Timo observes, you have imposed additional conditions that make the function non-differentiable, in contrtadiction to your problem statement.

 

You have something messed up in the problem statement.

 

in the text (http://arxiv.org/PS_cache/hep-th/pdf/0011/0011225v2.pdf eq2.13) says:

 

"in a compact internal space without boundary, the integral vanishes"

the inicial space is [math] z \in -\infty, \infty[/math], but [math] z = r \phi [/math] , whit r radio konstant

in the text (http://arxiv.org/PS_...1/0011225v2.pdf eq2.13) says:

 

"in a compact internal space without boundary, the integral vanishes"

the inicial space is [math] z \in -\infty, \infty[/math], but [math] z = r \phi [/math] , whit r radio konstant

 

 

What you are faced with is an application of (the general differential goemoetry version of) Stokes theorem. This is sraightforward, but you have missed the point that your problem statement is self-contradictory.

 

It is also clear that you are in way over yoir head. You need to master a great deal of more basic mathematics and physics before you attempt advanced M theory.