Integral Brane World: Why $\oint (A'e^A)' dy = 0$ but $\oint e^A dy \neq 0$?

[alejandrito29]

if the extra coordinate y \in [-\pi,\pi] with A(y)=A(y+2 \pi) and A' is non continuom in -\pi,0,\pi

why

\oint (A'e^A)' dy =0 but \oint e^A dy \neq 0?

 

[mitchell porter]

What is A?

 

[Finbar]

The first integral is a total derivative so

\oint ( A' e^A)' dy= A'(\pi) e^{A(\pi)} - A'(-\pi) e^{A(-\pi)}

which would vanish provided

A'(\pi) = A'(-\pi)

but you seem to imply this may not be the case?

The integral
\oint A' e^A dy= \oint ( e^A)' dy = e^{A(\pi)} - e^{A(-\pi)} =0

is also a total derivative and certainly does vanish due to the boundary conditions.


I see no reason for

\oint ( e^A) dy

to vanish as its not a total derivative so it depends on the explicit form of the function A(y).

 

[alejandrito29]

The first integral is a total derivative so

\oint ( A' e^A)' dy= A'(\pi) e^{A(\pi)} - A'(-\pi) e^{A(-\pi)}

which would vanish provided

A'(\pi) = A'(-\pi)

but A' is discontinuous in -\pi,0,\pi

 

[alejandrito29]

What is A?

A=|y|
where
A' = 1 , y \in )0,\pi(

A' = -1 , y \in )-\pi,0(

A ' = undefined , y =- \pi, 0,\pi