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

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

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