Integral of the second partial derivative of a periodic function

Question

The defined integral of the first partial derivative of a function $f=f(x,y,z)$ is

$\int_{a}^{b} \frac{\partial f}{\partial z} dz = f(x,y,b)-f(x,y,a)$.

So if $f$ is a periodic function in $z$ on the $[a,b]$ interval then $f(x,y,a)=f(x,y,b)$ and the result of the integral is 0. Please correct me if I am wrong.

The question is what happens for the integral of the second derivative

$\int_{a}^{b} \frac{\partial^2 f}{\partial z^2} dz$

for a periodic and a non-periodic function.

Answer 1

Yes, you are correct (assuming suitable regularity of $f$). If $f$ is periodic, then so are its derivatives. This follows from differentiating the periodic condition: $$f(z + p) = f(z)$$ The period $p$ is constant.

So if $f$ is periodic with period $b-a$, then so is $\frac{\partial f}{\partial z}$ Therefore you can apply your original result to $\frac{\partial f}{\partial z}$ instead of $f$ to get $$\int_{a}^{b} \frac{\partial^2 f}{\partial z^2} dz = 0$$

For a non periodic function, all you can say is

$$\int_{a}^{b} \frac{\partial^2 f}{\partial z^2} dz = \frac{\partial f}{\partial z}(x,y,b)-\frac{\partial f}{\partial z}(x,y,a)$$