1. ## change of variables

The problem is to show that $\int^{x+ct+L}_{x-ct-L} g(s) \ ds = - \int^{x+ct}_{x-ct} g(s) \ ds$

where g is an odd 2L periodic function and g(L-x)=g(x)

I don't understand the following change of variables: $\int^{x+ct+L}_{x-ct-L} g(s) \ ds = \int^{x+ct}_{x-ct} g(s+L) \ ds$

But if if the above is true, then it's easy to see that $\int^{x+ct}_{x-ct} g(s+L) \ ds = -\int^{x+ct}_{x-ct} g(-s-L) \ ds = - \int^{x+ct}_{x-ct} g(-s+L) \ ds$ $= -\int^{x+ct}_{x-ct} g(s) \ ds$

2. Originally Posted by Random Variable
The problem is to show that $\int^{x+ct+L}_{x-ct-L} g(s) \ ds = - \int^{x+ct}_{x-ct} g(s) \ ds$

where g is an odd 2L periodic function and g(L-x)=g(x)

I don't understand the following change of variables: $\int^{x+ct+L}_{x-ct-L} g(s) \ ds = \int^{x+ct}_{x-ct} g(s+L) \ ds$

But if if the above is true, then it's easy to see that $\int^{x+ct}_{x-ct} g(s+L) \ ds = -\int^{x+ct}_{x-ct} g(-s-L) \ ds = - \int^{x+ct}_{x-ct} g(-s+L) \ ds$ $= -\int^{x+ct}_{x-ct} g(s) \ ds$

Let $u=s-L \implies du=ds$ using this we get

$
\int^{x+ct+L}_{x-ct-L} g(s) \ ds =\int g(u+L)
$

To figure out the new limits of integration plug them into the above eqation for s. So we get

$u=\underbrace{(x+ct+L)}_{s}-L=x+ct$

Now just do the same thing for the bottom limit.

3. But then the bottom limit would be $x-ct-2L$.

What am I not understanding?

Wait. If g is 2L periodic, then then an antiderivative of g is also 2L periodic.

4. Originally Posted by Random Variable
But then the bottom limit would be $x-ct-2L$.

What am I not understanding?
The function is 2L periodic so you can can add any multiple of 2L as many times as you want

It is exactly like adding $2\pi$ to this integral

$\int_{0}^{2\pi}sin(x)dx=\int_{2\pi}^{4\pi}\sin(x)d x$

I hope this helps

5. Wow. That was a brain fart of epic proportions.