We define an even function to be a function such that whenever is in the domain of so is , and for all in the domain, we have
We define an odd function to be a function such that whenever is in the domain of so is , and for all in the domain, we have
Then, for integrable on prove the following.
- If is an even function then
- If is an odd function then
- Proof. First, since the integral is additive with respect to the interval of integration (Theorem 1.17 of Apostol), we have
Then, for the first integral we use the expansion/contraction of the interval of integration with to get
Since is an even function by assumption, we have for all . Since we then have,
So, putting this all together we have,
- Proof. Similar to part (a) we have,
But in this case, since is an odd function, we have for all . Thus,
I think before you come up with the additive property with respect to the interval of integration, you need to point out the integral of f from -b to 0 exists. That means you should write down the contraction / expansion with respect to the interval of integration with k = -1 first, and you can proceed as the rest of the solution.