Question

We have `g:RR->RR` a continous function with property that `int_(-x)^xg(t)dt=0` for each `x inRR`.How to demonstrate that g is an odd function?

Answer 1

Define `f(x) = int_0^x g(t) dt` and use the fact that if `f` is even then `f'` is odd.

Explanation:

Suppose that `g` is continuous on `RR` and for every real `x`, we have `int_-x^x g(t) dt = 0`.

Define `f(x) = int_0^x g(t) dt`.

Then for every `x`, `f'(x) = g(x)`

Then `int_-x^x g(t) dt = [ f(t) ]_-x^x = f(x) - f(-x) = 0`

Therefore, `f(-x) = f(x)` so `f` is even.

This implies that `f'` is odd, and since `f' = g`, we see that `g` is odd.

To see that `f` even implies `f'` odd, apply the chain rule to `f(-x)`

Answer 2

See below.

Explanation:

`int_(-x)^x g(t)dt=int_(-x)^x g(-t)(-dt)=-int_(-x)^xg(-t)dt`

so

`int_(-x)^x (g(t)+g(-t))dt=0 forall x` then `g(t) = -g(-t)` so `g(t)` is odd.