# periodic odd function integral

Let $$f$$ be a periodic function with period $$T$$ and $$f(x)+f(-x)=0$$ in the interval $$[-T/2,T/2]$$. Prove that $$\int_{a}^{x}f(t)dt$$ is a periodic function with periodic T.

My try: $$\int_{a}^{x+T}f(t)dt = \int_{a}^{x}f(t)dt + \int_{x}^{x+T} f(t)dt = \int_{a}^{x}f(t)dt + \int_{0}^{T}f(t)dt$$

The second term should ideally come out to be 0 but it clearly isn't necessary as f(x) is odd only in [-T/2,T/2].

Where is my mistake?

the function is periodic with period $$T$$ and $$f(x)+f(-x)=0$$, you can combine the fact that $$f(x+T)=f(x)$$ and then
$$\int_{-T/2}^{T/2}f(x)dx=0\\ \int_{-\frac{T}{2}}^{\frac{T}{2}}f(x)dx=\\ \int_{-\frac{T}{2}}^{0}f(x)dx+\int_{0}^{\frac{T}{2}}f(x)dx\\ \int_{-\frac{T}{2}}^{0}f(x+T)dx+\int_{0}^{\frac{T}{2}}f(x)dx\\ y=x+T\\ dy=dx\\ \int_{\frac{T}{2}}^{T}f(y)dy+\int_{0}^{\frac{T}{2}}f(x)dx=\\ \int_0^Tf(x)dx$$