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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?

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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

$$\int_{-T/2}^{T/2}f(x)dx=0\\ \int_{-\frac{T}{2}}^{\frac{T}{2}}f(x)dx=\\ \int_{-\frac{T}{2}}^{0}f(x)dx+\int_{0}^{\frac{T}{2}}f(x)dx\\ \int_{-\frac{T}{2}}^{0}f(x+T)dx+\int_{0}^{\frac{T}{2}}f(x)dx\\ y=x+T\\ dy=dx\\ \int_{\frac{T}{2}}^{T}f(y)dy+\int_{0}^{\frac{T}{2}}f(x)dx=\\ \int_0^Tf(x)dx$$

now you can conclude you argument.

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