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I have this function:

\begin{equation*} f(x)=% \begin{cases} 1 &x\in\left[ -\pi,-\pi/2\right[ \\ -1 &x\in\left[ -\pi/2,0\right[\\ 1 & x\in\left[ 0,\pi/2\right[ \\ -1 & x\in\left[ \pi/2,\pi\right]\\ \end{cases} \end{equation*}

First I thought that it was odd, but then I realized that $f(0) = 1 \neq - f(0) = -1$ was true. Does it matter when you calculate an integral and want to use the property of an odd function? For instance can I still deduce that:

\begin{equation*} \int_{-a}^{a} f(x) dx = 0 \end{equation*}

?

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up vote 2 down vote accepted

Changing the value of the function in a finite number of points (so that you get an odd function) does not affect the integral, so yes, the integral is zero.

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Redefining $f$ at $0$ (to be $0$) and $\frac\pi2$ (to be $1$) would make $f$ odd. Since only two points were changed (or any finite number of points), the integral is unaffected, so you can still deduce that $$ \int_{-a}^af(x)\,\mathrm{d}x=0 $$

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