# Prove that the integral of an odd step function on [-1,1] is 0

Let $\varphi\colon[-1,1]\to \mathbb R$ be an odd step function.Prove that: $\int_{-1}^1\! \varphi(t)\, dt = 0$

Thanks!

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## 1 Answer

Let $f$ be an odd integrable function, that is, $f(x)=-f(-x)$ for all $x\in\mathbb{R}$. Then:

$\int_{-R}^R f(x) dx=\int_{-R}^0f(x)dx+\int_0^Rf(x)dx=-\int_0^Rf(x)dx+\int_0^Rf(x)dx=0$

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1
Note that this doesn't even use the assumption that $\varphi$ is a step function -- the conclusion holds for every odd integrable function. – Henning Makholm Apr 16 '12 at 14:16
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