Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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


share|cite|improve this question

1 Answer 1

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

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.