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I know if $f$ is an odd function then

$$\int_{-L}^L f(x)\:dx = 0$$

my question is, is the converse necessarily true? Intuitively, I feel it should be that by assuming that the integral with those symmetric bounds is zero I can somehow show $f(-x) = -f(x)$, but I'm not sure how to go about it.

My attempt:

$$\int_{-L}^L f(x)\:dx = \int_{-L}^{0} f(x) \: dx + \int_{0}^{L} f(x) \: dx = 0$$.

So,

$$\int_{-L}^{0} f(x) \: dx = - \int_{0}^{L} f(x) \: dx$$

How do I use this to show $f(-x) = -f(x)$, or am I wrong to think the converse is necessarily true?

Can I just do this from my last step,

$$\int_{0}^{L} f(-x) \: dx = - \int_{0}^{L} f(x) \: dx$$.

Thus $f(-x) = -f(x)$, or is this not allowed?

Eddie
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  • For a continous function, when you make the interval of integration very short, the value of the integral approaches the value of the function times the length of the interval. Subtract everything except a short interval from the last equality you have. – ploosu2 Dec 02 '14 at 22:21
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    The fundamental theorem of calculus lets you differentiate both sides with respect to $L$. – Arthur Dec 02 '14 at 22:21
  • If $\int_{-L}^{L}f(x)dx=0$ for all $L>0$, then $f$ is an odd function. – mjqxxxx Dec 02 '14 at 22:30
  • Suppose L = 1, and we know it holds true for that case, then can we conclude $f$ is odd on the interval $[-1,1]$? This is essentially what I am taking for granted for a problem I'm working on. – Eddie Dec 02 '14 at 22:34

1 Answers1

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First, suppose that $L$ is fixed. Without losing generality, we can take $L=1$.

Then just take $$f(x)=\begin{cases}1,&x\in [-1,0],\\1-4x,&x\in[0,1].\end{cases}$$ Obviously, $\int_{-1}^1f(x)dx=0$, yet $f$ is not odd.

Second, if we put the condition that $\int_{-L}^Lf(x)dx=0$ for every $L>0$, the situation changes. Suppose for simplicity that $f$ is continuous, then the antiderivative $F:x\to\int_{0}^xf(s)ds$ is a $C^1$ function.

We can say that $$\int_{-L}^Lf(x)dx=0\Rightarrow \forall L>0\ \ F(L)-F(-L)=0,$$hence the derivative with respect to $L$ is also zero:$$\forall L>0\ \ F'(L)+F'(-L)=0\Rightarrow \forall L>0\ \ f(L)+f(-L)=0,$$therefore $f$ is odd.

TZakrevskiy
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