# Is there a $g$ such that $\int\limits_{-a}^a \frac{f(x)}{g(x)} \mathrm dx = \int\limits_0^a f(x) \mathrm dx$ for *odd* function $f$?

It was shown that if $f$ is continuous and even on $[−a,a], a>0$, then $\int\limits_{-a}^a \frac{f(x)}{1+e^{x}} \mathrm dx = \int\limits_0^a f(x) \mathrm dx$.

I wonder are there any similar property holds if $f$ is odd? So my question is as follow:

If $f$ is continuous and odd on $[−a,a], a>0$, is there a function $g$ such that $\int\limits_{-a}^a \frac{f(x)}{g(x)} \mathrm dx = \int\limits_0^a f(x) \mathrm dx$?

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Well, there's an almost trivial example: Let $g: \mathbb{R} \rightarrow \mathbb{R}$ be defined by $g(x) = \left\{ \begin{array}{lr} -2 &: x \in (-\infty, 0)\\ 2 &: x \in [0, \infty). \end{array} \right.$

Then note that, since $f$ is odd, $\int_{-a}^0 f = -\int_0^a f$. So we have: $\int_{-a}^a \frac{f}{g} = \int_{-a}^0 \frac{f}{g} + \int_{0}^a \frac{f}{g}\\ = -\frac{1}{2} \int_{-a}^0 f + \frac{1}{2} \int_{0}^a f = \int_0^a f.$

So yes, we can always find such a $g$. I can look for a more complicated example if you wanted something fancier, but the given $g$ works for now.

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You could let $g(x) = 2$ if $x\geq 0$, and $g(x)=-2$ if $x<0$. Then $$\int\limits_{-a}^a \frac{f(x)}{g(x)}\,dx = \int\limits_{-a}^0 \frac{f(x)}{-2}\,dx + \int\limits_{0}^a \frac{f(x)}{2}\,dx = 2\int\limits_{0}^a \frac{f(x)}{2}\,dx = \int\limits_{0}^a f(x)\,dx.$$

The condition actually isn't very hard to satisfy: all you need is for $|g(x)|$ to be bounded below and for $\frac{1}{g(x)}-\frac{1}{g(-x)}$ to equal $1$ identically; then the exact same reasoning goes through.

In the even case, the similar condition $\frac{1}{g(-x)}+\frac{1}{g(x)}=1$ is satisfied with $g(x)=1+e^x$, since $$\frac{1}{1+e^{-x}} + \frac{1}{1+e^x} = \frac{(1+e^x) + (1 + e^{-x})}{(1+e^x)(1+e^{-x})} = \frac{1+e^x + 1 + e^{-x}}{1+e^x + e^{-x} + 1} = 1.$$

If you want something similar for the odd case, there's no such nonzero $g(x)$ that's also continuous at $x=0$. That's because if $\frac{1}{g(x)}-\frac{1}{g(-x)}=1$ (which you can prove using bump functions to be necessary, and not just sufficient), then letting $x\to 0$ we have $0=\frac{1}{g(0)}-\frac{1}{g(0)} = 1.$ So you might as well use the discontinuous $g$ described above.

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