What functions $g$ satisfy $\int_{-L}^{L} \frac{f(x)}{1 + g(x)}\:dx = \int_{0}^{L} f(x)\:dx$ for every even function $f$? As has been covered in a number of questions on this site, there is a well know property of single variable real continuous even functions $f(x)$:
\begin{equation}
 \int_{-L}^{L} \frac{f(x)}{1 + e^x}\:dx = \int_{0}^{L} f(x)\:dx
\end{equation}
for $L \in \mathbb{R}^+$ being either finite or infinite. 
When you evaluate the proof, there is a fundamental property of $g(x) = e^x$ that allows for this to occur and that is:
\begin{equation}
g(-x) = \frac{1}{g(x)}
\end{equation}
We see this holds not only for $e$ but for any $a \in \mathbb{R}^+$
My question: outside of $a^x$ are there any real valued functions the satisfy this condition?
 A: Assume that $g: \Bbb R \to \Bbb R \setminus \{ -1 \}$ is a continuous function with
$$ \tag 1
\int_{-L}^{L} \frac{f(x)}{1 + g(x)}\,dx = \int_{0}^{L} f(x)\,dx
$$
for all $L > 0$ and all even continuous functions $f: [-L, L]\to \Bbb R$.
Then in particular (choosing $f(x) = 1$)
$$
\int_{-L}^{L} \frac{1}{1 + g(x)}\,dx = L
$$
for all $L > 0$, and differentiating this with respect to $L$ gives
$$
 \frac{1}{1 + g(L)} + \frac{1}{1 + g(-L)} = 1 
\iff g(L) g(-L) = 1 \, .
$$
Therefore
$$ \tag 2
 g(x) g(-x) = 1 
$$ must hold for all $x \in \Bbb R$. 
It is clear now that $g$ can have no zeros. Also $g(0)^2 = 1$ and $g(0) \ne -1$, therefore $g(0) = 1$. Since we assumed $g$ to be continuous, $g(x)> 0$ for all $x \in \Bbb R$ follows.
So we can define $h(x) = \log g(x)$. Substituting this in $(2)$ gives
$$
 h(x) + h(-x) = 0
$$
so that

$$ \tag 3 g(x) = e^{h(x)} \text{ for some odd continuous function $h$.}$$

On the other hand, every function $g$ defined by $(3)$ satisfies $(2)$, and consequently $(1)$, so that is the most general (continuous) solution.
