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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?

Did
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    Function $f(x)$ should be even. – Kemono Chen Dec 17 '18 at 05:21
  • Yes, thanks, I forgot that in my question. I will amend now. –  Dec 17 '18 at 05:21
  • There are many. How about $e^{x^{3}}$? You can start with all kinds of functions on $[0,\infty)$ with $g(0)=1$ an define $g(x)=\frac 1 {g(-x)}$ for $x<0$. – Kavi Rama Murthy Dec 17 '18 at 05:52
  • @KaviRamaMurthy - I'm curious in compiling a list of functions (hence the 'Big List' tag). If you could please post up families of functions that satisfy the condition spoken to, I would be very appreciative. –  Dec 17 '18 at 05:54
  • Assume $g\in C^\omega(\mathbb{R})$, we can expand $g(-x)g(x)=1$ at $x=0$ and get the series representation of $g$ with some free variables. – Kemono Chen Dec 17 '18 at 05:56
  • @KemonoChen - I will have to give that a go. Thanks for the suggestion. –  Dec 17 '18 at 08:41
  • @KemonoChen 1. Not needed. 2. Not every $C^\infty$ function is equal to its series expansion at every point. – Did Dec 17 '18 at 10:26
  • @Did To the second point: Yes but I wrote $C^{\large\omega}$, not $\infty$. It denotes real analytic functions. – Kemono Chen Dec 18 '18 at 02:34

1 Answers1

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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.

Martin R
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