$\int_{- \infty}^{\infty} \frac{f(x)}{1+\exp{g(x)}}dx=\int_{0}^{\infty} f(x) dx$ for $f(x)=f(-x),~g(x)=-g(-x)$ - are there other formulas like that? If $f(x)$ any even function, integrable on $(0,\infty)$ and $g(x)$ any odd function, then we have:

$$\int_{- \infty}^{\infty} \frac{f(x)}{1+e^{g(x)}}dx=\int_{0}^{\infty} f(x) dx \tag{1}$$

The proof is elementary:
$$I(a)=\int_{- \infty}^{\infty} \frac{f(x)}{a+e^{g(x)}}dx$$
$$I(1/a)=\int_{- \infty}^{\infty} \frac{f(x)}{1/a+e^{g(x)}}dx=a \int_{- \infty}^{\infty} \frac{e^{-g(x)}f(x)}{e^{-g(x)}+a}dx= \\ = a \int_{- \infty}^{\infty} f(x)dx-a^2\int_{- \infty}^{\infty} \frac{f(x)}{a+e^{g(x)}}dx$$
$$\frac{1}{a} I(1/a)+aI(a)=\int_{- \infty}^{\infty} f(x)dx$$
$$I(1)=\int_{0}^{\infty} f(x)dx$$

With this formula we can write some crazy looking integrals to scare people, like:
$$\int_{- \infty}^{\infty} \frac{e^{-x^2}}{1+e^{\sin (\sinh x)+x^3-\arctan x}}dx=\frac{\sqrt{\pi}}{2}$$
To be fair, it might also be useful for some quatum statistics applications (i.e. Fermi-Dirac distribution).


I want to know, what other formulas like $(1)$ exist? Maybe with the exponential function, or some other functions

I also know of Glasser's Theorem, but I wonder if some more interesting cases exist.
To be more specific, I mean the non-trivial formulas of the following kind:
$$\int_{a}^b g(x) f(x) dx=k \int_{A}^B f(x) dx$$
With $k$ being some constant, independent on $f(x)$, $f(x)$ is a general function (with some restricitions), $g(x)$ is some interesting function. $A,B$ might be different from $a,b$, but also should not depend on $f(x)$.
 A: Here is a slightly different variation to OPs example, followed by another one.

Suppose $p$ is an even function, i.e. $p(x)=p(-x)$ and $q(x)q(-x)=1$.
Then
  \begin{align*}
\int_{-a}^{a}\frac{p(x)}{1+q(x)}\,dx=\int_{0}^{a}p(x)\,dx
\end{align*}

A proof of the statement together with an application can be found in this answer.
Note: This technique can be found e.g. in Inside Interesting Integrals  by P.J. Nahin. He applies it to the seemingly complicated integral
\begin{align*}
 \int_{-1}^{1}\frac{\cos(x)}{1+e^{(1/x)}}\,dx
 \end{align*}
which becomes easy if you know the trick.

Suppose $p$ satisfies the functional equation $p\left(\frac{1}{x}\right)=p(x)$.
Then the following integral is independent of $s$:
  \begin{align*}
\int_{0}^{\infty}\frac{p(x)}{x^s+1}\,\frac{dx}{x}=\int_{0}^{1}p(x)\,\frac{dx}{x}
\end{align*}

Proof: Splitting the LHS integral into two pieces on $[0,1]$ and $[1,\infty)$ and substituting $x\rightarrow \frac{1}{x}$ in the second one gives
\begin{align*}
\int_{0}^{\infty}\frac{p(x)}{x^s+1}\,\frac{dx}{x}&
=\int_{0}^{1}\frac{p(x)}{x^s+1}\,\frac{dx}{x}+\int_{1}^{\infty}\frac{x^s p(1/x)}{x^s+1}\,\frac{dx}{x}\\
&=\int_{0}^{1}p(x)\,\frac{dx}{x}
\end{align*}
Note: This integral can be found e.g. in Irresistible Integrals  by G. Boros and V.H. Moll.
A: Any function $g(x)$ such that $g(c+a)+g(c-a)=k$ for all $a$ on the interval $(0,b)$ with any function $f(x)$ such that $f(c-a)=f(c+a)$ for all $a$ on the interval $(0,b)$ will satisfy the equation
$$\int_{c-b}^{c+b}{f(x)g(x)dx}=k*\int_{c}^{c+b}{f(x)dx}$$
because, using a trapezoidal Riemann sum after splitting the integrals into
$$\int_{c-b}^{c}{f(x)g(x)dx} + \int_{c}^{c+b}{f(x)g(x)dx}$$
and using $\Delta x= \frac bn$
$$\lim_{n \to \infty}\Delta x *\sum_{i=1}^{n-1}f(c-b+i*\Delta x)*g(c-b+i*\Delta x)+f(c+b-i*\Delta x)*g(c+b-i*\Delta x)$$
$$+\frac{\Delta x}2*(f(b-a)*g(b-a)+f(b-c)*g(b-c))$$
The second part of the Riemann sum has the indices going backwards for the sake of the "proof" and we only look at one specific index in this part.
$$\lim_{n \to \infty}\Delta x*(f(c-(b-i*\Delta x))*g(b-(a-i*\Delta x))+f(b+a-i*\Delta x)*g(b+a-i*\Delta x))$$
Letting $w=b-i*\Delta x$
$$\lim_{n \to \infty}\Delta x*(f(c-w)*g(c-w)+f(c+w)*g(c+w))$$
$$f(c+w)=f(c-w)$$
$$\lim_{n \to \infty}\Delta x*(f(c+w)*g(c-w)+f(c+w)*g(c+w))$$
$$\lim_{n \to \infty}\Delta x*(f(c+w)*(g(c-w)+g(c+w)))$$
$$g(c+w)+g(c-w)=k$$
$$\lim_{n \to \infty}\Delta x*(f(c+w)*k)$$
Going back, we have
$$k*\lim_{n \to \infty}\Delta x*\sum_{i=1}^n f(c+b-i*\Delta x)+\frac{\Delta x}2*(f(c)+f(c+b))$$
which is the Trapezoid Rule for the integral
$$k*\int_c^{c+b}f(x)dx$$
You had $c=0$ (which coincidentally made $f(x)$ an even function), $b=\infty$, and $g(x)=\frac 1{1+e^{h(x)}}$.
I do not know if this is an actual theorem, corollary, etc. I also don't know if the logistic function is the only solution to $g(c+a)+g(c-a)=k$ for all $a$ on the interval $(0,b)$.
Although I tried to avoid any errors, if you see any, let me know.
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
In Statistical Physics we use a lot the relation:
\begin{align}
{1 \over \expo{x} + 1} & =
\ds{\left\lbrace\begin{array}{lcl}
\ds{\Theta\pars{-x} + {\,\mathrm{sgn}\pars{x} \over \expo{\verts{x}} + 1}}
& \mbox{if} & \ds{x \not= 0}
\\[2mm]
\ds{\half} & \mbox{if} & \ds{x = 0}
\end{array}\right.}
\end{align}
$\ds{\Theta}$ and $\ds{\,\mathrm{sgn}}$ are the Heaviside Step Function and the Sign Function, respectively. For example,
\begin{align}
\int_{-\infty}^{\infty}{\,\mathrm{F}\pars{x} \over \expo{x} + 1}\,\dd x & =
\int_{-\infty}^{0}\mathrm{F}\pars{x}\,\dd x +
\int_{-\infty}^{\infty}{\,\mathrm{sgn}\pars{x}\,\mathrm{F}\pars{x} \over \expo{\verts{x}} + 1}\,\dd x
\\[5mm] & = 
\int_{-\infty}^{0}\mathrm{F}\pars{x}\,\dd x +
\int_{0}^{\infty}{\,\mathrm{F}\pars{x} - \,\mathrm{F}\pars{-x} \over
\expo{x} + 1}\,\dd x
\end{align}

I can imagine your statement is a general form of the present one. There are similar relations for $\ds{1 \over \expo{x} - 1}$ too.

