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

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.
