solution to some expectation equation We have an equation: for some $c>0$ and $g(x)$ such that $g(x)+g(-x)=1$,
\begin{align*}
\int_{\infty}^{\infty} \frac{1}{\sqrt{2\pi}} \left[ \exp(-\frac{(x-1)^2}{2}) - \exp(-\frac{(x+1)^2}{2}) \right] g(x) = c
\end{align*}
Then, some engineering probability paper says
\begin{align*}
g(x) = \left[ 1+\exp \left( \lambda \frac{1-e^{-2x}}{1+e^{-2x}} \right)\right]^{-1}
\end{align*}
is the solution with some $\lambda < 0$.
Why is this true? I spent two days on this, but cannot figure it out unfortunately. Any hint or help will be appreciated.
 A: I don't buy the solution in the paper. See below for a counterexample.
Let's recast this a bit. Given the integral equation
\begin{align*}
\int_{\infty}^{\infty} \frac{1}{\sqrt{2\pi}} \left[ \exp(-\frac{(x-1)^2}{2}) - \exp(-\frac{(x+1)^2}{2}) \right] g(x) = c
\end{align*}
You may recognize that the terms in brackets are gaussian densities $\phi(x;\mu,\sigma)$ so we can recast this as the difference in two expected values:
\begin{align*}
\int_{\infty}^{\infty} g(x)\phi(x;1,1)dx - \int_{\infty}^{\infty} g(x) \phi(x;-1,1)dx = c
\end{align*}
Which we can represent as
$$E_{X}[g(X)] - E_{Y}[g(Y)]=c\text{ where } X\sim N(1,1),\;Y\sim N(-1,1)$$
Since $g(x)+g(-x) = 1$ we know $g(0)=\frac12$ and $g'(x) -g'(-x) =0$ which implies $g'$ is even and therefore $g(x)-\frac12$ is odd.
Let $g(x)=\frac12$, then $g(-x) + g(x) = 1$ and $g(x) - \frac12$ is the zero function on $\mathbb{R}$ which happens to be both even and odd!
Therefore,
$$E_{X}[g(X)] - E_{Y}[g(Y)]=\frac12 - \frac12  = 0$$
Here we see that $g(x) = \frac12 \iff \lambda=0$ because
$$\left[ 1+\exp \left( \lambda \frac{1-e^{-2x}}{1+e^{-2x}} \right)\right]^{-1} = \frac12 \iff \lambda = 0$$
More generally, let $c>0$. If $\lambda < 0$ then $g(x;\lambda)$ is a monotonic increasing function of $x$ that approaches a constant function as $\lambda$ approaches zero from below:
$$\lim_{\lambda \to 0^-} g(x;\lambda) = \frac12 \implies E_X[g(X)]-E_Y[g(Y)] = 0$$
And the Heaviside step function as you approach $-\infty$:
$$\lim_{\lambda \to -\infty} g(x;\lambda) = H(x)\equiv \mathbb{1}_{\geq 0}(x) \implies E_X[H(X)] - E_Y[H(Y)] = $$
$$P(X>0) - P(Y>0) = \Phi(1) - \Phi(-1) \approx 0.68$$
Since $g(x;\lambda)$ is strictly increasing in $x$ for all $\lambda < 0$  and $\text{sgn } \left[\frac{\partial}{\partial \lambda}g(x;\lambda)\right]= \text{sgn }x \;\;\forall \lambda < 0$, we see that as $\lambda \to 0^-$ we put increasing weight on $x>0$ and less on $x<0$ monotonically and continuously as a function of $\lambda$, hence $\lambda \to -\infty$ and $\lambda \to 0^-$ will bracket the range of this integral equation.
Taking these two extremes we see that
$$ 0 < E_X[g(X;\lambda) - E_Y[g(Y;\lambda)] < 0.68 \;\;\forall \lambda < 0$$
So the proposed function for $g$ can only accommodate $c \in (0,0.68)$, hardly $c>0$.
EDIT: OP confirmed that $c \in (0,0.68)$ was the actual range considered in the paper, and not simply $c>0$, in which case there is a 1-to-1 relationship between $c$ and $\lambda$.
The question that remains is how to derive this. I contend that the paper's proposed functional form for $g(x)$ is a solution, but not the solution, as there is a broader class of single-parameter functions whose parameter $\theta$ is determined by $c\in (0,0.68)$ (i.e., $\theta_{c} = f(c)$), where $f$ is determined by the choice of $g$.
From the above, we see that as long as $g\geq 0,\;g(x)+g(-x)=1$ then we are in good shape. From the way the authors set $\lambda < 0$, we can throw in that $\theta \in (a,b) \subset \mathbb{R}$ and $g$ is monotonic increasing in $x$ and
$$\lim_{\theta \to b^-} g(x;\theta) = \frac12,\;\; \lim_{\theta \to a^+}g(x;\theta) = H(x)$$
So, for example, this also works for $g$:
$$g(x;\theta) = \frac12 -\theta[1-H(x)] + \theta H(x) = \frac12 -\theta + 2\theta H(x),\;\;\theta \in \left(0,\frac12\right)$$
We can see that
$$g(x;\theta) + g(-x;\theta)= \left[\frac12 -\theta + 2\theta\right] + \left[ \frac12-\theta \right] =  1$$
And the function approaches $\frac12$ and $H(x)$ as $\theta \to 0^+,\;\;\theta \to \frac12^-$, respectively.
In fact, unlike the proposed solution, we can analytically identify $c$ produced by a given $\theta$:
$$E_X[g(X;\theta)] - E_Y[g(Y;\theta)] = $$
$$\left(\frac12 - \theta \right)P(X\leq 0)+ \left(\frac12 + \theta \right)P(X>0) -\left(\frac12 - \theta \right)P(Y\leq 0)-\left(\frac12 + \theta \right)P(Y>0)=$$
$$\left(\frac12 - \theta \right)\Phi(-1)+ \left(\frac12 + \theta \right)\Phi(1) -\left(\frac12 - \theta \right)\Phi(1)-\left(\frac12 + \theta \right)\Phi(-1)=$$
$$\Phi(1)\left[\frac12 + \theta - \frac12 + \theta\right] +   \Phi(-1)\left[\frac12 - \theta -\frac12-\theta\right] = 2\theta\left[\Phi(1)-\Phi(-1)\right]$$
$$\approx 2(.68)\theta = 1.36\theta\;\;\theta \in \left(0,\frac12\right)$$
So for this version of $g$, we get a nice linear relationship between $c$ and $\theta$, and this satisfies the requirements of the paper.
I think there must have been other considerations (e.g., smoothness or continuity) that the authors felt was necessary. But even then, there are several other options (like the sigmod).
So, the reason you can't derive their answer, is because you can't -- getting to the form in the paper requires additional desiderata apart from $g(x) + g(-1)=1$ and the need for the integral formula to evaluate to some $c\in (0,0.68)$.
