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For every real $x>0$ there is a unique $\lambda_x>0$ such that

$$F(\lambda_x)=x,\qquad F(c)=\int_0^1\frac{dy}{y(1+e^{c/y})} =\int_c^\infty\frac{dt}{t(1+e^t)}.$$

Define

$$c_x=\int_0^1h\left(\frac1{1+e^{\lambda_x/y}}\right)\,dy.$$

Then $0<c_x<1$, $c_x$ is continuous and strictly increasing, $c_x\to0$ as $x\downarrow0$, and $c_x\to1$ as $x\to\infty$. With $c_0=0$, one also has

$$\sup_{x>0}\bigl(c_x-c_{(1-\eta)x}\bigr)\longrightarrow0 \qquad(\eta\downarrow0).$$

Source: published PDF, p. 2, Theorem 1, and the continuity step used on p. 11. The uniform scaling statement expands the latter step. The printed decimal $c_1\approx0.91117$ is not numerically certified here.

Bears on. Problem 297.

Proof

The substitution $t=c/y$ gives the two formulas for $F$. For $c>0$ the integrand is positive and integrable. Differentiation of the second formula yields

$$F'(c)=-\frac1{c(1+e^c)}<0.$$

Moreover $F(c)\to0$ as $c\to\infty$. On $(0,1]$, $(1+e^t)^{-1}=1/2+O(t)$, so $F(c)=\frac12\log(1/c)+O(1)$ as $c\downarrow0$. Thus $F$ maps $(0,\infty)$ continuously and strictly decreasingly onto $(0,\infty)$. This proves existence, positivity, uniqueness, and continuity of $\lambda_x$, and its limits $\infty$ at $x=0$ and 0 at $x=\infty$. A nonpositive real multiplier could not solve the first integral: the integral would diverge at zero.

For every $y>0$, increasing $x$ decreases $\lambda_x$ and strictly increases $(1+e^{\lambda_x/y})^{-1}$ within $(0,1/2)$. Binary entropy strictly increases there. Integration gives strict increase of $c_x$. The integrands lie between 0 and 1, so dominated convergence proves continuity and the endpoint limits, and also $0<c_x<1$.

For uniform scaling continuity fix $\zeta>0$. Choose $a>0$ with $c_a<\zeta$ and $b>a$ with $1-c_b<\zeta$. If $x\le a$, the loss is at most $c_a$. If $x\ge2b$ and $\eta\le1/2$, both arguments are at least $b$, and the loss is at most $1-c_b$. For $a\le x\le2b$, both arguments lie in $[a/2,2b]$ when $\eta\le1/2$, and their difference is at most $2b\eta$. Uniform continuity on that compact interval makes the remaining loss less than $\zeta$ for sufficiently small $\eta$.