Is there a general family of curves that satisfies the following conditions? Is there a general family of curves $f(x,c)$ that satisfies the following conditions?


*

*$f(x,c)$ is strictly increasing for $x \ge 0$

*$f(0,c) = 0$

*$f(1,c) = 1$

*$f(x,c) \to \infty$ as $x \to \infty$

*For every positive constant $0 < c < 1$,
$$
\int_{0}^{\infty} c^{f(x,c)}dx = 1.
$$


I would like to know if there is a general family of curves satisfying the above conditions and if not, what additional conditions should we add to obtain the required family of curves.
 A: The naming and scaling of the parameter $c$ is irrelevant. We just want a family of strictly increasing functions $(f_q)_{q>0}$ such that for each $q$ we have
$$f_q(0)=0,\quad f_q(1)=1,\quad \lim_{x\to \infty}f_q(x)=\infty\ ,\tag{1}$$
and such that for a certain number $c=c(q)\in\ ]0,1[\ $ depending smoothly and invertibly on $q$ the condition
$$\int_0^\infty c^{\>f_q(x)}\ dx=1\tag{2}$$
is fulfilled.
The condition $(1)$ is satisfied by the functions
$$f_q(x):= x^q\qquad(q>0)\ .$$
Write $c:=e^{-p}$, $\>p>0$. Then we want
$$J_q:=\int_0^\infty\exp(-p x^q)\ dx=1\ .$$
Substituting
$$x:=\left({u\over p}\right)^{1/q}$$
we obtain
$$J_q={p^{-1/q}\over q}\int_0^\infty e^{-u}\ u^{{1\over q}-1}\ du={p^{-1/q}\over q}\Gamma\left({1\over q}\right)=p^{-1/q}\ \Gamma\left({q+1\over q}\right)\ .$$
The condition $J_q=1$ leads to
$$p=\Gamma^q\left({q+1\over q}\right)>0\ ,$$
and then via $c=e^{-p}$ to a value $c$ such that $(2)$ holds.
A: Correct me if I am being stupid here, but I don't see that $x^a$ actually works. Let $a=1$ and then note that $$\int_0^\infty c^{x}dx = \frac{-1}{\ln(c)}$$ which is not always $1$. As for your question, the family is very large and one way to see this is by using the transformation $$f(x,c) = \log_c(g_c(x))$$ which introduces a new function $g_c(x)$ with the restrictions:


*

*$g_c(x)>0$ is strictly decreasing.

*$g_c(0) = 1$.

*$g_c(1) = c$.

*$\int_0^\infty g_c(x)dx = 1$.


A strictly decreasing function which must pass through only two points and is required to have an integral over $\mathbb{R}^+$ of $1$ is not much of a restriction at all.  One can even cook up all kinds of piecewise defined functions that could easily satisfy this.  Technically, just the family of functions $g_c(x)$ which satisfies the above $4$ conditions is a general family and there is no reason to expect that there is a closed from for such functions.
As for placing more restrictions on these functions, I would have to know more about why you are looking for these functions in the first place. But I hope these thoughts at least help :)
