# Find monotonic functions going from $0$ to $+\infty$ for $x \in (-\infty,+\infty)$ (similar to $e^x$)

How can we find functions on $\mathbb{R}$ with exponential-like properties, namely:

• $f(x)$ is infinitely differentiable;
• $f(x)$ and all its derivatives are monotonic;
• $f(x)$ and all its derivatives obey the following limits:

$$\lim_{x \to -\infty}f(x)=0$$

$$\lim_{x \to +\infty}f(x)=+\infty$$

One such function is obviously the exponent itself ($a,b$ - real positive constants):

$$f(x)=ae^{bx}$$

Another function which seems to have these properties (I don't know how to prove it) is the 'Sophomore's function':

$$s(x)=\int_0^1 u^{-u~x} du=\sum_{k=1}^{\infty} \frac{x^{k-1}}{k^k}$$

For the proof of the integral formula see this answer by Sangchul Lee.

The derivatives are easy to find (both for the series and the integral formula) and they all seem to obey the above properties:

How can we find other such functions?

And (related) how to prove that $s(x)$ has these properties?

• I think a nice restriction to the problem would be to only look for analytic functions. Then you could try to characterize these functions by their power series coefficients. – mathematician Apr 16 '16 at 20:59
• @mathematician, could you provide an example of an infinitely differentiable function which is not analytic? – Yuriy S Apr 16 '16 at 21:02
• The classic example is the function $f$ defined by $f(x) = exp(-1/x^2)$ for $x > 0$ and $f(x) = 0$ elsewhere. – Vik78 Apr 16 '16 at 21:43

If $g(x) \ge 0$, $\lim_{x \to -\infty} g(x) = 0$, and $\int_{-\infty}^{\infty} g(t) dt = \infty$, then $f(x) =\int_{-\infty}^x g(t) dt$ is such a function.
If you want all the derivatives to be monotonic, impose that restriction on $g$.
• I really think the condition $g(x) \geq 0$ should be clarified. Otherwise, $g(x)=C$ fits the first two conditions for any $C \geq 0$, but $f(x)$ doesn't work the way it should – Yuriy S Apr 16 '16 at 21:42
• Otherwise, a good suggestion, but you could at least offer some explicit $g(x)$ – Yuriy S Apr 16 '16 at 21:45
• I added the condition $\lim_{x \to -\infty} g(x) = 0$. – marty cohen Apr 16 '16 at 21:58