Sigmoid function that approaches infinity as x approaches infinity. The function I'm looking for looks like an error function, but instead of having asymptotes $1$ and $-1$, the function I'm looking for does not have asymptote. It increases to infinity.
The derivative of this function looks like a Gaussian function; it also approaches zero (as x approaches infinity) but at slower rate.
$$\lim_{x\to \infty} f(x) = \infty$$
$$\lim_{x\to -\infty} f(x) = -\infty$$
$$f''(0) = 0$$
($f''(x) = 0$ at only one point)
The derivative of this function $f'(x)$ looks like a Gaussian function; it also approaches zero (as $x$ approaches infinity) but at slower rate.
$$\lim_{x\to \infty} f'(x) = 0$$
$$\lim_{x\to -\infty} f'(x) = 0$$
I think $\ln(x)\mathrm{erf}(x)$ is close, but the maximum gradient or $f'(x)$ is not at $x=0$.

I think I got the answer.
I started by assuming the function $f(x)$ derivative $f'(x)$ kinda looks like the Gaussian function, and its double derivative $f''(x)$ looks like the original function $f(x)$.
$$f''(x)=f(x)$$
So I asked Wolfram|Alpha's help (yup I cheated) and the general solution is (ignoring constants)
$$f(x)=e^x±e^{-x}$$
The $f(x)=e^x-e^{-x}$ really looks like the function I want, except it needs to be reflected at $y=x$ axis.
Solving $x$, I get
$$f(x)=\ln\left(\frac{x+\sqrt{x^2+4}}2\right)$$

