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 $$f(x)=\ln\left(\frac{x+\sqrt{x^2+4}}2\right)$$