Density function and Integration to $1$

Question

I have a function that's continuous and strictly positive on $\mathbb R$(it's also a density function w.r.t lebesgue to a probability measure), how do I go about defining it if I have the following equalities:

$$g(x)=\log\frac{f(x)}{f(0)} \quad \text{ and }\quad g(x)=p x^2$$ with $p \in\mathbb{R}$. I've tried to throw both of them into the function $x\mapsto e^x$ and obtained $$f(x)=f(0)e^{px^2}$$ And since it's a density to a probability measure the integration of it over $\mathbb R$ has to equal $1$. Now the integration divergeres for $p\ge0$. So the question is now, for $p<0$, what is $f(0)$ so that it integrates to $1$?

Answer 1

One approach is to observe that $$e^{-x^2}$$ is the "core" (i.e. the part where $x$ appears) of the normal distribution, which has density $$f(x)=\frac{1}{σ\sqrt{2\pi}}\mathrm e^{-(x-μ)^2/(2σ^2)}$$ where $μ$ and $σ$ are the parameters of the distribution. In your case $μ=0$ and $f(0)$ and $p$ should be such that $$\begin{cases}f(0)=\dfrac{1}{σ\sqrt{2\pi}} \\[0.2cm] p=-\dfrac{1}{2σ^2}\end{cases}$$ Observe that there is no $x$ in these equations.

Answer 2

Integral:

$$\int_{\mathbb{R}}f(0)e^{px^2}dx=f(0)\int_{\mathbb{R}}e^{px^2}dx$$

is Gaussian integral, see here how to calculate this integral, so:

$$\int_{\mathbb{R}}e^{px^2}dx=\sqrt{\frac{\pi}{-p}}$$

So $p=\sqrt{\frac{-p}{\pi}}$.