Gaussian distribution and proportionality I have a very simple question regarding normal distribution, which nonetheless bugs me a bit. In my reference book related to Bayesian data analysis the author seems to use the following results of the book:
$$\int_{-\infty}^{\infty} x \,g(x)\,dx = c\mu\sqrt{2\pi\sigma^2}\;\;\;\;\;\;(1)$$
$$\int_{-\infty}^{\infty} (x-\mu)^2 \,g(x)\,dx = c\sigma^2\sqrt{2\pi\sigma^2}\;\;\;\;\;\;(2)$$
$$\int_{-\infty}^{\infty} \,g(x)\,dx = c\sqrt{2\pi\sigma^2}\;\;\;\;\;\;(3)$$
where $c, \mu, \sigma^2$ are constant values to write $g(x)$ as the Gaussian distribution: 
$$p(x)=\frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right).$$
I'm not fully certain how the author got from $(1)$, $(2)$ and $(3)$ to $p(x)$, but I am guessing proportionality plays a part here? Do you think the author defines (without explicitly stating it) that: 
$$p(x) = \frac{g(x)}{c\sqrt{2\pi\sigma^2}}?$$
To answer user @Stockfish the function $g(x)$ is actually defined as: 
$$g(x) = \int\exp\left(-\frac{\beta}{2}\{x-\mu-\textbf{g}^T\textbf{v}\}^2-\frac12\textbf{v}^T\textbf{A}\textbf{v}\right)\,d\textbf{v},$$
where $\beta$ is a constant, $\textbf{g}, \textbf{v}\in R^W$ and $\textbf{A}$ is a symmetric $W\times W$ real matrix. 
UPDATE: 
This post is actually a follow up question to my previous post (if it helps): 
How to evaluate the integral $\int\mathbf{g}^T\mathbf{v} \exp{(-\frac{1}{2}\mathbf{v}^T \mathbf{A}\mathbf{v})}\,d\mathbf{v}$
 A: 
Do you think the author defines (without explicitly stating it) that:
  $$p(x) = \frac{g(x)}{c\sqrt{2\pi\sigma^2}}$$

Yes, you are right, as you can see from the first three integrals, 
$$\int_{-\infty}^{\infty} x \,g(x)\,dx  = k E(X)$$
$$\int_{-\infty}^{\infty} (x-\mu)^2 \,g(x)\,dx = k \text{var}(X) $$
$$\int_{-\infty}^{\infty} \,g(x)\,dx = k(1)$$
where $k = c\sqrt{2\pi\sigma^2}$.
That being said, it is now easy to see why $g(x) = k p(x)$, where $p(x)$ is the PDF of $\mathcal{N}(\mu,\sigma^2)$
A: I think I got it myself also, it seems to be as I suspected. If you evaluate the integral I gave above ($g(x)$ that is), you will get: 
$$g(x) = (2\pi)^{W/2}|(\textbf{A}+\beta\textbf{g}\textbf{g}^T)|^{-1/2} \exp\left(-\frac{1}{2\sigma^2} (x-\mu)^2 \right),$$
where $\sigma^2 = \frac{1}{\beta-\beta^2\textbf{g}^T(\textbf{A}+\beta\textbf{g}\textbf{g}^T)^{-1}\textbf{g}}=\frac1\beta+\textbf{g}^T\textbf{A}^{-1}\textbf{g}$ and $c=(2\pi)^{W/2}|(\textbf{A}+\beta\textbf{g}\textbf{g}^T)|^{-1/2}.$
So we have $p(x)$ as: 
$$p(x) = \frac{g(x)}{c\sqrt{2\pi\sigma^2}}=\frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right).$$
