Distinguish Normal Distribution, Gaussian Distribution and Normalised Gaussian Distribution? As I understood it, the 'normal distribution' is $$\frac{1}{\sqrt{2\pi}}\exp\left(\frac{-(x-\mu)^2}{2{\sigma}^2}\right)$$
Now according to this the 'normal probability density function' is $$f(x)=\frac{1}{\sigma\sqrt{2\pi}}\exp\left(\frac{-(x-\mu)^2}{2{\sigma}^2}\right)$$ and according to this at the top of page 466 (or 436 if you own the book) the 'normalised Gaussian distribution' is
$$f(t)=\frac{1}{\tau \sqrt{2\pi}}\exp\left(\frac{-t^2}{2{\tau}^2}\right)$$
In the cases above, $\mu =$ mean, $\sigma =$ standard deviation and $\tau =\Delta t$
I am very confused; each time I search the web for answers I get different words and formulae, this post is designed to try to dispel this confusion.
Could someone please explain the meaning and difference between the formulae given above and the similarity and/or differences between the phrases used in the title and body? 
Thank you.
 A: The second formula is the standard expression for the probability density function (PDF) corresponding to the normal (or Gaussian) distribution with mean $\mu$ and standard deviation $\sigma$. As it is a PDF, it is normalised to 1, i.e., its integral over admissible values of $x$ is $1$. The first formula is missing the $1/\sigma$ factor, thus it is not a PDF. Finally, the third formula can be obtained from the second one with direct substitution $\sigma\rightarrow\tau$, $x\rightarrow t$, and $\mu\rightarrow0$.
A: A gaussian distribution is the same as a normal distribution. The standard gaussian or standard normal distribution is the gaussian distribution with $\mu = 0$, $\sigma = 1$.
BTW, the first equation above is incorrect, as its integral over the reals is $\sigma$, and not 1. As the probability of the sample space, by definition, has to be 1, any distribution must integrate to, well, 1. 
You could normalise this by dividing it by $\sigma$ (to get the second equation). Note that this normalisation has nothing to do with the normal in normal distribution. The distribution's name comes from it being normal in the sense of usual, and the process of making the probability of $\mathbb{R}$ (or whatever your sample space it) equal 1 comes from making it fit to the norm, or rule.
A: The normal distibution, also called the Gaussian distribution is the probability distribution that assigns to every measurable set $A$ of real numbers the probability
$$
\int_A \frac 1 {\sqrt{2\pi}} \exp\left( \frac{-1} 2 \left( \frac{x-\mu} \sigma \right)^2 \right) \, \frac{dx} \sigma.
$$
In particular, there is the cumulative probability distribution function of the normal distribution:
$$
x\mapsto \int_{-\infty}^x {\sqrt{2\pi}} \exp\left( \frac{-1} 2 \left( \frac{w-\mu} \sigma \right)^2 \right) \, \frac{dw} \sigma.
$$
The normal probability density function, also called the Gaussian probability density function, is
$$
x\mapsto \frac 1 {\sqrt{2\pi}} \exp\left( \frac{-1} 2 \left( \frac{x-\mu} \sigma \right)^2 \right) \cdot \frac 1 \sigma.
$$
If you have any trouble remembering which is which, then remember the meanings of the terms probability distribution and probability density function.  As with all density function, whether we're talking about probability density or mass density or energy density or population density, the value of the density is an intensive quantity (density) and its integral is an extensive quantity (probability or mass or energy or population).
I suspect the word "normalised" is intended to indicate that it has a standard deviation depending on a parameter.  That is $\tau$.  But that has nothing to do with the word "normal" as used above.  The word "normal" is somewhat overburdened.
"Gaussian" is something of a misnomer as well.  Abraham de Moivre identified the importance of this particular probability distribution in the first half of the 18th century, well before Carl Gauss was born (Gauss was born in the 1770s).
