Why does $\log p(\mu, \log \sigma \mid y)=\text{constant} -n\log\sigma-\frac{1}{2\sigma^2}\left((n-1)s^2+n(\overline{y}-\mu)^2\right)$? I'm studying Bayesian data analysis and I came across to the following derivation in my study material:

Given normally distributed i.i.d samples $y_1, y_2, \ldots, y_n$,
$y_i\sim N(\mu, \sigma^2)$ and a uniform prior on $(\mu, \log\sigma)$,
i.e. $p(\mu, \log\sigma) \propto 1$ then the log-posterior $\log
 p(\mu, \log \sigma \mid y)$ is:
$$\log p(\mu, \log \sigma \mid y)=\text{constant}
 -n\log\sigma-\frac{1}{2\sigma^2}\left((n-1)s^2+n(\overline{y}-\mu)^2\right),$$
where $s^2$ is the sample variance
$s^2=\frac{1}{n-1}\sum_{i=1}^n(y_i-\overline{y})^2$ and
$\overline{y}=\frac{1}{n}\sum_{i=1}^n y_i$ is the sample average.

My question is: how do we arrive in the log-posterior shown above? Can you give the main steps explicitly?
This problem is from Gelman's et al. Bayesian Data analysis book.
 A: In the usual way of writing the posterior up to some unknown normalising constant we have
$$
p(\mu, \log \sigma \mid \mathbf{y} ) \propto p(\mathbf{y} \mid \mu, \log \sigma)p(\mu, \log \sigma), \tag{1}
$$
then letting $p(\mu, \log \sigma) \propto 1$ be an improper prior $(1)$ becomes
$$
p(\mu, \log \sigma \mid \mathbf{y} ) \propto p(\mathbf{y}\mid\mu, \log \sigma) \cdot 1 \qquad \text{or} \qquad p(\mu,\log \sigma \mid \mathbf{y} ) = \frac{1}{Z}p(\mathbf{y} \mid \mu, \log \sigma),
$$
for some unknown normalising constant, which may not actually exist and is something that must be checked for after arriving at a candidate form for the posterior. Now taking the logarithm of both sides we have
$$
\begin{align*}
\log p(\mu,\log \sigma\mid\mathbf{y} ) &= \log \left(\frac{1}{(2\pi )^{n/2} \sigma^n}e^{-\frac{\sum_i (y_i-\mu)^2}{2\sigma^2}}\right)+ \log(1) + \text{constant} \\
&=-\frac{1}{2\sigma^2}\sum_i\left(y_i - \mu\right)^2-n\log\sigma + \mbox{constant} \tag{2}.
\end{align*}
$$
Now the following is a common decomposition
$$
\begin{align*}
\sum_i (y_i -\mu)^2 &= \sum_i(y_i-\bar{y} +\bar{y}-\mu)^2 \\
&=\sum_i (y_i - \bar{y})^2+\sum_i(\bar{y}-\mu)^2+2\underbrace{\sum_i(y_i-\bar{y})(\bar{y}-\mu)}_{=0} \\
&=(n-1)s^2+n(\bar{y}-\mu)^2. \tag{3}
\end{align*}
$$
Then putting $(3)$ into $(2)$ gives the result. You can then if you like check this does indeed normalise
