Why does MLE not include the integral for joint probability of a continuous random variable

I understand that the MLE considers the joint probability of the observed data. So consider, for example, we are trying to estimate the mean $$\mu$$ from a random sample distributed iid according to the normal density function $$\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$$ given $$X, X=\{x_1,x_2,\cdots,x_n\}$$ with $$\mu, \sigma^2$$ unknown. We then consider the joint probability as $$P\{X\} = \prod_i^n P\{x_i\} = \prod_i^n \frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x_i-\mu)^2}{2\sigma^2}}$$ by independence. The we use the log, set the equation $$=0$$ and solve. But I have always wondered why the integral is not at all present in the formulation. Perhaps I have written it wrong but what am I not understanding.

• The joint probability of observing $(X_1, X_2,\ldots, X_n) = (x_1,x_2,\ldots, x_n)$ is $0$. Instead, the value of the likelihood is defined to be the joint density of $(X_1, X_2,\ldots, X_n)$ evaluated at $(x_1,x_2,\ldots, x_n)$, and considered as a function of the parameter $\mu$. That is, $$L(\mu; x_1,x_2,\ldots,x_n) = f(x_1,x_2,\ldots, x_n; \mu).$$ And for heaven's sake don't murmur the usual shibboleths of "take logs, set derivative$=0$" etc. $L(\mu)$ is in the form of a normal density function (times a constant) and has a maximum at the mean of this normal density function. Sep 13 '15 at 22:05

First let me touch up a few things in your description of MLE. On the left hand side what we have is $P(X | \mu)$. This is important because we are thinking about this as a function of $\mu$.
Then we take a log (often) because it makes the expression easier to deal with because it breaks the product into a sum and brings down powers of exponential a where parameters (such as $\mu$ often reside). The reason we can do this is that log is monotone increasing so $log(f(x))$ is maximized at the same $x$ as $f(x)$.
The product is indexed over the countable list of your random sample, as indicated by the superscript $$n$$ over the Pi operator.