To illustrate, here is a textbook example for deriving the formula for the expectation of two independent random variables:
If a random variable $x_1$ has probability density $\rho_1(x_1)\mathrm{d}x_1$ and another random variable $x_2$ has probability density $\rho_2(x_2)\mathrm{d}x_2$. Then the joint probability density is the product of the two separate probability densities:
$$\rho(x_1,x_2)\mathrm{d}x_1\mathrm{d}x_2=\rho_1(x_1)\mathrm{d}x_1\rho_2(x_2)\mathrm{d}x_2$$
So the expectation is $$E(X)= \langle x_1+ x_2\rangle=\iint(x_1+x_2)\rho(x_1,x_2)\mathrm{d}x_1\mathrm{d}x_2$$ $$=\int x_1\rho_1(x_1)\mathrm{d}x_1\int \rho_2(x_2)\mathrm{d}x_2+\int \rho_1(x_1)\mathrm{d}x_1\int x_2 \rho_2(x_2)\mathrm{d}x_2$$ $$=\langle x_1\rangle+\langle x_2\rangle=E_1+E_2$$
Now, since by definition $$\int_{-\infty}^{\infty} x_1\rho_1(x_1)\mathrm{d}x_1=E_1$$ and $$\int_{-\infty}^{\infty} x_2\rho_2(x_2)\mathrm{d}x_2=E_2$$ and $$\int_{-\infty}^{\infty} \rho_1(x_1)\mathrm{d}x_1=\int_{-\infty}^{\infty} \rho_2(x_2)\mathrm{d}x_2=1$$
So it must be the case that $$\underbrace{\int x_1\rho_1(x_1)\mathrm{d}x_1}_{=E_1}\underbrace{\int \rho_2(x_2)\mathrm{d}x_2}_{=1}+\underbrace{\int \rho_1(x_1)\mathrm{d}x_1}_{=1} \underbrace{\int x_2 \rho_2(x_2)\mathrm{d}x_2}_{=E_2}$$
So why have the limits been omitted?
In response to the comments below:
Is the word expectation enough context to imply $\displaystyle\int_{-\infty}^{+\infty}$?
No. For a single absolutely continuous random variable, yes.
Consider the first expectation:
$$E(X)= \langle x_1+ x_2\rangle=\iint(x_1+x_2)\rho(x_1,x_2)\mathrm{d}x_1\mathrm{d}x_2$$
This is supposed to be:
$$E(X)= \langle x_1+ x_2\rangle=\iint_{\mathbb R^2}(x_1+x_2)\rho(x_1,x_2)\mathrm{d}x_1\mathrm{d}x_2$$
or
$$E(X)= \langle x_1+ x_2\rangle=\int_{\mathbb R}\int_{\mathbb R}(x_1+x_2)\rho(x_1,x_2)\mathrm{d}x_1\mathrm{d}x_2$$
Going back to the one variable case:
$$E[X] = \int_{\mathbb R} xf_X(x)dx$$
In the case where the expectation is not well-defined because of a certain subset $A \subseteq \mathbb R$, then
$$E[X] = \int_{\mathbb R} xf_X(x)dx = \int_{\mathbb R \setminus A} xf_X(x)dx,$$
provided the integral is well-defined.
If $$\int_{B} xf_X(x)dx = 0$$ for $B \subseteq \mathbb{R}$, then $$\int_{\mathbb{R}} xf_X(x)dx = \int_{\mathbb{R} \setminus B} xf_X(x)dx$$
For a single continuous random variable, the pdf may not exist.
For a single discrete random variable.
$$E[X] = \sum_{x \in Range(X)} xf_X(x)dx$$
If $Range(X) =\mathbb{R}$, $X$ is not discrete.