注意：$\displaystyle\lim_{a\to\infty}\int_{-a}^{a} f(x)\; dx$

Recall that to evaluate improper integrals, we first break the integral up so it is improper for only one reason. So, for example, we write $$\int_{-1}^1\frac{1}{x}=\int_{-1}^0\frac{1}{x}+\int_0^{1}\frac{1}{x}$$ But we can evaluate the improper integrals on the RHS (which are now improper for only a single reason), and they diverge. Thus $\int_{-1}^1\frac{1}{x}$ diverges, even though it ``seems'' like it the integral should be $0$ by symmetry. Beware of this issue! Similarly, $\int_{-\infty}^\infty\frac{1}{x}$ is also undefined.

The idea of symmetry comes up in Cauchy Principal Values, however. This is a way to give a value to improper integrals like the ones here. We write $$\operatorname{CPV}\int_{-1}^1\frac{1}{x}=\lim_{\varepsilon\to0^+}\left(\int_{-1}^{-\varepsilon}\frac{1}{x}+\int_{\varepsilon}^1\frac{1}{x}\right)=0$$ This is, however, not the value of the improper integral, which we showed earlier is undefined. It's just a way to assign a value to the expression, which can sometimes be useful.

We also have to be a little careful when taking limits like above. Otherwise, we can get weird things like $$\lim_{\varepsilon\to0^+}\left(\int_{-1}^{-\varepsilon}\frac{1}{x}+\int_{2\varepsilon}^1\frac{1}{x}\right)=\lim_{\varepsilon\to0^+}(\log(\varepsilon)-\log(2\varepsilon))=-\log 2.$$

The Cauchy principal value limit $\lim_{a\to\infty}\int_{-a}^{a} f(x)\,dx$ can exist even if the improper integral $\int_{-\infty}^{\infty}f(x)\,dx$ diverges. For instance, take an odd function, like $f(x) = x$; then $\int_{-a}^{a} f(x)\,dx = 0$ for all $a$, so $\int_{-a}^{a} f(x)\,dx \to 0$ as $a \to \infty$, even though the doubly-improper integral $\int_{-\infty}^{\infty}x\,dx$ does not converge.