Improper integral with $-\infty$ and $+\infty$ as bounds Let $f : \mathbb{R} \to \mathbb{R}$ be a continuous function. We say that the improper integral
$$
\int_{-\infty}^{+\infty} f \,\ dx
$$
converges if the improper integrals
$$
\int_{-\infty}^{0} f \,\ dx \text{ and } \int_{0}^{+\infty} f \,\ dx
$$
both converge and in that case we define
$$
\int_{-\infty}^{+\infty} f \,\ dx := \int_{-\infty}^{0} f \,\ dx + \int_{0}^{+\infty} f \,\ dx.
$$
I would like to know : when $\int_{-\infty}^{+\infty} f \,\ dx$ is defined, is the Cauchy principal value
$$
\lim_{a \to \infty} \int_{-a}^{a} f \,\ dx
$$
equal to $\int_{-\infty}^{+\infty} f \,\ dx$ ?
My guess : I think it is. In that case, haven't we
\begin{align*}
\int_{-a}^{a} f \,\ dx - \int_{-\infty}^{\infty} f \,\ dx = - \int_{a}^{\infty} f \,\ dx - \int_{-\infty}^{-a} f \,\ dx \to 0 + 0 = 0 \quad (a \to \infty) \space ?
\end{align*}
 A: Convergence of $\int_{-\infty}^\infty f(x)\,dx$ implies $\text{PV}\int_{-\infty}^\infty f(x)\,dx$ exists by a modification of the argument you give (let the relevant limits happen independently of one another):
\begin{align*}
&PV\int_{-\infty}^\infty f(x)\,dx-\int_{-\infty}^\infty f(x)\,dx\\
%&=\lim_{a\to\infty}\int_{-a}^{a} f(x)\,dx - \left(\lim_{b\to \infty}\int_{-b}^{0} %f(x)\,dx+\lim_{c\to\infty}\int_0^c f(x)\,dx\right)\\
&=\lim_{a\to \infty} \left(\int_{-a}^{0} f(x)\,dx+\int_0^a f(x)\,dx\right)-\left(\lim_{b\to \infty}\int_{-b}^{0} f(x)\,dx+\lim_{c\to\infty}\int_0^c f(x)\,dx\right)\\
&=\lim_{a\to\infty}\lim_{b\to\infty}\lim_{c\to\infty}\left(\int_{-a}^{-b} f(x)\,dx+\int_c^a f(x)\,dx\right)\\
&=0.
\end{align*}
The hypothesis that the improper integral converges says that the pieces in the second line above are finite and hence sidesteps the issue of doing arithmetic with indeterminate forms (e.g., $\infty-\infty$) above.
The converse is of the statement above is of course false: the PV can exist without the improper integral converging (e.g., take $f(x)=x$).
Another condition is: if $f(x)$ is an even function, then the value of the improper integral equals the Cauchy principal value of the integral.
A: Yes, you are correct. And, as you realize, the converse is, in general, false.
