Does the principal value of $\int_{-\infty}^{\infty} e^{ix} dx$ vanish? The integral $\int_{-\infty}^{\infty} e^{ix} dx$ diverges. I have read (in a wikipedia article) that the principal value of this integral vanishes: $P \int_{-\infty}^{\infty} e^{ix} dx = 0$. How can one see that?
Thank you for your effort!

The integral can be regularized using the integral analogue of Cesàro summation.  By definition we have
$$
\begin{align}
\textrm{C} \int_{-\infty}^{\infty} e^{ix}\,dx &= \lim_{a \to \infty} \frac{1}{a} \int_0^a \int_{-y}^{y} e^{ix}\,dx\,dy \\
&= \lim_{a \to \infty} \frac{2}{a} \int_0^a \sin y\,dy \\
&= \lim_{a \to \infty} \frac{2}{a} (1 - \cos a) \\
&= 0.
\end{align}
$$
Perhaps this is what the article was referring to?
