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!
 A: 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?
A: You might want to consider the this as a the distribution:
$$P.V.\left(\int_{-\infty}^{\infty}e^{ixt}dx\right)_{t=1}$$
Now the distribution needs to be tested against test functions as:
$$\int_{-\infty}^{\infty}P.V.\left(\int_{-\infty}^{\infty}e^{ixt}dx\right)\phi(t)dt$$
Which is, because of the definition of Principal Value:
$$\int_{-\infty}^{\infty}\left[\lim_{N\to\infty}\frac{e^{ixt}}{it}\right]_{-N}^{N}\phi(t)dt=\int_{-\infty}^{\infty}\lim_{N\to\infty}\left(\frac{e^{iNt}-e^{-iNt}}{it}\right)\phi(t)dt$$
Now we change variables to $y=Nt$ so we have, $dt=dy/N$:
$$\int_{-\infty}^{\infty}\frac{2sen(y)}{y}\lim_{N\to\infty}\phi(y/N)dy$$
Finally, because $\phi$ is a test function, it is continuos:
$$\int_{-\infty}^{\infty}P.V.\left(\int_{-\infty}^{\infty}e^{ixt}dx\right)\phi(t)dt=2\pi\phi(0)$$
Which tells you that:
$$P.V.\left(\int_{-\infty}^{\infty}e^{ixt}dx\right)_{t=1}=\left(2\pi\delta(t)\right)_{t=1}=0$$
Now the Dirac Delta is a Distribution whose converges to $0$ at every point but zero.
