Here you can see that Cauchy's principle value ($PV$) of the complex exponential is zero, i.e.,
$$
PV\left[\int_{-\infty}^{+\infty} e^{iax}dx\right] = 0
\tag1 $$
I'm trying to prove this a follows:
$$
PV\left[\int_{-\infty}^{+\infty} e^{iax}dx\right] = \lim_{T\rightarrow +\infty}\int_{-T}^{+T} e^{iax}dx = \frac{1}{a} \lim_{T\rightarrow +\infty} \left[ \sin(T) – \sin(-T)\right] \tag2
$$
In principle, one would write $\sin(T) – \sin(-T) = 2\sin(T)$ and would realize that the limit does not exist so the only thing I can try to prove the result in Eq. (1) is to work out both sines separatedly and assume that
$$
\lim_{T\rightarrow +\infty} \sin(T) = \lim_{T\rightarrow +\infty}\sin(-T) \tag3
$$
so finally you arrive to Eq. (1). Nevertheless, this is not very convincing from my point of view because RHS limit is undetermined as well as LHS, so you should not assume Eq. (3) is right. Then, what's the right approach?
Best Answer
Note that
$$e^{iax}=\cos(ax)+i\sin(ax)$$
So, the integral is equal to
$$I=2\int_0^\infty \cos ax\, dx + i \int_0^\infty \sin ax\, dx - i \int_0^\infty \sin ax\, dx$$
All three integrals are Cesaro-summable:
$$\int_0^\infty \cos ax\, dx = 0$$
$$\int_0^\infty \sin ax\, dx = 1$$
So, we have
$$I=2\cdot0+i\cdot1-i\cdot1=0$$