Meaning of principal value integrals

Question

Consider the integral

$I = [PV]\int_{-\infty}^{\infty} \frac{exp(iax)}{x} dx$

where $a$ real and positive, and $[PV] $ denotes 'the principal value of'.

Using a semicirle contour in the upper half plane one can show

$I = [PV]\int_{-\infty}^{\infty} \frac{cos(ax)}{x} dx + i\int_{-\infty}^{\infty} \frac{sin(ax)}{x} dx = i\pi$

I am ok with how all these work, but now I have a rather naive question - does that mean we cannot find the integral

$J = \int_{-\infty}^{\infty} \frac{exp(iax)}{x} dx$,

that it does not exist at all?

In general, $[PV]\int_{-\infty}^{\infty} f(x) dx \neq \int_{-\infty}^{\infty} f(x) dx$, is that right? And surely they cannot both exist?

Answer 1

The integral defining $J$ has no meaning unless you specify how you integrate around $0$ : for instance the limit $$ \lim_{\epsilon\to 0} \int_{-\infty}^{-\epsilon}\frac{e^{iax}}{x}dx + \int_{2\epsilon}^{\infty}\frac{e^{iax}}{x}dx $$ will not converge to the same value as $$ I = \lim_{\epsilon\to 0} \int_{-\infty}^{-\epsilon}\frac{e^{iax}}{x}dx + \int_{\epsilon}^{\infty}\frac{e^{iax}}{x}dx .$$

Answer 2

It is true that $\int_{-\infty}^\infty \frac{e^{iax}}x dx$ doesn't exist -- for small enough $\varepsilon$ the real part of $\int_0^\varepsilon \frac{e^{iax}}x dx $ is infinity, and then certainly the integral cannot exist over a larger interval (such as the entire real line).

But you cannot reason "the principal value exists; therefore the integral itself doesn't". On the contrary, when the integral itself exists -- such as for $\int_{-\infty}^\infty \frac{\sin x}{x} dx$, then the principal value also exists and equals the ordinary value of the integral.