Problem computing a Cauchy principle value For $x\in\mathbb{R}$, I compute
$$\begin{aligned}
\text{P.V. }\int_{\mathbb{R}}\frac{1}{\xi}e^{ix\xi}d\xi&=\lim_{\epsilon\to 0^{+}}\int_{\mathbb{R}\setminus[-\epsilon,\epsilon]}\frac{1}{\xi}e^{ix\xi}d\xi
\\
&=\lim_{\epsilon\to 0^{+}}\int_{-\infty}^{-\epsilon}+\int_{\epsilon}^{\infty}\frac{1}{\xi}e^{ix\xi}d\xi
\\
&=\lim_{\epsilon\to 0^{+}}\left(\frac{-i}{x\xi}e^{ix\xi}\bigg|_{\xi=-\infty}^{-\epsilon}-\frac{i}{x\xi}e^{ix\xi}\bigg|_{\xi=\epsilon}^{\infty}\right)
\\
&=-\frac{i}{x}\lim_{\epsilon\to 0^{+}}\frac{1}{\epsilon}(e^{ix\xi}+e^{-ix\xi}),
\end{aligned}$$
which just isn't going to work out.
I think the answer I should eventually get should be $-i\pi\text{ sgn }\xi$
 A: I'll change the notation a bit. We want to look at
$$\int_{|t|>h} \frac{\cos (xt) + i\sin (xt)}{t}\, dt$$
for small $h>0.$ First note that each such integral converges from Dirichlet's test. Using the evenness, oddness of the cosine, sine, we see the above equals
$$\int_{|t|>h} \frac{i\sin (xt)}{t}\, dt = 2i\int_{h}^\infty \frac{\sin (xt)}{t}\, dt.$$
Suppose $x>0.$ Let $t=s/x.$ Then the last integral equals
$$\int_{hx}^\infty \frac{\sin (s)}{s}\, ds.$$
The limit of this as $h\to 0^+$ is well known: It is $\pi/2.$
So your PV integral equals $i\pi$ if $x>0.$ Now check that if $x=0,$ the PV integral is $0,$ and if $x< 0,$ using the oddness of the sine again, you get $-i\pi.$ The total answer is then $i\pi\text { sgn } (x).$
A: From Fourier analysis:
$$
        \lim_{r\uparrow\infty}\int_{0}^{r}\frac{\sin(u)}{u}du= \int_{0}^{\infty}\frac{\sin(u)}{u}du=\frac{\pi}{2}.
$$
There is no issue concerning convergence near $u=0$, but the integral is not absolutely convergent near $u=\infty$, even though the improper integral exists. For $x \ne 0$, the following exists as an improper integral near $\infty$,
\begin{align}
    \int_{|\xi|\ge \epsilon}\frac{e^{ix\xi}}{\xi}d\xi
  & = \int_{-\infty}^{-\epsilon}\frac{e^{ix\xi}}{\xi}d\xi+\int_{\epsilon}^{\infty}\frac{e^{ix\xi}}{\xi}d\xi \\
  & = -\int_{\epsilon}^{\infty}\frac{e^{-ix\xi}}{\xi}d\xi+\int_{\epsilon}^{\infty}\frac{e^{ix\xi}}{\xi}d\xi \\
  & = \int_{\epsilon}^{\infty}\frac{e^{ix\xi}-e^{-ix\xi}}{\xi}d\xi \\
  & = 2i\int_{\epsilon}^{\infty}\frac{\sin(x\xi)}{\xi}d\xi.
\end{align}
If $x > 0$, let $\xi=u/x$ in order to obtain
$$
   \int_{|\xi|\ge \epsilon}\frac{e^{ix\xi}}{\xi}d\xi=2i\int_{\epsilon x}^{\infty}\frac{\sin(u)}{u}du \rightarrow i\pi \mbox{ as } \epsilon\downarrow 0.
$$
If $x < 0$, let $\xi=-u/x$ in order to obtain
$$
   \int_{|\xi|\ge \epsilon}\frac{e^{ix\xi}}{\xi}d\xi=-2i\int_{-\epsilon x}^{\infty}\frac{\sin(u)}{u}du \rightarrow -i\pi \mbox{ as } \epsilon\downarrow 0.
$$
In other words,
$$
          \mbox{P.V.}\int_{-\infty}^{\infty}\frac{e^{ix\xi}}{\xi}d\xi = i\pi\cdot\mbox{sgn}(x),\;\; x\ne 0.
$$
