Problem computing a Cauchy principle value

Question

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 $-i\pi\text{ sgn }\xi$

Answer 1

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, $$ \mbox{P.V.}\int_{-\infty}^{\infty}\frac{e^{ix\xi}}{\xi}d\xi = i\pi\cdot\mbox{sgn}(x),\;\; x\ne 0. $$

Answer 2

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 $i\pi\text { sgn } (x).$