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Prove $\lim_{t\to\infty} \sin(tx) \text{P.V.}\frac{1}{x} = \pi \delta$ in the distributional sense.

This was the work I had put into the question:

For $\phi \in \mathcal{D}(\mathbb{R})$, \begin{align*} \langle \sin(tx) \text{P.V.}\frac{1}{x}, \phi \rangle &= \langle \text{P.V.}\frac{1}{x}, \sin(tx) \phi \rangle\\ &= \lim_{\epsilon \to 0^+} \int_{|x| \geq \epsilon} \frac{\sin(tx)}{x} \phi(x) dx. \end{align*}

We know that for any interval $[-R,R]$, the DCT applies and \begin{align*} \lim_{t\to \infty} \int_{-R}^R \frac{\sin(y)}{y}\phi(y/t) dy &= \int_{-R}^R \lim_{t\to\infty} \frac{\sin(y)}{y} \phi(y/t) dy \\ &= \phi(0) \int_{-R}^R \frac{\sin(y)}{y} dy. \end{align*}

Thus, we need only consider the tails, i.e. we need to show that $\lim_{t\to \infty} \int_R^\infty \frac{\sin(y)}{y}[\phi(y/t) - \phi(0)] dy$ and $\lim_{t\to\infty} \int_{-\infty}^{-R} \frac{\sin(y)}{y}[\phi(y/t)-\phi(0)]dy$ go to $0$ as $R$ increases. So \begin{align*} \lim_{t\to \infty} \int_R^\infty \frac{\sin(y)}{y} [\phi(y/t)-\phi(0)] dy = \lim_{t\to \infty} \int_{R/t}^\infty \frac{\sin(tx)}{x} [\phi(x) - \phi(0)] dx. \end{align*}

The problem with this though is the fact that with u-substitution the limits of integration become impossible to get a grip on, hence we cannot utilize the first result. You can approach this problem in various ways using these same techniques and ideas, but in the end you land back with the same dilemma of being unable to properly control the limits of integration. Any help would be appreciated.

mathishard.butweloveit
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  • I think at some point my professor found the solution in a book, but I don't have the book (thought I did) and I don't remember what it was. So if you have a book reference + solution, that would be great. – mathishard.butweloveit Jul 29 '18 at 21:11

1 Answers1

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Let $\phi_e(x)=\frac12(\phi(x)+\phi(-x))$ denote the even part of $\phi(x)$. Then, recognizing that $\phi(0)=\phi_e(0)$ we can write

$$\begin{align} \text{PV}\int_{-\infty}^\infty \frac{\sin(tx)}{x}\,\phi(x)\,dx&=\pi\phi(0)+2\int_0^\infty \frac{\sin(tx)}{x}(\phi_e(x)-\phi_e(0))\,dx\\\\ \tag1 \end{align}$$

Inasmuch as $\phi(x)$ is a suitable test function, then for $x\sim0$, $\phi_e(x)-\phi_e(0)=O(x^2)$ and $\phi_e'(x)=O(x)$. Moreover, $\phi(x)$ has compact support.

Hence, upon integrating by parts the integral on the right-hand side of $(1)$ with $u=\displaystyle \frac{\phi_e(x)-\phi_e(0)}{x}$ and $\displaystyle v=-\frac{\cos(tx)}{t}$, we find that

$$\begin{align} \int_0^\infty \left(\frac{\phi_e(x)-\phi_e(0)}{x}\right)\,\sin(tx)\,dx&=\frac1t \int_0^\infty \left(\frac{\phi_e(x)-\phi_e(0)}{x^2}\right)\,\cos(tx)\,dx\\\\ &-\frac1t \int_0^\infty \left(\frac{\phi'_e(x)}{x}\right)\,\cos(tx)\,dx\tag2 \end{align}$$

Both integrals on the right-hand side of $(2)$ converge, owing to the aforementioned small argument behavior and compact support of $\phi_e(x)$. Therefore, letting $t\to \infty$ reveals

$$\lim_{t\to\infty}\int_0^\infty \frac{\sin(tx)}{x}(\phi_e(x)-\phi_e(0))\,dx=0\tag3$$

Finally, substituting $(3)$ into $(1)$ we find that

$$\lim_{t\to\infty}\text{PV}\int_{-\infty}^\infty \frac{\sin(tx)}{x}\,\phi(x)\,dx=\pi\phi(0)$$

as was to be shown!

Mark Viola
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  • That's a crazy neat argument, very clever trick identifying the even portion. Is there anything in particular that made you think of this particular trick? Also out of curiosity, do you think there might be another way to approach this problem without defining $\phi_e$? – mathishard.butweloveit Jul 29 '18 at 21:54
  • I'd speculate that there is a "slicker" way to go that streamlines the analysis. I was motivated by the fact that the kernel $\frac{\sin(tx)}{x}$ is even in $x$ and therefore the PV renders the integral over the odd part of $\phi$ to be $0$. That left only the even part, which has the small argument properties I exploited upon integrating by parts. – Mark Viola Jul 29 '18 at 21:59
  • Interesting, yes I am not as familiar with thinking about things like this. Thank you. – mathishard.butweloveit Jul 29 '18 at 22:01
  • You're quite welcome. It was my pleasure. – Mark Viola Jul 29 '18 at 22:03