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