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

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!