I thought that it might be instructive to present a way forward that uses a regularization of the Dirac Delta. To that end we proceed.
PRELIMINARIES:
Let $\displaystyle \delta_L(k)=\frac1{2\pi}\int_{-L}^Le^{ikx}\,dx$. Then, we can write
$$\delta_L(k)=\begin{cases}\frac{\sin(kL)}{\pi k}&,k\ne0\\\\\frac L\pi&,k=0\tag1\end{cases}$$
The function $\delta_L(k)$ has the following properties:
- For each $L$, $\delta_L(k)$ is an analytic function of $k$.
- $\lim_{L\to \infty} \delta_L(0)= \infty$
- $\left|\int_{-\infty}^x \delta_L(k')\,dk'\right|$ is uniformly bounded.
- $\lim_{L\to \infty}\int_{-\infty}^{k}\delta_L(k')\,dk'=u(k)$, where $u$
is the unit step funciton.
- For each $L>0$, $\int_{-\infty}^\infty \delta_L(k)\,dk=1$
While heuristically $\delta_L(k)$ "approximates" a Dirac Delta when $L$ is "large," the limit of $\delta_L(k)$ fails to exist. However, if we interpret this limit in the distributional sense, then $\lim_{L\to\infty}\delta_L(k)\sim\delta(k)$. We will now show that this is indeed that case.
ANALYSIS:
Let $\phi(k)\in S$ where $S$ is the Schwarz Space of functions.
We will now evaluate the limit
$$\begin{align}
\lim_{L\to \infty}\int_{-\infty}^\infty \delta_L(k)\phi(k)\,dk=\lim_{L\to \infty}\int_{-\infty}^\infty \frac{\sin(kL)}{\pi k}\phi(k)\,dk\tag1
\end{align}$$
Integrating by parts the integral on the right-hand side of $(1)$ with $u=\phi(k)$ and $v=\int_{-\infty}^{kL}\frac{\sin(x)}{\pi x}\,dx$ reveals
$$\begin{align}
\lim_{L\to \infty}\int_{-\infty}^\infty \delta_L(k)\phi(k)\,dk&=-\lim_{L\to \infty}\int_{-\infty}^\infty \phi'(k)\int_{-\infty}^{kL}\frac{\sin(x)}{\pi x}\,dx\,dk\tag2
\end{align}$$
Using Property 3 in the Preliminaries section, there exists a number $C$ such that $\left|\phi'(k)\int_{-\infty}^{kL}\frac{\sin(x)}{x}\,dx\right|\le C\,|\phi'(k)|$. Inasmuch as $C|\phi'(k)|$ is integrable, the Dominated Convergence Theorem guarantees that
$$\begin{align}
\lim_{L\to \infty}\int_{-\infty}^\infty \delta_L(k)\phi(k)\,dk&=-\lim_{L\to \infty}\int_{-\infty}^\infty \phi'(k)\int_{-\infty}^{kL}\frac{\sin(x)}{\pi x}\,dx\,dk\tag3\\\\
&=-\int_{-\infty}^\infty \phi'(k)\lim_{L\to \infty}\left(\int_{-\infty}^{kL}\frac{\sin(x)}{\pi x}\,dx\right)\,dk\\\\
&=- \int_{-\infty}^\infty \phi'(k)\underbrace{u(k)}_{\text{Unit Step}}\,dx\\\\
&=-\int_0^\infty \phi'(k)\,dk\\\\
&=\phi(0)
\end{align}$$
Therefore, in the sense of distributions as given by $(3)$, we assert that $\lim_{L\to\infty}\delta_L(k)\sim \delta(k)$ whereby rescaling yields the distributional limit
$$\bbox[5px,border:2px solid #C0A000]{\lim_{L\to \infty}\int_{-L}^Le^{ikx}\,dx\sim 2\pi \delta(k)}$$
as was to be shown!