Plancherel's Theorem states that for $f \in L^{2}(\mathbb{R})$ we have
$$f(x) = \frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty}F(k)e^{ikx}dk \Longleftrightarrow F(k) = \frac{1}{\sqrt{2 \pi}}\int^{\infty}_{\infty}f(x)e^{-ikx}dx.$$
If we consider $f(x) := \delta(x)$ (the delta function) then using this theorem it follows simply that $$\delta(x) = \frac{1}{2 \pi}\int_{-\infty}^{\infty}e^{ikx}dk.$$
Clearly then for $x = 0$ and $x \neq 0$ the integral is divergent. Also, apparently $\delta(x)$ is not square integrable, hence I'm not sure that we can even use Plancherel's Theorem. But having said that I understand that this result does hold. Is it incorrect to show that this result is true using Plancherel's Theorem?
Thanks.
Yes: it is incorrect to use Plancherel to prove this. Plancherel's theorem is stated as: "...the integral of a function..." but Dirac's delta is not a function, but a tempered distribution.
Note that the relation you are trying to prove can also be written as $\mathcal F[1]=\delta(x)$, where $\mathcal F$ means Fourier Transform. In physics we use this relation all the time, and it's been around for centuries, but the framework in which the relation can be proved was formalised around 1950, by Schwartz.
I really encourage you to read about tempered distributions, for example, in this paper I found online. The truth is: $\mathcal F[1]=\delta(x)$ is true, but it doesn't mean what you think: it is a statement on distributions, not on functions. The Fourier Transform is well defined for distributions, but not through Riemann integrals. Long story short: the integral you wrote is divergent for any value of $x$, but can we can make sense of it if we extend the meaning of integration and the meaning of the Fourier Transform.
The take home message is: the Dirac delta is not a function but a distribution and the integral you wrote is right but it is not a Riemann integral. The actual meaning of the expression you wrote is a bit subtle at first, but if you read about Schwartz distributions you'll see that it all makes sense and is very well-defined.