fourier transform - Using Plancherel's theorem on delta function

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?