The Fourier transform of any distribution is defined to satisfy the self-adjoint property with any function from the Schwartz's class, $\mathscr{S}$ i.e. if $\delta$ is the Dirac Delta distribution and $f \in \mathscr{S}$, we have $$\langle\delta, \tilde{f} \rangle = \langle\tilde{\delta}, f \rangle $$ where $\tilde{g}$ denotes the Fourier transform of $g$ and $$\langle h,k \rangle = \int_{-\infty}^{\infty} h(y) k(x-y) dy$$ for $h,k \in \mathscr{S}$ and $\langle\delta, \tilde{f} \rangle = \tilde{f}(0)$. We have that $$\int_{-\infty}^{\infty} f(x) \exp(2 \pi i \xi x) dx = \tilde{f}(\xi)$$
Hence, $$\tilde{f}(0) = \int_{-\infty}^{\infty} f(x) dx$$ Since $$\langle\delta, \tilde{f} \rangle = \langle\tilde{\delta}, f \rangle $$ we get that
$$\langle 1,f \rangle = \int_{-\infty}^{\infty} f(x) dx = \langle \tilde{\delta},f \rangle$$ for all $f \in \mathscr{S}$. Hence, $\tilde{\delta} = 1$.