How to prove that inverse Fourier transform of "1" is delta function?

Question

$\mathscr{F}\{\delta(t)\}=1$, so this means inverse fourier transform of 1 is dirac delta function so I tried to prove it by solving the integral but I got something which doesn't converge.

Answer 1

$$\mathscr{F^{-1}}\{1\}=\int_{-\infty}^{\infty}e^{2\pi ixy}dy=\lim_{M\to\infty}\frac{\sin{2\pi Mx}}{\pi x}$$ Now we need to consider 2 cases:

1) $x=0$, then $\lim_{M\to\infty}\frac{\sin{2\pi Mx}}{\pi x}=\infty$

2) $x\ne0$, then $\lim_{M\to\infty}\frac{\sin{2\pi Mx}}{\pi x}=0$

Hence combining these 2 cases, we obtain

$$\mathscr{F^{-1}}\{1\}=\delta(x)$$