How to prove that inverse Fourier transform of "1" is delta funstion? $\mathscr{F}\{\delta(t)\}=1$, 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.
In the following $\langle f, \cdot \rangle$ denotes the linear functional on Schwartz space induced by $f$ and $f^\lor$ stands for the inverse Fourier transform of $f$. By definition, for any Schwartz function $\varphi$
\begin{align*}
\langle 1^\lor, \varphi \rangle=\langle 1, \varphi^\lor \rangle=&amp;\int_\mathbb{R} \left(\int_\mathbb{R} e^{2\pi ixy}\varphi(y) dy\right)dx
=\lim_{M\to\infty}\int_{-M}^M \left(\int_\mathbb{R} e^{2\pi ixy}\varphi(y) dy\right)dx.
\end{align*}
By Fubini's theorem we have
\begin{align*}
\int_{-M}^M \left(\int_\mathbb{R} e^{2\pi ixy}\varphi(y) dy\right)dx=&amp;\int_\mathbb{R} \varphi(y)\left(\int_{-M}^M e^{2\pi ixy}dx\right) dy
=\pi^{-1}\int_\mathbb{R} \varphi\left(y\right)\frac{\sin (2\pi My)}{y} dy.
\end{align*}
Since $\varphi$ is differentiable at $y=0$, we have $|\varphi(y)-\varphi(0)|\le C|y|$ for some constant $C$. Thus
$\left(\varphi(y)-\varphi(0)\right)/y\in L^1_{loc}(\mathbb{R}).$
Then by Riemann-Lebesgue Lemma we have
\begin{align*}
\lim_{M\to\infty}\int_{-1}^1 \left(\varphi(y)-\varphi(0)\right)\frac{\sin (2\pi My)}{y} dy=0,
\end{align*}
which means
\begin{align*}
\lim_{M\to\infty}\int_{-1}^1 \varphi\left(y\right)\frac{\sin (2\pi My)}{y} dy=&amp;\varphi(0)\lim_{M\to\infty}\int_{-1}^1\frac{\sin (2\pi My)}{y} dy
\\
=&amp;\varphi(0) \int_{-\infty}^\infty \frac{\sin y}{y}dy=\pi\varphi(0).
\end{align*}
Note that $\varphi\left(y\right)/y$ is integrable on $\mathbb{R}\setminus [-1,1]$. Thus by Riemann-Lebesgue Lemma we have
\begin{align*}
\lim_{M\to\infty}\int_{1}^\infty \varphi\left(y\right)\frac{\sin (2\pi My)}{y} dy=\lim_{M\to\infty}\int_{-\infty}^{-1} \varphi\left(y\right)\frac{\sin (2\pi My)}{y} dy=0.
\end{align*}
To sum up the above argument we have for all Schwartz functions $\varphi$,
\begin{align*}
\langle 1^\lor, \varphi \rangle=\pi^{-1}\lim_{M\to\infty}\int_\mathbb{R} \varphi\left(y\right)\frac{\sin (2\pi My)}{y} dy=\varphi(0)=\langle \delta, \varphi \rangle.
\end{align*}
Therefore $1^\lor=\delta$.
$$\mathscr{F^{-1}}\{1\}=\int_{-\infty}^{\infty}e^{2\pi ixy}dy=\lim_{M\to\infty}\frac{\sin{2\pi Mx}}{\pi x}$$
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

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