From the duality theorem if: $$ FT(x(t)) = X(w) $$ Then $$ FT(X(t)) = 2\pi x(-w) $$
If I choose $X(t) = 1$, then $X(w) = 1$ thus $x(t) = \delta(t)$.
Now plugging back into the second equation: $$ FT(1) = 2\pi \delta(-w) = 2\pi \delta(w) $$
Which is wrong, the general shape of the result is correct, indeed a constant in time domain will be an impulse at zero in frequency domain, but why is it scaled by $2\pi$? this $2\pi$ has to be removed somehow.
Now I usually use the duality theorem for finding inverse fourier transform, but I was trying to find a way to find fourier transform of $sinc(t)$ and I expect it to be a square pulse but I can't get the scaling right.
I am now taking a communications course and have successfully got to know the issue, simply speaking Fourier Transform is defined as
$$ G(f) = \int_{-\infty}^{\infty} g(t) e^{-j2\pi ft} \,dt $$
Which can be re-written in terms of $\omega$ as following
$$ G(\omega) = G(j\omega) = G(e^{j\omega}) = \int_{-\infty}^{\infty} g(t) e^{-j\omega t} \,dt $$
Now the Inverse-Fourier Transform is defined as
$$ g(t) = \int_{-\infty}^{\infty} G(f) e^{j2\pi ft} \,df $$
And from a quite simple observation by adjusting exponential to have a negative power and swapping variables $t$ and $f$
$$ g(-t) = \int_{-\infty}^{\infty} G(f) e^{-j2\pi ft} \,df $$
$$ g(-f) = \int_{-\infty}^{\infty} G(t) e^{-j2\pi ft} \,dt $$
This in a simple form means
$$ \text{If} ~ g(t)\Leftrightarrow G(f) ~ \text{, then} ~ G(t) \Leftrightarrow g(-f) $$
Now in my example I choose $G(t) = 1$ which by comparing to the following and plugging back into the theorem means
$$ g(t) = \delta (t) ~\text{then} ~ G(f) = 1 \\ G(t) = 1 ~ \text{then} ~ g(-f) = \delta (f) ~ \text{and since dirac function is even} ~ g(f) = \delta (f) $$
The mistake I had is that I was solving assuming the frequency version of Inverse-Fourier Transform.