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After substituting $t=0$ and $F(\omega) = 2 \frac{\sin \omega}{\omega}$ into the inverse Fourier transform formula, the equation becomes $1 = \frac{1}{2\pi} \int_{-\infty}^{\infty} 2 \frac{\sin \omega}{\omega} d\omega$. The constant 2 in the numerator of the integrand cancels with the constant 2 in the denominator of the prefactor $\frac{1}{2\pi}$.

Conditions: The inverse Fourier transform formula is used: $f(0) = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) d\omega$.; $f(0) = 1$.; $F(\omega) = 2 \frac{\sin \omega}{\omega}$.

After substituting $t=0$ and $F(\omega) = 2 \frac{\sin \omega}{\omega}$ into the inverse Fourier transform formula, the equation becomes $1 = \frac{1}{2\pi} \int_{-\infty}^{\infty} 2 \frac{\sin \omega}{\omega} d\omega$. The constant 2 in the numerator of the integrand cancels with the constant 2 in the denominator of the prefactor $\frac{1}{2\pi}$.

Conditions: The inverse Fourier transform formula is used: $f(0) = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) d\omega$.; $f(0) = 1$.; $F(\omega) = 2 \frac{\sin \omega}{\omega}$.