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The function $g(\omega) = \frac{\sin \omega}{\omega}$ is an even function because $\sin(-\omega) = -\sin(\omega)$ and $-\omega$ in the denominator cancels the negative sign, resulting in $g(-\omega) = g(\omega)$. For any even function, the integral over the symmetric interval $(-\infty, \infty)$ is equal to twice the integral over the half-interval $(0, \infty)$.

Conditions: The integrand is $\frac{\sin \omega}{\omega}$.; The integration interval is symmetric about the origin.; The value of the full-axis integral is known to be $\pi$.

The function $g(\omega) = \frac{\sin \omega}{\omega}$ is an even function because $\sin(-\omega) = -\sin(\omega)$ and $-\omega$ in the denominator cancels the negative sign, resulting in $g(-\omega) = g(\omega)$. For any even function, the integral over the symmetric interval $(-\infty, \infty)$ is equal to twice the integral over the half-interval $(0, \infty)$.

Conditions: The integrand is $\frac{\sin \omega}{\omega}$.; The integration interval is symmetric about the origin.; The value of the full-axis integral is known to be $\pi$.