What would you like to understand?

The Fourier transform of the rectangular pulse function $f(t)$ is computed by substituting its piecewise definition into the integral definition of the Fourier transform. Because $f(t)$ is zero outside the interval $[-1, 1]$ and equals 1 inside, the infinite integral reduces to a finite integral over $[-1, 1]$.

Conditions: The function is defined as $f(t) = 1$ for $|t| \le 1$ and $0$ otherwise.; The Fourier transform is defined as $F(\omega) = \int_{-\infty}^{\infty} f(t) e^{-i\omega t} dt$.; $\omega$ is a real number.

The Fourier transform of the rectangular pulse function $f(t)$ is computed by substituting its piecewise definition into the integral definition of the Fourier transform. Because $f(t)$ is zero outside the interval $[-1, 1]$ and equals 1 inside, the infinite integral reduces to a finite integral over $[-1, 1]$.

Conditions: The function is defined as $f(t) = 1$ for $|t| \le 1$ and $0$ otherwise.; The Fourier transform is defined as $F(\omega) = \int_{-\infty}^{\infty} f(t) e^{-i\omega t} dt$.; $\omega$ is a real number.

The sine term is eliminated because the integrand $\sin(\omega t)$ is an odd function with respect to $t$, and the integration interval $[-1, 1]$ is symmetric about the origin. The definite integral of an odd function over a symmetric interval is always zero.

Conditions: The integration interval is $[-1, 1]$.; The function being integrated includes $\sin(\omega t)$.; $\omega$ is a constant with respect to the integration variable $t$.

The sine term is eliminated because the integrand $\sin(\omega t)$ is an odd function with respect to $t$, and the integration interval $[-1, 1]$ is symmetric about the origin. The definite integral of an odd function over a symmetric interval is always zero.

Conditions: The integration interval is $[-1, 1]$.; The function being integrated includes $\sin(\omega t)$.; $\omega$ is a constant with respect to the integration variable $t$.