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The value of the Dirichlet integral $\int_{0}^{\infty} \frac{\sin x}{x} dx$ is $\frac{\pi}{2}$. The video also states that the squared form $\int_{0}^{\infty} \left(\frac{\sin x}{x}\right)^2 dx$ yields the same result, $\frac{\pi}{2}$.

Conditions: The integral is an improper Riemann integral (or Lebesgue integral).; The lower limit is 0 and the upper limit is $\infty$.; The integrand is $\frac{\sin x}{x}$.

The value of the Dirichlet integral $\int_{0}^{\infty} \frac{\sin x}{x} dx$ is $\frac{\pi}{2}$. The video also states that the squared form $\int_{0}^{\infty} \left(\frac{\sin x}{x}\right)^2 dx$ yields the same result, $\frac{\pi}{2}$.

Conditions: The integral is an improper Riemann integral (or Lebesgue integral).; The lower limit is 0 and the upper limit is $\infty$.; The integrand is $\frac{\sin x}{x}$.

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 $\pi$.

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.

Substituting $t = 0$ into the inverse Fourier transform formula simplifies the exponential term $e^{i\omega t}$ to 1, removing the oscillatory component from the integrand. This transforms the inverse transform equation into a direct relationship between the value of the original function at the origin, $f(0)$, and the integral of its Fourier transform $F(\omega)$ over the entire real line.

Conditions: The inverse Fourier transform formula is $f(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) e^{i\omega t} d\omega$.; The function $f(t)$ is the rectangular pulse with $f(0)=1$.; $F(\omega)$ is the previously computed Fourier transform $2 \frac{\sin \omega}{\omega}$.

Substituting $t = 0$ into the inverse Fourier transform formula simplifies the exponential term $e^{i\omega t}$ to 1, removing the oscillatory component from the integrand. This transforms the inverse transform equation into a direct relationship between the value of the original function at the origin, $f(0)$, and the integral of its Fourier transform $F(\omega)$ over the entire real line.

Conditions: The inverse Fourier transform formula is $f(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) e^{i\omega t} d\omega$.; The function $f(t)$ is the rectangular pulse with $f(0)=1$.; $F(\omega)$ is the previously computed Fourier transform $2 \frac{\sin \omega}{\omega}$.