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How does the evenness of sin⁡(w)/w\sin (w)/w allow converting the full-axis integral to the half-axis integral?

The function g(ω)=sin⁡ωωg(\omega) = \frac{\sin \omega}{\omega} is an even function because sin⁡(−ω)=−sin⁡(ω)\sin(-\omega) = -\sin(\omega) and −ω-\omega in the denominator cancels the negative sign, resulting in g(−ω)=g(ω)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,∞)(0, \infty). Therefore, once the full-axis integral ∫−∞∞sin⁡ωωdω\int_{-\infty}^{\infty} \frac{\sin \omega}{\omega} d\omega is determined to be π\pi, the half-axis integral ∫0∞sin⁡ωωdω\int_{0}^{\infty} \frac{\sin \omega}{\omega} d\omega is simply π2\frac{\pi}{2}.

Conditions

  • The integrand is sin⁡ωω\frac{\sin \omega}{\omega}.
  • The integration interval is symmetric about the origin.
  • The value of the full-axis integral is known to be π\pi.

Reasoning, step by step

  1. Verify the parity of the function h(ω)=sin⁡ωωh(\omega) = \frac{\sin \omega}{\omega}.
  2. Calculate h(−ω)=sin⁡(−ω)−ωh(-\omega) = \frac{\sin(-\omega)}{-\omega}.
  3. Use the odd property of sine: sin⁡(−ω)=−sin⁡(ω)\sin(-\omega) = -\sin(\omega).
  4. Substitute back: h(−ω)=−sin⁡(ω)−ω=sin⁡(ω)ω=h(ω)h(-\omega) = \frac{-\sin(\omega)}{-\omega} = \frac{\sin(\omega)}{\omega} = h(\omega).
  5. Conclude that h(ω)h(\omega) is an even function.
  6. Apply the property of integrals of even functions: ∫−∞∞h(ω)dω=2∫0∞h(ω)dω\int_{-\infty}^{\infty} h(\omega) d\omega = 2 \int_{0}^{\infty} h(\omega) d\omega.
  7. Substitute the known value of the full-axis integral: π=2∫0∞sin⁡ωωdω\pi = 2 \int_{0}^{\infty} \frac{\sin \omega}{\omega} d\omega.
  8. Solve for the half-axis integral: ∫0∞sin⁡ωωdω=π2\int_{0}^{\infty} \frac{\sin \omega}{\omega} d\omega = \frac{\pi}{2}.
  9. Identify this result as the Dirichlet integral.

Example

The video derives: '∫−∞∞sin⁡ωω\int_{-\infty}^{\infty}\frac{\sin\omega}{\omega}\,dω=πd\omega=\pi' and then states 'π2=∫0∞sin⁡ωωdω\frac{\pi}{2}=\int_0^{\infty}\frac{\sin\omega}{\omega}d\omega' based on the evenness of the integrand.

Common misconceptions

  • Confusing the parity of sin⁡ω\sin \omega (odd) with the parity of sin⁡ωω\frac{\sin \omega}{\omega} (even).
  • Assuming the integral from 0 to infinity is the same as from negative infinity to infinity; it is exactly half.
  • Forgetting that the singularity at ω=0\omega=0 is removable and does not affect the symmetry argument for the improper integral.

Watch the explanation

4:52 – 5:10Watch this moment ↗

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