How does the constant 2 cancel out in the inverse Fourier transform equation?

Conditions

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

Reasoning, step by step

1. Start with the substituted equation: $1 = \frac{1}{2\pi} \int_{-\infty}^{\infty} 2 \frac{\sin \omega}{\omega} d\omega$.
2. Identify the constant factor 2 inside the integral and the constant factor 2 in the denominator of $\frac{1}{2\pi}$.
3. Move the constant 2 from the integral to the front: $1 = \frac{2}{2\pi} \int_{-\infty}^{\infty} \frac{\sin \omega}{\omega} d\omega$.
4. Simplify the fraction $\frac{2}{2\pi}$ to $\frac{1}{\pi}$: $1 = \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{\sin \omega}{\omega} d\omega$.
5. Multiply both sides by $\pi$ to isolate the integral: $\pi = \int_{-\infty}^{\infty} \frac{\sin \omega}{\omega} d\omega$.
6. Conclude that the cancellation of 2 is a direct algebraic simplification of the constants.

Example

The video shows the transition: '$1=\frac{1}{2\pi}\int_{-\infty}^{\infty}2\frac{\sin\omega}{\omega}$\,$d\omega$' to '$1=\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{\sin\omega}{\omega}$\,$d\omega$' and finally '$\pi=\int_{-\infty}^{\infty}\frac{\sin\omega}{\omega}$\,$d\omega$'.

Common misconceptions

• Thinking the 2 cancels with the $\pi$; it cancels with the 2 in $2\pi$.
• Forgetting that the constant can be pulled out of the linear integral.
• Assuming the integral itself changes value during this step; only the algebraic coefficient changes.

Watch the explanation

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