# Fourier Transform of 1

## Theorem

Let:

$\map f x = 1$

Then:

$\map {\hat f} s = \map \delta s$

where $\map {\hat f} s$ is the Fourier transform of $\map f x$.

## Proof

By the definition of a Fourier transform:

 $\ds \map {\hat f} s$ $=$ $\ds \int_{-\infty}^\infty e^{-2 \pi i x s} \map f x \rd x$ $\ds$ $=$ $\ds \int_{-\infty}^\infty e^{-2 \pi i x s} 1 \rd x$ $\ds$ $=$ $\ds \int_{-\infty}^\infty \paren {\map \cos {2 \pi s x} - i \map \sin {2 \pi s x} } \rd x$ Euler's Formula/Corollary $\ds$ $=$ $\ds \int_{-\infty}^\infty \map \cos {2 \pi s x} \rd x - i \int_{-\infty}^\infty \map \sin {2 \pi s x } \rd x$ Linear Combination of Definite Integrals $\ds$ $=$ $\ds \dfrac 1 {2 \pi s} \paren {\paren {\lim_{\gamma \mathop \to +\infty} \bigintlimits {\map \sin {2 \pi s x } } {-\gamma} \gamma} - i \paren {\lim_{\gamma \mathop \to +\infty} \bigintlimits {-\map \cos {2 \pi s x } } {-\gamma} \gamma} }$ Primitive of Sine Function and Primitive of Cosine Function $\ds$ $=$ $\ds \dfrac 1 {2 \pi s} \paren {\paren {\lim_{\gamma \mathop \to +\infty} \bigintlimits {\map \sin {2 \pi s x } } {-\gamma} \gamma} + i \paren {\lim_{\gamma \mathop \to +\infty} \bigintlimits {\map \cos {2 \pi s x } } {-\gamma} \gamma} }$ $\ds$ $=$ $\ds \dfrac 1 {2 \pi s} \paren {\paren {\lim_{\gamma \mathop \to +\infty} \bigintlimits {\map \sin {2 \pi s x } } {-\gamma} \gamma} + 0 }$ Cosine of Conjugate Angle: $\map \cos {-x} = \map {\cos} x$ $\ds$ $=$ $\ds \dfrac 1 {2 \pi s} \lim_{\gamma \mathop \to +\infty} 2 \map \sin {2 \pi s \gamma}$ Sine of Conjugate Angle: $\map \sin {-x} = -\map \sin x$ $\ds$ $=$ $\ds \dfrac 1 {\pi s} \lim_{\epsilon \mathop \to 0} \map \sin {\frac {2 \pi s} \epsilon}$ Let $\epsilon = \dfrac 1 \gamma$ $\ds$ $=$ $\ds \map \delta s$ Definition of Dirac Delta Function: Limit 5

$\blacksquare$