# Even Impulse Pair is Fourier Transform of Cosine Function

## Theorem

Consider the (real) cosine function $\map \cos x: \R \to \R$.

$\map f x = \map \cos {\pi x}$

Then:

 $\ds \map {\hat f} s$ $=$ $\ds \dfrac 1 2 \map \delta {s - \dfrac 1 2} + \dfrac 1 2 \map \delta {s + \dfrac 1 2}$ $\ds$ $=$ $\ds \map {\operatorname {II} } s$

where:

$\map {\hat f} s$ is the Fourier transform of $\map f x$.
$\operatorname {II}$ denotes the even impulse pair function.

## Proof

By the definition of a Fourier transform:

 $\ds \map {\hat f} s$ $=$ $\ds \int_{-\infty}^\infty e^{-2 \pi i x s} \map f x \rd x$ $\ds$ $=$ $\ds \int_{-\infty}^\infty e^{-2 \pi i x s} \map \cos {\pi x} \rd x$ $\ds$ $=$ $\ds \int_{-\infty}^\infty e^{-2 \pi i x s} \dfrac 1 2 \paren {e^{i \pi x} + e^{-i \pi x} } \rd x$ Euler's Cosine Identity $\ds$ $=$ $\ds \dfrac 1 2 \int_{-\infty}^\infty e^{-2 \pi i x s} e^{i \pi x} \rd x + \dfrac 1 2 \int_{-\infty}^\infty e^{-2 \pi i x s} e^{-i \pi x} \rd x$ Linear Combination of Definite Integrals $\ds$ $=$ $\ds \dfrac 1 2 \int_{-\infty}^\infty e^{-2 \pi i x \paren {s - \frac 1 2 } } \rd x + \dfrac 1 2 \int_{-\infty}^\infty e^{-2 \pi i x \paren {s + \frac 1 2 } } \rd x$ $\ds$ $=$ $\ds \dfrac 1 2 \map \delta {s - \dfrac 1 2 } + \dfrac 1 2 \map \delta {s + \dfrac 1 2 }$ Fourier Transform of 1

$\blacksquare$