The Fourier Integral Theorem

So far we have looked at expressing functions - particularly $2\pi$-periodic functions, in terms of their Fourier series. Of course, not all functions are $2\pi$-periodic and it may be impossible to represent a function defined on, say, all of $\mathbb{R}$ by a Fourier series. The next best alternativ would be representing such functions as an integral. Fortunately, under suitable conditions, a function $f \in L(\mathbb{R})$ can be represented as an integral as the following theorem known as the Fourier Integral Theorem states.

Theorem 1 (The Fourier Integral Theorem): If $f \in L(\mathbb{R})$ and if there exists an $x \in \mathbb{R}$ and a $\delta > 0$ such that either: 1) $f$ is of bounded variation on $[x - \delta, x + \delta]$. 2) $f(x+)$ and $f(x-)$ both exist, and $\displaystyle{\int_{0}^{\delta} \frac{f(x + t) - f(x+)}{t} \: dt}$ and $\displaystyle{\int_0^{\delta} \frac{f(x + t) - f(x-)}{t} \: dt}$. Then $\displaystyle{\frac{f(x+) + f(x-)}{2} = \frac{1}{\pi} \int_0^{\infty} \left [ \int_{-\infty}^{\infty} f(u) \cos v (u - x) \: du \right ] \: dv}$ where $\displaystyle{\int_0^{\infty} ... \: dv}$ is an improper Riemann integral.

Before we look at any examples regarding the Fourier integral theorem we will first establish the following two equalities for even and off functions satisfying the Fourier integral theorem:

Theorem 2: If $f \in L(\mathbb{R})$ satisfies either of the conditions of the Fourier integral theorem then: a) If $f$ is an even function then $\displaystyle{\frac{f(x+) + f(x-)}{2} = \frac{2}{\pi} \int_0^{\infty} \cos vx \left [ \int_0^{\infty} f(u) \cos vu \: du \right ] \: dv}$. b) If $f$ is an odd function then $\displaystyle{\frac{f(x+) + f(x-)}{2} = \frac{2}{\pi} \int_0^{\infty} \sin vx \left [ \int_0^{\infty} f(u) \sin vu \: du \right ] \: dv}$.

• Proof of a) Suppose that $f \in L(\mathbb{R})$ satisfies either of the conditions of the Fourier integral theorem. Then:

(1)

\begin{align} \quad \frac{f(x+) + f(x-)}{2} &= \frac{1}{\pi} \int_0^{\infty} \left [ \int_{-\infty}^{\infty} f(u) \cos v(u - x) \: du \right ] \: dv \\ &= \frac{1}{\pi} \int_0^{\infty} \left [ \int_{-\infty}^{\infty} f(u)[\cos vu \cos vx + \sin vu \sin vx \right ] \: dv \\ &= \frac{1}{\pi} \int_0^{\infty} \left [ \int_{-\infty}^{\infty} f(u) \cos vu \cos vx \: du + \int_{-\infty}^{\infty} f(u) \sin vu \sin vx \: du \right] \: dv \end{align}

• Since $f$ is an even function of $u$ and $\sin vu$ is an odd function of $u$ we have that $f(u) \sin vu$ is an odd function of $u$ and so on the symmetric interval $(-\infty, \infty)$, $\displaystyle{\int_{-\infty}^{\infty} f(u) \sin vu \sin vx \: du = 0}$ so:

(2)

\begin{align} \quad \frac{f(x+) + f(x-)}{2} &= \frac{1}{\pi} \int_0^{\infty} \left [ \int_{-\infty}^{\infty} f(u) \cos vu \cos vx \: du \right ] \: dv \\ &= \frac{1}{\pi} \int_0^{\infty} \cos vx \left[ \int_{-\infty}^{\infty} f(u) \cos vu \: du \right ] \: dv \end{align}

• Since $f$ is an even function of $u$ and $\cos vu$ is an even function of $u$ we have that $f(u) \cos vu$ is an even function of $u$ and so on the symmetric interval $(-\infty, \infty)$, $\displaystyle{\int_{-\infty}^{\infty} f(u) \cos vu \: du = 2 \int_0^{\infty} f(u) \cos vu \: du}$, so:

(3)

\begin{align} \quad \frac{f(x+) + f(x-)}{2} &= \frac{2}{\pi} \int_0^{\infty} \cos vx \left [ \int_0^{\infty} f(u) \cos vu \: du \right ] \: dv \quad \blacksquare \end{align}

• Proof of b) Suppose that $f \in L(\mathbb{R})$ satisfies either of the conditions of the Fourier integral theorem. Then once again we have that:

(4)

\begin{align} \quad \frac{f(x+) + f(x-)}{2} &= \frac{1}{\pi} \int_0^{\infty} \left [ \int_{-\infty}^{\infty} f(u) \cos vu \cos vx \: du + \int_{-\infty}^{\infty} f(u) \sin vu \sin vx \: du \right] \: dv \end{align}

• Since $f$ is an odd function of $u$ and $\cos vu$ is an even function of $u$ we have that $f(u) \cos vu$ is an odd function of $u$ and similarly, since $\sin vu$ is an odd function of $u$ we have that $f(u) \sin vu$ is an even function of $u$. So, like in the proof of (a) above:

(5)

\begin{align} \quad \frac{f(x+) + f(x-)}{2} &= \frac{2}{\pi} \int_0^{\infty} \sin vx \left [ \int_0^{\infty} f(u) \sin vu \: du \right ] \: dv \quad \blacksquare \end{align}