I know that the following theorem holds
Fourier's Integral Theorem. Suppose $f:\mathbb{R}\to\mathbb{R}$ is piece-wise continuous on $\mathbb{R}$ and at every point $x$ the right and left derivatives exist. Also, assume that $f$ satisfies $\int_{-\infty}^{+\infty}|f(t)|dt<+\infty$. Then the following holds
\begin{align*}
\frac{1}{2}\big(f(t^+)+f(t^-)\big)&=\int_{0}^{+\infty}(A(\omega)\cos(\omega t)+B(\omega)\sin(\omega t))dt \\
A(\omega)&=\frac{1}{\pi}\int_{-\infty}^{+\infty}f(t)\cos(\omega t)dt \\
B(\omega)&=\frac{1}{\pi}\int_{-\infty}^{+\infty}f(t)\sin(\omega t)dt
\end{align*}
Assume that $H(t)$ is the usual Heaviside or unit-step function. I saw in some text that the Fourier integral representation of
$$f(t):=H(t)-\frac{1}{2}=
\begin{cases}
+\frac{1}{2},\quad t\gt 0 \\
-\frac{1}{2},\quad t\lt 0
\end{cases}$$
can be given by
$$f(t)=\frac{1}{\pi}\int_{0}^{+\infty} \frac{\sin \omega t}{\omega} d\omega \tag{1}$$
however, the function $f(t)$ does not satisfy the condition $\int_{-\infty}^{+\infty}|f(t)|dt<+\infty$ of the Fourier integral theorem and when one tries to compute its Fourier integral representation will see that $A(\omega)=0$ but $B(\omega)=\lim_{R\to+\infty}\frac{1}{\pi\omega}\sin\omega R$ which does not exist! So, here are my questions
How can I derive $(1)$? I heard that this can be derived by using the generalized functions but I have no idea what this really means. Can someone explain the main idea of this for me? Please put it simple for someone with little knowledge of elementary analysis! 🙂
Best Answer
Note that
$$f(t):=H(t)-\frac{1}{2}= \begin{cases} +\frac{1}{2},\quad t\ge 0 \\ -\frac{1}{2},\quad t\lt 0 \end{cases}$$
is essentially $\dfrac{1}{2}\mathrm{sgn}(t)$.
The Inverse Fourier Transform, in the limit, of $\dfrac{1}{\omega}$ is (some constant times) $\mathrm{sgn}(t)$. Often, solving a(n Inverse) Fourier Transform in the limit does involve generalized functions, which are sequences of functions in a limit. However, this particular Inverse Fourier Transform can be solved with a contour integration in the complex plane.
Starting with an integral similar to your integral (1), but which looks more like a standard Inverse Fourier Transform:
$$\begin{align*}g(t)&=-\frac{1}{2\pi i}\int_{-\infty}^{+\infty} \frac{e^{i \omega t}}{\omega} d\omega \tag{A}\\ \end{align*}$$
This integral, like your integral (1), has a problem with the infinite discontinuity of $\dfrac{1}{\omega}$ at $\omega = 0$.
To deal with that infinite discontinuity, rewrite the integral as
$$g(t) = -\dfrac{1}{2\pi i}\lim_{\epsilon \rightarrow 0}\left(\int_{-\infty}^{-\epsilon} \frac{e^{i \omega t}}{\omega} d\omega + \int_{+\epsilon}^{+\infty} \frac{e^{i \omega t}}{\omega} d\omega \right)$$
To solve the above expression, use the following contour integral
$$\int_{C} \dfrac{e^{i z t}}{z} dz$$
where the contour $C$ is a semicircle of radius $R$ in the complex plane, with its diameter along the real axis ($\omega$), and with a small indentation of radius $\epsilon$ at the origin. The value of the contour integral is $0$, since there are no enclosed poles.
Breaking the contour integral into 4 segments:
$$\int_{-R}^{-\epsilon}\dfrac{e^{i\omega t}}{\omega}d\omega + \int_{\pi}^{0} i\epsilon e^{i\theta}\dfrac{e^{it\epsilon e^{i\theta}}}{\epsilon e^{i\theta}} d\theta + \int_{+\epsilon}^{+R}\dfrac{e^{i\omega t}}{\omega}d\omega + \int_{0}^{\pi} iR e^{i\theta}\dfrac{e^{itR e^{i\theta}}}{R e^{i\theta}} d\theta =0$$
Taking the limit as $R \rightarrow \infty$, the fourth integral goes to $0$.
Taking the limit as $\epsilon \rightarrow 0$, the second integral goes to $i\pi\, \mathrm{sgn}(t)$
This leaves
$$\lim_{\epsilon \rightarrow 0}\left(\int_{-\infty}^{-\epsilon}\dfrac{e^{i\omega t}}{\omega}d\omega + \int_{+\epsilon}^{+\infty}\dfrac{e^{i\omega t}}{\omega}d\omega \right)+ i\pi\, \mathrm{sgn}(t)=0$$
So
$$ g(t) = -\dfrac{1}{2\pi i}\lim_{\epsilon \rightarrow 0}\left(\int_{-\infty}^{-\epsilon}\dfrac{e^{i\omega t}}{\omega}d\omega + \int_{+\epsilon}^{+\infty}\dfrac{e^{i\omega t}}{\omega}d\omega \right) = \dfrac{1}{2}\mathrm{sgn}(t)$$
I'm not happy with that negative sign out front. I think I have a sign mistake. When I manipulate my integral (A) to try and get to your integral (1) I end up with
$$\begin{align*}g(t)&=-\frac{1}{2\pi i}\int_{-\infty}^{+\infty} \frac{e^{i \omega t}}{\omega} d\omega\\ &= -\frac{1}{2\pi i}\int_{-\infty}^{+\infty} \frac{\cos(\omega t)}{\omega} +\frac{i\sin(\omega t)}{\omega} d\omega \\ &= -\frac{1}{2\pi i}\int_{-\infty}^{+\infty} \frac{i\sin(\omega t)}{\omega} d\omega \\ &= -\frac{1}{\pi}\int_{0}^{+\infty} \frac{\sin(\omega t)}{\omega} d\omega \\ \end{align*}$$
Which is close to your integral (1) except for the minus sign.