It is well known that the fourier transform for unit step $U(t)$ is

\begin{equation} F(U(t))=\frac{1}{j\omega}+\pi \delta(\omega) \end{equation}

When I try to arrive to this expression from the definition of fouriet transform, I got

\begin{equation} \lim_{T\rightarrow \infty}\int_{-T}^{T} U(t) e^{-j\omega t} dt=\lim_{T\rightarrow \infty}\int_{0}^{T}e^{-j\omega t} dt= \lim_{T\rightarrow \infty} \left[ \frac{e^{-j\omega t}}{-j\omega}\right]_0^T \\ = \frac{1}{j\omega} - \lim_{T\rightarrow \infty} \left[\frac{e^{-j\omega T}}{j\omega}\right]=\frac{1}{j\omega} - \lim_{T\rightarrow \infty}\left[ \frac{\cos(\omega T)}{j\omega}\right] +\lim_{T\rightarrow \infty}\left[\frac{\sin(\omega T)}{\omega} \right] \end{equation}

We know that

\begin{equation} \lim_{T\rightarrow \infty}\left[\frac{\sin(\omega T)}{\omega} \right] = \pi \delta(\omega) \end{equation}

My question is, how can we show that

\begin{equation} x(\omega)=\lim_{T\rightarrow \infty}\left[ \frac{\cos(\omega T)}{j\omega}\right] = 0\\ \end{equation}

in some distributional sense?

Given a compact-support test function $\phi(\omega)$, we can write

\begin{equation} \int x(\omega) \phi(\omega) d\omega = \int_{|\omega|<\epsilon} x(\omega) \phi(\omega) d\omega + \int_{|\omega|>\epsilon} x(\omega) \phi(\omega) d\omega \end{equation}

it is not hard to see that as $T \rightarrow \infty$ the 2nd integral on the right-hand side of the above equation will become zero by Riemann-Lebesgue lemma. But what about the 1st integral? Can we say that as $\epsilon \rightarrow 0$, $\phi(\omega)$ will become "almost constant" and will come out of the integral?, ie

\begin{equation} \lim_{\epsilon \rightarrow 0}\int_{|\omega|<\epsilon} x(\omega) \phi(\omega) d\omega = \phi(0) \lim_{\epsilon \rightarrow 0}\int_{|\omega|<\epsilon} x(\omega) d\omega=0 \end{equation} as $x(\omega)$ is an odd function?

If so, what will happen if the boundary of the test function is at $\omega=0$?

  • $\begingroup$ Technical remark: we work with tempered distributions here, so test functions are not necessarily compactly supported: they come from the Schwartz class. $\endgroup$ – user147263 Jul 4 '14 at 15:55

Given a test function $\phi$, decompose it into even part $\phi_e(\omega)=\frac12(\phi(\omega)+\phi(-\omega))$ and odd part $\phi_o(\omega)=\frac12(\phi(\omega)-\phi(-\omega))$ and odd part. So, $\phi=\phi_e+\phi_o$. We have $$ \int \frac{\cos \omega T}{j \omega} \phi_e(\omega)\,d\omega = 0 $$ by symmetry.

Since $\phi_o(0)=0$, we can write $\phi_o(\omega)=\omega \psi(\omega)$ where $\psi$ is another test function. Hence $$ \int \frac{\cos \omega T}{j \omega} \phi_o(\omega)\,d\omega = \frac{1}{j} \int \cos (\omega T) \psi(\omega)\,d\omega $$ which tends to zero by the Riemann-Lebesgue lemma.

  • $\begingroup$ I believe this solves the problem. An extremely clean and neat solution.. $\endgroup$ – user1992175 Jul 4 '14 at 21:03

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