I´m reading a paper and there is one step that I don´t know how to get there, so, any help will be good.

Let \begin{equation} U=\int\limits_0^tu(\tau)d\tau. \end{equation}

Then, they say that the Fourier transform of $U$ (in relation to time $t$) is given by


When I tried to calculate the Fourier transform of $U$ I couldn´t reproduce that result. My attempt:

$$\hat{U}(w)=\frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{+\infty}e^{-itw}\int\limits_0^tu(\tau)d\tau dt,$$

separating variables I get,

\begin{align} \hat{U}(w)&=\left.\frac{1}{\sqrt{2\pi}}\left(\int\limits_0^tu(\tau)d\tau\frac{e^{-itw}}{-iw}\right)\right|_{-\infty}^{+\infty}+\frac{1}{iw}\frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{+\infty}e^{-itw}u(t)dt\\ &=\left.\frac{1}{\sqrt{2\pi}}\left(\int\limits_0^tu(\tau)d\tau\frac{e^{-itw}}{-iw}\right)\right|_{-\infty}^{+\infty}+\frac{\hat{u}(w)}{iw}. \end{align} I also know that the $\delta$ function can be expressed as


but I don´t know how to use that, and so... I stop here


Let \begin{equation} f(t) \stackrel{a}= \int_{0}^t u(\tau) \mathrm d\tau \stackrel{b}= \int_{-\infty}^t u(\tau) \mathrm d\tau \stackrel{c}= \int_{-\infty}^{+\infty} u(\tau)u(t-\tau) \mathrm d\tau \stackrel{d}=u(t)*u(t) \end{equation}

  • $(a)$ Denote the integral you're seeking as $f(t)$.
  • $(b)$ Since $u(t) = 0$ for $t<0$, then this does not change the integral
  • $(c)$ Using $u(t-\tau) = 0$ for $t-\tau < 0$ and $1$ otherwise (i.e. $t > \tau$, which is the domain of integration.)
  • $(d)$ Using the definition of convolution for continuous functions.

Using the convolution theorem and $U(w) = \frac{1}{iw }+\pi \delta( w)$, which is the Fourier transform of $u(t)$. \begin{equation} F(w )= U(w)U(w) = \frac{U(w)}{i w}+\pi U (w)\delta(w) \end{equation}

Since $\delta(0)$ is the only non-zero point, then \begin{equation} F(\omega)=\frac{U(\omega)}{i\omega}+\pi U(0)\delta(\omega) \end{equation}

  • $\begingroup$ Thanks for the answer, but can you do a more detailed explanation? Because I don´t understand your answer $\endgroup$ – Fernando Sep 18 '18 at 19:16
  • $\begingroup$ which part don't you understand\ $\endgroup$ – Ahmad Bazzi Sep 18 '18 at 19:16
  • $\begingroup$ for example, in the second line, the second and the third equalities $\endgroup$ – Fernando Sep 18 '18 at 19:19
  • $\begingroup$ No, I mean, $$\int_{0}^t u(\tau) \mathrm d\tau= \int_{-\infty}^t u(\tau) \mathrm d\tau$$ and $$\int_{-\infty}^t u(\tau) \mathrm d\tau = \int_{-\infty}^{+\infty} u(\tau)u(t-\tau) \mathrm d\tau$$ $\endgroup$ – Fernando Sep 18 '18 at 19:22
  • $\begingroup$ I have edited, please recheck $\endgroup$ – Ahmad Bazzi Sep 18 '18 at 19:25

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