# Solving integral of sinusoid involving unit step and dirac delta function

I am trying to integrate the following function of $\tau$ from $-\infty$ to $t$. The expected result is $cos(\Omega_0*t)*u(t)$, where $u(t)$ is the unit step function. I'm getting close to the expected result, but not quite. Where am I going wrong?

• Delta is 1 at x = 0 only and 0 elsewhere. Unit step is 1 at x >= 0 and 0 elsewhere. So ignore the integral where functions are zero. Take the limits of integrals properly. – Bhaskar Sep 16 '15 at 17:19
• @L16H7 That does not address the issue unfortunately. To be precise, we first remark that the "integral signs" do not represent integrals, but represent linear functionals acting on distributions. Second, the Dirac Delta is not a function, it is a distribution and it is meaningless, therefore, to assign a value at a point. Third, there is no limit to be taken since the integral sign is not an integral. The correct way to proceed is outlined in the posted answer. – Mark Viola Sep 16 '15 at 18:25
• FWIW, my main mistake above was in the integration of the dirac function - which should integrate to u(t), not 1. Once that's fixed the integration method above "works" even though as per Dr. MV it may not be entirely sound mathematically. – user2398029 Sep 17 '15 at 3:04

## 1 Answer

We have

\begin{align} \int_{-\infty}^t-\Omega_0\sin(\Omega_0\tau)u(\tau)\,d\tau&=\int_{-\infty}^t\frac{d\cos(\Omega_0\tau)}{d\tau}u(\tau)\,d\tau \\\\ &=\int_{-\infty}^{\infty}\frac{d\cos(\Omega_0\tau)}{d\tau}u(t-\tau)u(\tau)\,d\tau \tag 1\\\\ &=-\int_{-\infty}^{\infty}\frac{du(\tau)u(t-\tau)}{d\tau}\cos(\Omega_0\tau)\,d\tau \tag 2\\\\ &=-\int_{-\infty}^{\infty}\delta(\tau)u(t-\tau)\cos(\Omega_0\tau)\,d\tau+\int_{-\infty}^{\infty}\delta(t-\tau)u(\tau)\cos(\Omega_0\tau)\,d\tau\\\\ &=-u(t)+\cos(\Omega_0t)u(t) \end{align}

as expected! Note that the development is in the context of Generalized Functions or a Distribution Theoretic Approach. In that setting, the notation using the integral sign is interpreted as a linear functional and therefore Equations $(1)$ and $(2)$ are more appropriately written as

\begin{align} \left\langle \frac{d\cos(\Omega\tau)}{d\tau},u(\tau)u(t-\tau)\right \rangle&=-\left \langle \cos(\Omega_0\tau),\frac{du(\tau)u(t-\tau)}{d\tau}\right \rangle\\\\ &=-u(t)+\cos(\Omega t)u(t) \end{align}

• Thank you for your help. What about the + 1 part from the derivative of the dirac function? – user2398029 Sep 16 '15 at 23:47
• I would like to accept but the solution (see question) doesn't involve this additional u(t) - that's part of what I don't understand. – user2398029 Sep 17 '15 at 0:34
• Also, the solution I am given doesn't have a negative sign. – user2398029 Sep 17 '15 at 0:37
• @louism I've edited. The solution seems to match the one you were given. The extra $-u(t)$ cancels the term $\int_{-\infty}^t\delta(\tau)\,d\tau$. – Mark Viola Sep 17 '15 at 2:04
• Excellent, thanks! – user2398029 Sep 17 '15 at 2:11