Convolution of sine and unit step function

I started studying signal convolution recently and the first sample problem I got is to find convolution of sine and unit step function (Heaviside function). Here is what I have right now.

$y(t)=H(t)*sin(t)=\int_{-\infty}^{\infty} H(\tau) sin(t - \tau)d\tau = \int_{0}^{\infty}sin(t - \tau)d\tau$

Now when I apply substitution, I have $\lambda = t - \tau, d\lambda = -d\tau$. Then I have

$y(t) = -\int_{t}^{t-\infty} sin(\lambda)d\lambda = \int_{t - \infty}^{t} sin(\lambda)d\lambda = -cos(t) + cos(t - \infty)$.

This, of course, makes no sense. Wolfram Alpha says the answer is simply $-cos(t)$. What am I doing wrong? My textbook does not have the solution for this sample problem. Am I calculating the integral in a wrong way?

The two functions you are trying to convolve are both not integrable. The natural domain of the convolution is $L^1(\mathbf R)$, to this class neither of your two functions belong. The convolution can still be done but you need to view both, $H(t)$ and $g(t) = \sin(t)$ as distributions. I do not want to go into detail about the existence of the convolution in this case (this requires a careful analysis of the domain of definition of $H$ and $g$). If we accept that the two distributions can be convolved we need three ingredients (note that all the operations are now on distributions and they need to be well defined and explained in that context):

• $\partial f * g = f*\partial g$
• $\delta * f = f$, where $\delta$ is the Dirac distribution
• $\partial H = \delta$

Now we can perform the following computation:

$$H * \sin = H * \partial (-\cos) = \partial H * (-\cos) = \delta * (-\cos) = -\cos$$

• Thanks for the answer, Chris. Can you just explain what $L^{1}(R)$ refers to and why did you differentiate $sin$ and after than switched things and used derivative of $H$? Or maybe just a link where I can find out more about these cases? Thanks in advance. Nov 28, 2014 at 13:18
• My bad, I wanted to say, why did you write $sin$ as derivative of $-cos$ and then switched place for the derivative sign? Nov 28, 2014 at 13:27
• well, if you are not a mathematician you might not care too much about the technicalities, such as the definition of Lebesgue Spaces, to which $L^1(\mathbb R)$ belongs. I wrote $\sin$ as the derivative of $-\cos$ because I knew that I could get the derivative across to the other side of the convolution and that $\partial H = \delta$ and further that $\delta * f = f$. If you understand each of the steps in the equation I wrote you should see that it is indeed a proof for the relation you were interested in. Nov 30, 2014 at 13:59
• Two more links you might find helpful: convolutions, distributions. While in the convolution link you will find the fact that $\partial f * g = f * \partial g$ and in the other link you will find information on distributions, i.e. a generalization of functions. Hope that helps... Nov 30, 2014 at 14:02
• Thank you very much for the links. I was confused for a bit because the first bullet in your answer is not consistent with your last comment. Namely, I was confused about why were you using derivatives at all because of the commutativity in convolution, but now I see what you mean. I will gladly accept your answer, but I just want to make sure we're on the same page. Thanks. Nov 30, 2014 at 17:20