In this link I found an interesting document about the properties of the Dirac delta function.
Then I checked for three identities that puzzled me:
\begin{equation} \begin{array} f\delta(t)e^t=\delta(t) \\ \delta(t)\cos t=\delta(t) \\ \delta(t)\sin t=0 \end{array} \end{equation}
How can I prove this?
I tried using integration
\begin{equation} \int_{-\infty}^\infty\delta(t)e^tdt=\delta(t)e^t-\int_{-\infty}^\infty \delta(t)'e^tdt \end{equation}
\begin{equation} \int_{-\infty}^\infty \delta(t)'e^tdt=\delta(t)'e^t-\int_{-\infty}^\infty \delta(t)''e^tdt \end{equation}
But this seems to no avail.
Any suggestions?
Thanks
You have to perform all calculations in the sense of distributions, i.e., take a test function $\psi$ and "integrate" against it. $\delta(t) e^t = \delta(t)$ is not a pointwise equality. It's an equality between two distributions. Two distributions $f, g$ are said to be equal if: $\int f(t) \psi(t)\ dt = \int g(t) \psi(t) \ dt $ for any test function $\psi$.
By definition: $\int \delta(t) \psi(t) \ dt = \psi(0)$. Thus: $$ \int \delta(t) e^t \psi(t) \ dt = e^0 \psi(0) = \psi(0) = \int \delta(t) \psi(t) \ dt $$ Thus, $\delta(t) e^t = \delta(t)$ in the sense of distributions. Same logic applies to the remaining identities.