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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

Luthier415Hz
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    Your title and question body do not match. I suspect all these identities come from the fact that $\int_{\mathbb{R}} f(x) \delta(x) dx = f(0)$, so that in all meaningful places $f(x)\delta(x)$ can be replaced by $f(0) \delta(x)$. – preferred_anon Mar 04 '22 at 11:57

1 Answers1

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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.

Saleh
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