How do I find the Laplace Transform of
$$ \delta(t-2\pi)\cos(t) $$
where $\delta(t) $ is the Dirac Delta Function.
I know that it boils down to the following integral
$$ \int_{0}^\infty e^{-st}\delta(t-2\pi)\cos(t)dt $$
and I know that the answer is simply
$$ e^{-2\pi s} $$
and I know how to integrate just the Dirac Delta Function. I don't know how to integrate it as the product of another function of the same variable. It doesn't explain anywhere in the chapter in my text, either, so I'm a bit confused. I assume I'll have to integrate by parts, or something, but I'm not to sure what to do with that Delta Function in there. Any help would be appreciated. Thanks!
Let $f(t) = \delta(t-2\pi)$.
You can write $f(t)\cos(t) = \frac{1}{2}(f(t)e^{it} + f(t)e^{-it})$ by Euler's formula.
So $L\{f(t)\cos(t)\}(s) = \frac{1}{2}(F(s+i)+F(s-i)))$, (since you are doing a frequency shift by multiplying by an exponential in $t$) where $F(s) = L\{f\}(s) = e^{-2\pi s}$.
Thus, you have $L\{f(t)\cos(t)\}(s)= \frac{1}{2}[e^{-2\pi s +2\pi i} + e^{-2\pi s - 2\pi i}] = \frac{1}{2}[e^{-2\pi s}e^{2\pi i} + e^{-2\pi s}e^{-2\pi i}] =\frac{1}{2}\cdot 2 \cdot e^{-2\pi s}$, since $e^{-2\pi i} = e^{2 \pi i} = 1$.