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!
Best Answer
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 $e^{-2\pi i} = e^{2 \pi i} = 1$.