What is Laplace of $\cos t \log t \ \delta(t-π)$?

Question

I know the Laplace of $\delta(t-\pi)$. But I don't know how to find Laplace of $\cos t \log t \ \delta(t-π)$.

How can I find Laplace of $\cos t \log t \ \delta(t-\pi)$ ?

Answer 1

In general $$\int_{-\infty}^\infty\delta(t-c)f(t)dt=f(c)$$ the trick is to notice that $$\int_{-\infty}^\infty\delta(t-c)f(t)dt=\int_{c-\epsilon}^{c+\epsilon}\delta(t-c)f(t)dt$$ for $\epsilon>0$. If we can assume that $f(t)$ is continuous in the interval $(c-\epsilon,c+\epsilon)$, that is, for every $\eta>0$ there exists an $\epsilon$ such that $|f(t)-f(c)|<\eta$ whenever $|t-c|<\epsilon$, then $$\int_{c-\epsilon}^{c+\epsilon}\delta(t-c)f(t)dt=f(c)\int_{c-\epsilon}^{c+\epsilon}\delta(t-c)dt=f(c)$$ This tells you that $$\boxed{\mathcal L\{\delta(t-c)f(t)\}=f(c)e^{-cs}}$$

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In your case $f(t)=\cos(t)\log(t)\delta(t-\pi)$, with $c=\pi$. Thus $$\mathcal L\{\cos(t)\log(t)\delta(t-\pi)\}=-\log(\pi)e^{-\pi s}$$