Inverse Laplace transform of convolution

Question

$$\int_{0}^t \sin(2\pi(t-T)) \delta(t-5) \, dt$$

Wouldn't you just replace the $T$ in $\sin(2\pi(t-T))$ with a $5$ and that would be the answer?

Answer 1

Since the delta function is involved in the integrand the integral is not necessarily of the convolution type. This is seen by: \begin{align} I &= \int_{0}^{t} \sin(2\pi(t-T)) \, \delta(t-5) \, dt \\ &= \begin{cases} \sin(2\pi(5-T)) & t>5 \\ 0 & t<5 \end{cases}\\ &= \begin{cases} - \sin(2\pi \, T)) & t>5 \\ 0 & t<5 \end{cases} \end{align}