integral of Dirac delta function with sine It is well known that the Dirac delta function has the following property:
$\int_{-\infty}^{\infty}\delta(t-a)f(t)dt=f(a)$
If $g(t)=\int_{0}^{t}\sin(t-\tau)\delta(\tau-\pi)d\tau$
then 
$g(t) = \left\{ \begin{array}{ll}
0, & \textrm{if}\quad t<\pi\\
\sin(t-\pi), & \textrm{if}\quad t\geq\pi
\end{array} \right.$
How can you show this result?
 A: Surely this is no the "optimal" way to show it and a simple shift of variables will do, but...
Since $\delta (\tau -\pi )=\frac{\mathrm d \theta (\tau -\pi )}{\mathrm d \tau }$ (where $\theta (\tau -\pi )$ is the Heaviside theta function) you can set it like that:
$$g(t)=\int_{0}^{t}\sin(t-\tau)\delta(\tau-\pi)\mathrm d\tau=\int_{0}^{t}\sin(t-\tau)\mathrm d\theta (\tau -\pi )$$
Then integrate it by parts and use the definition of Heaviside theta function:
$$
\require{cancel}
\begin{eqnarray}
g(t)&=&\cancelto{0}{\sin(t-\tau)\theta (\tau -\pi )\bigg|_0^t}-\int_{0}^{t}\frac{\partial \theta (\tau -\pi )}{\partial \tau }\mathrm d\sin(t-\tau)=\\
&=&-\int_{0}^{t}\theta (\tau -\pi )\mathrm d\sin(t-\tau)=\int_{0}^{t}\cos(t-\tau)\theta (\tau -\pi )\mathrm d\tau=\\
&=&\cases{\int_{0}^{t}\cos(t-\tau)\cdot 0 \ \mathrm d\tau \quad \mbox{if} \quad  t<\pi\\\int_{t}^{t}\cos(t-\tau)\cdot \frac{1}{2} \ \mathrm d\tau \quad \mbox{if} \quad  t=\pi\\\int_{\pi}^{t}\cos(t-\tau)\cdot 1 \ \mathrm d\tau \quad \mbox{if} \quad  t>\pi }\\&=&\cases{0\quad  \quad  \quad  \quad  \ \  \mbox{if} \quad  t<\pi\\0 \quad  \quad  \quad  \quad  \ \ \mbox{if} \quad  t=\pi\\\sin(t-\pi) \quad \mbox{if} \quad  t>\pi }  
\end{eqnarray}
$$
A: The sifting property in the differential form:
$\sin \left ( t-\tau  \right )\delta \left ( \tau -\pi  \right )=\sin \left ( t-\pi  \right )\delta \left ( \tau -\pi  \right )$
$$g(t)=\int_{0}^{t}\sin(t-\tau)\delta(\tau-\pi)\mathrm d\tau=\sin(t-\pi)\int_{0}^{t}\delta(\tau-\pi)\mathrm d\tau=\sin \left ( t-\pi  \right )\theta  \left ( t -\pi  \right )$$
where
$$\sin \left ( t-\pi  \right )\theta  \left ( t -\pi  \right )=\left\{\begin{matrix}0,& t<\pi \\ \sin \left( t-\pi \right ),& t\geq\pi\end{matrix}\right.$$
