It is well known that the Dirac delta function has the following property:


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?

  • 2
    $\begingroup$ Reminds one of convolution. A shift of variables, perhaps. $\endgroup$ – Torsten Hĕrculĕ Cärlemän Dec 24 '13 at 15:32
  • $\begingroup$ I would use $t>\pi$ in the second case. But since the two formulas $0$ and $\sin(t-\pi)$ agree at $t=\pi$, you are OK. $\endgroup$ – GEdgar Dec 24 '13 at 16:30

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: $$ \require{cancel} \begin{eqnarray} g(t)&=&\cancelto{0}{\sin(t-\tau)\theta (\tau -\pi )\bigg|_0^t}-\int_{0}^{t}\frac{\partial \theta (\tau -\pi )}{\partial \tau }\mathrm d\sin(t-\tau)=\\ &=&-\int_{0}^{t}\theta (\tau -\pi )\mathrm d\sin(t-\tau)=\int_{0}^{t}\cos(t-\tau)\theta (\tau -\pi )\mathrm d\tau=\\ &=&\cases{\int_{0}^{t}\cos(t-\tau)\cdot 0 \ \mathrm d\tau \quad \mbox{if} \quad t<\pi\\\int_{t}^{t}\cos(t-\tau)\cdot \frac{1}{2} \ \mathrm d\tau \quad \mbox{if} \quad t=\pi\\\int_{\pi}^{t}\cos(t-\tau)\cdot 1 \ \mathrm d\tau \quad \mbox{if} \quad t>\pi }\\&=&\cases{0\quad \quad \quad \quad \ \ \mbox{if} \quad t<\pi\\0 \quad \quad \quad \quad \ \ \mbox{if} \quad t=\pi\\\sin(t-\pi) \quad \mbox{if} \quad t>\pi } \end{eqnarray} $$


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 $$\sin \left ( t-\pi \right )\theta \left ( t -\pi \right )=\left\{\begin{matrix}0,& t<\pi \\ \sin \left( t-\pi \right ),& t\geq\pi\end{matrix}\right.$$


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