How can i prove Delta function property with derivative?

Question

How can i proof below delta function property?

$$\int_{-\infty}^{\infty}f(t)\delta^{(n)}(t-a)dt=(-1)^{n}f^{(n)}(a)$$, where (n) denotes nth derivate.

Answer 1

If you don't know the formal definition of distributions and derivatives of such, just do integration by parts $n$ times. You will end up with $\int_{-\infty}^{\infty} f^{(n)}(t) \, \delta(t-a) \, dt,$ which equals $f^{(n)}(a).$

Answer 2

Let $\Lambda$ be a distribution and $\varphi$ be a test function. The very definition of the derivative of $\Lambda$ is \begin{align} \langle \Lambda',\varphi\rangle = - \langle \Lambda,\varphi'\rangle \end{align} This is because this matches the integration by part formula when $\Lambda$ is a true function. Apply this $n$ times and you get \begin{align} \langle \Lambda^{(n)},\varphi\rangle = (-1)^n \langle \Lambda,\varphi^{(n)}\rangle \end{align} Just replace $\Lambda$ by $\delta_a$ and you have your answer.

Notice $\langle \Lambda,\varphi\rangle $ stands for $\Lambda(\varphi)$ or $\int \Lambda\cdot \varphi$.