-1

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.

Ethan Bolker
  • 95,224
  • 7
  • 108
  • 199
JiHea Lim
  • 3
  • 1
    What is your definition of the derivative of a distribution? – Didier May 26 '20 at 12:21
  • Important point, often not appreciated by students: Any time you're thinking seriously about math you need to have the definitions of all the terms in mind! Seem worth pointing out here, because that equation is nothing but the definition of $\delta^{(n)}$. – David C. Ullrich May 26 '20 at 12:58
  • Or to be fair, it's the definition for $n=1$, and the case $n>1$ obviously follows (officially by induction). – David C. Ullrich May 26 '20 at 12:59

2 Answers2

1

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 $\int \Lambda\cdot \varphi$.

md2perpe
  • 26,770
Didier
  • 19,132
0

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 $f^{(n)}(a).$

md2perpe
  • 26,770