Proof for this

  • Thread starter Reshma
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  • #1
Reshma
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Show that [tex]x \frac{d(\delta (x))}{dx} = -\delta (x)[/tex]
where [itex]\delta (x)[/itex] is a Dirac delta function.

My work:

Let f(x) be a arbitrary function. Using integration by parts:
[tex]\int_{-\infty}^{+\infty}f(x)\left (x \frac{d(\delta (x))}{dx}\right)dx = xf(x)\delta (x)\vert _{-\infty}^{+\infty} - \int_{-\infty}^{+\infty}d\left (\frac{xf(x) \delta (x)}{dx}\right)dx[/tex]

The first term is zero, since [itex]\delta (x) = 0 [/itex]
at [itex]-\infty, +\infty[/itex].
How is the second term evaluated?
 

Answers and Replies

  • #2
arildno
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Well, if we are to do this symbol manipulation properly, you should have:
[tex]\int_{-\infty}^{\infty}xf(x)\delta{'}dx=xf(x)\delta\mid_{-\infty}^{\infty}-\int_{-\infty}^{\infty}(f(x)+xf'(x))\delta(x)dx=-(f(0)+0*f'(0))=-f(0)[/tex]
 
  • #3
Reshma
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Wow, thanks Arildno! I got it!
 

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