- #1
Reshma
- 749
- 6
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?
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?