Evaluating the Second Term of Integration by Parts for $\delta (x)$

[Reshma]

Show that x \frac{d(\delta (x))}{dx} = -\delta (x)
where \delta (x) is a Dirac delta function.

My work:

Let f(x) be a arbitrary function. Using integration by parts:
\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

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

 

[arildno]

Well, if we are to do this symbol manipulation properly, you should have:
\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)

 

[Reshma]

Wow, thanks Arildno! I got it!