Dirac Delta, higher derivatives with test function

• I

$$x^m \delta^{(n)}(x) = (-1)^m \frac {n!} {(n-m)!} \delta^{(n-m)}(x) , m \leq n$$

I understand the case where m=n and m>n but not this. Just testing the left hand side with m=3 and n=4 and integrating by parts multiple times, I get -6. With the same values, the right hand side evaluates to -24.

To elaborate, what I'm doing on the left hand side is integration by parts multiple times:

$$\int x^3\delta^{(4)} (x) dx=x^3\delta^{(3)}(x)-\int 3x^2\delta^{(3)}(x)dx$$

Bounds are -infinity to +infinity (Latex ignorance). The non integral term evaluates to zero because of the way test functions are defined nonzero only inside a finite interval. So,

$$\int x^3\delta^{(4)} (x) dx=-\int 3x^2\delta^{(3)}(x)dx$$

Then the idea is to repeat the integration by parts until the dirac delta derivative order is 1.

$$\int 3x^2\delta^{(3)}(x)dx=-\int6x\delta^{(2)}(x)dx$$
$$\int6x\delta^{(2)}(x)dx=-\int6\delta(x)dx=-6$$

Being mindful of the way the negative sign flip flops each iteration. I know there is a formula for this but I'm just showing my understanding.

jasonRF
Gold Member
You forgot to include a test function in your integral. Let ##\phi(x)## be a test function and try starting with
$$\int \, \phi(x) \, x^3 \, \delta^{(4)}(x) \, dx$$
and I think you will get the correct result.

Jason

mishima
stevendaryl
Staff Emeritus

$$x^m \delta^{(n)}(x) = (-1)^m \frac {n!} {(n-m)!} \delta^{(n-m)}(x) , m \leq n$$

I understand the case where m=n and m>n but not this. Just testing the left hand side with m=3 and n=4 and integrating by parts multiple times, I get -6. With the same values, the right hand side evaluates to -24.
For one thing, the equals sign there isn't correct. It's not that the two sides are equal, but that they integrate to the same thing.

For another thing, I don't understand why it isn't just:

$$\int x^m \frac{d^n}{dx^n} \delta(x) dx = m! \int \frac{d^{n-m}}{dx^{n-m}} \delta(x) dx$$

jasonRF
Gold Member
For one thing, the equals sign there isn't correct. It's not that the two sides are equal, but that they integrate to the same thing.
The equality sign is fine since the two sides are equal as distributions (generalized functions). The presence of the delta functions makes that pretty clear, I think. But distributions are not functions, which map numbers to numbers. Instead, they are functionals, which map functions to numbers. Using our standard non-rigorous physics/ engineering approach, this means that when you multiply them by an arbitrary test function and "integrate", they produce a number that in general depends on the particular test function. For the expression in the original post, both sides produce the same number when given the same test function.

All of this is also why the test function (##\phi(x)## in my example) had to be included in the integral. If you don't includ it, the explicit assumption is that ##\phi(x)=1##, and since all of the derivatives of ##\phi## are zero, the original expresion simply says ##0=0##. EDIT (again!): both sides are zero only for the ##m<n## case.

jason

Last edited:
mishima
stevendaryl
Staff Emeritus
The equality sign is fine since the two sides are equal as distributions (generalized functions).
Okay. I was thinking that they weren't equal as distributions, because doing integration by parts will produce a derivative of the test function. However, that derivative is multiplied by a power of ##x##, so it will evaluate to 0 at ##x=0##.

##\int (x \frac{d}{dx} \delta(x)) \phi(x) dx = \int \frac{d}{dx} (x \delta(x) \phi(x)) dx - \int \delta(x) \frac{d}{dx} (x \phi(x)) dx = \int \frac{d}{dx} (x \delta(x) \phi(x)) dx - \int \delta(x) (\phi(x) + x \frac{d}{dx} \phi(x)) dx##
##= 0## (surface integral) ##- \phi(0) - (x \frac{d}{dx} \phi(x))|_{x=0} = - \phi(0)##

jasonRF