Why is $\theta \delta $ (or $H \delta $) not well defined as a distribution

53 Views Asked by At

This is the statement I read in a paper. By $\delta$, I mean the Dirac delta function and by $\theta$ (or $H$) I mean the Heaviside step function.

I do not know the reason behind, but I would without hesitation write

$$ \int_{-\infty}^{\infty } \delta(x) H(x) dx = 1/2 ,$$

and

$$ \int_{-\infty}^{\infty } \delta(x) H(x) f(x) dx = f(0)/2 $$

for any smooth function $f$. This worked for several problems I encountered in quantum mechanics. They lead to right answers. That is why I am shocked by the statement above.

Could anyone point out the error for me?

1

There are 1 best solutions below

1
On

Why $\frac12$?

Using $\delta=H'$ and $H^n = H$ we get $$ \int_{-\infty}^{\infty} \delta(x) \, H(x) \, dx = \int_{-\infty}^{\infty} H'(x) \, H(x)^n \, dx = \int_{-\infty}^{\infty} \frac{1}{n+1} (H(x)^{n+1})' \, dx \\ = \frac{1}{n+1} \left[H(x)^{n+1}\right]_{-\infty}^{\infty} = \frac{1}{n+1}. $$ Only for $n=1$ we get $\frac12$. Where's the error?