Integrate the derivative of dirac

Question

I am trying to compute $$ \int_{-\infty}^{\infty}\frac{d\delta(t)}{dt} dt=\lim_{t\rightarrow\infty}\delta(t)-\lim_{t \rightarrow -\infty}\delta(t)=0-0=0$$

Is that valid?

Answer 1

In the language of distribution theory the first expression could be interpreted as $\langle \delta', \mathbf 1 \rangle,$ where $\mathbf 1$ is the function having value $1$ everywhere. This expression is valid since $\delta'$ has compact support and $\mathbf 1$ is smooth.

By the definition of derivatives of distribution we have $$ \langle \delta', \mathbf 1 \rangle = -\langle \delta, \mathbf 1' \rangle = -\langle \delta, \mathbf 0 \rangle = 0, $$ since the derivative of $\mathbf 1$ of course is the zero function.

So your result is valid although the calculation is not mathematically rigorous.

Answer 2

I think the answer is correct and reasoning not totally out of line, but to make things a bit more rigorous, you might want to get rid of the derivative before integrating.

$\int_{-\infty}^\infty f(x) \delta '(x) dx$

$u=f(x). dv=\delta'(x)$

$du=f'(x). v= \delta(x)$

$G(f)=f(x)\delta(x)|_{-\infty}^\infty-\int_{-\infty}^\infty f'(x)\delta(x) dx=-f'(0)$

Suppose $f(x)=e^x$.

Then $G(e^x)=0-0-1=-1. $

In your case $f(x)=1$, so $G(1)=0.$

Generally, you want to do whatever substitutions you can to the arguments of integration before applying the principle that $\int f(x)\delta(x)dx=f(0).$