Integration by part of a dirac function : something I don't understand

Question

Consider the following integral :

$$ \int_{\tau'}^{\tau''} d\tau f(\tau) \frac{d}{dt}[\delta(t-\tau)] $$

We can put out of the integral the derivative with respect to time $t$ and thus we have :

$$ \int_{\tau'}^{\tau''} d\tau f(\tau) \frac{d}{dt}[\delta(t-\tau)]=\frac{d}{dt}[f(t)] $$

But I also have :

$$ \frac{d}{dt}[\delta(t-\tau)] = -\frac{d}{d \tau}[\delta(t-\tau)]$$

And if I plug this I have :

$$ \int_{\tau'}^{\tau''} d\tau f(\tau) \frac{d}{dt}[\delta(t-\tau)] = - \int_{\tau'}^{\tau''} d\tau f(\tau) \frac{d}{d\tau}[\delta(t-\tau)]$$

And after doing an integration by part I will have the same result as before plus a surface term.

So the two way of computing the integral don't give the same results.

Where is the problem in what I have done ?

Answer 1

In THIS ANSWER and THIS ONE, I provided primers on the Dirac Delta.

---

The notation $\int_{\tau'}^{\tau''} f(\tau)\delta(\tau-t)\,d\tau$ is interpreted to mean the functional $\langle \delta_t, fp_{[\tau',\tau'']}\rangle$.

Here, $p_{[\tau',\tau'']}$ is the "rectangular pulse" function, $p_{[\tau',\tau'']}(x)=u(x-\tau')-u(x-\tau'')$, and $u$ is the unit step (or Heaviside Function) where

$$u(x)=\begin{cases}1&,x>0\\\\0&,x<0\end{cases}$$

---

Note that there are various conventions for the value $u(0)$.

---

Therefore, we have

$$\begin{align} \int_{\tau'}^{\tau''} f(\tau)\delta(\tau-t)\,d\tau&=\langle \delta_t,fp_{[\tau',\tau'']}\rangle\\\\ &=\begin{cases}f(t)&,t\in(\tau',\tau'')\\\\0&,t\notin [\tau',\tau'']\end{cases}\\\\ &=f(t)p_{[\tau',\tau'']}(t)\tag1 \end{align}$$

If $t=\tau'$ or if $t=\tau''$, then the value of the functional $\langle\delta_t, fp_{[\tau',\tau'']}\rangle$ is not well-defined. That is to say that as a function of $t$, the functional is (jump) discontinuous at $t=\tau'$ and $t=\tau''$.

---

Note, if $g(\tau)=f(-\tau)$, then the notation $\int_{\tau'}^{\tau''} f(\tau)\delta(t-\tau)\,d\tau$ is interpreted to mean the functional $\langle \delta_{-t},gp_{[-\tau'',-\tau']}\rangle$, which by virtue of $(1)$ shows that $$\int_{\tau'}^{\tau''} f(\tau)\delta(t-\tau)\,d\tau=\int_{\tau'}^{\tau''} f(\tau)\delta(\tau-t)\,d\tau$$.

---

Next, we have the notation

$$\begin{align} \frac{d}{dt}\langle \delta_t ,fp_{[\tau',\tau'']}\rangle&=\langle \frac{d}{dt}\delta_t ,fp_{[\tau',\tau'']}\rangle\\\\ &=\int_{\tau'}^{\tau''}f(\tau)\frac{d}{dt}\delta(t-\tau)\,d\tau\\\\ & =\begin{cases}f'(t)&,t\in (\tau',\tau'')\\\\0&,t\notin [\tau',\tau'']\end{cases}\\\\ &=f'(t)(p_{[\tau',\tau'']})(t) \end{align}$$

where the derivative is undefined at $t=\tau'$ and $t=\tau''$ due to the jump discontinuities.

---

If we interpret the derivative of the Heaviside function as the Dirac Delta distribution, then we can extend the definition of $\frac{d}{dt}\langle \delta_t ,fp_{[\tau',\tau'']}\rangle$ to include the jumps at $t=\tau'$ and $t=\tau''$. We then have in distribution

$$\langle \frac{d}{dt}\delta_t ,fp_{[\tau',\tau'']}\rangle=f'(t)p_{[\tau',\tau'']}(t)+f(\tau')\delta(t-\tau')-f(\tau'')\delta(t-\tau'')$$

which can be converted notationally to read

$$\bbox[5px,border:2px solid #C0A000]{\int_{\tau'}^{\tau''}f(\tau)\frac{d}{dt}\delta(t-\tau)\,d\tau=f'(t)p_{[\tau',\tau'']}(t)+f(\tau')\delta(t-\tau')-f(\tau'')\delta(t-\tau'')}\tag 2$$

---

Inasmuch as the distributional derivative is defined by

$$\langle \delta'_{t},fp_{\tau',\tau''}\rangle =-\langle \delta_{t},\left(fp_{\tau',\tau''}\right)'\rangle $$

we see that in distribution

$$\begin{align} \langle -\delta'_{t},fp_{\tau',\tau''}\rangle&=\langle \delta_{t},\left(fp_{\tau',\tau''}\right)'\rangle\\\\ &=f'(t)p_{[\tau',\tau'']}(t)+f(t)\frac{d}{dt}p_{[\tau',\tau'']}(t)\\\\ &=f'(t)p_{[\tau',\tau'']}(t)+f(\tau')\delta(t-\tau')-f(\tau'')\delta(t-\tau'') \end{align}$$

Hence, we have notationally

$$\bbox[5px,border:2px solid #C0A000]{\int_{\tau'}^{\tau''}\left(-\frac{d}{d\tau}\right)\delta(t-\tau)f(\tau)\,d\tau=f'(t)p_{[\tau',\tau'']}(t)+f(\tau')\delta(t-\tau')-f(\tau'')\delta(t-\tau'')}\tag 3$$

---

Comparing $(2)$ and $(3)$, we arrive at the equality that

$$\langle \frac{d}{dt}\delta_t ,fp_{[\tau',\tau'']}\rangle=\langle -\delta'_{t},fp_{\tau',\tau''}\rangle=f'(t)p_{[\tau',\tau'']}(t)+f(\tau')\delta(t-\tau')-f(\tau'')\delta(t-\tau'')$$

or using the alternative notation

$$\bbox[5px,border:2px solid #C0A000]{\int_{\tau'}^{\tau''} f(\tau)\left(-\frac{d}{d\tau}\right)\delta(t-\tau)\,d\tau=\int_{\tau'}^{\tau''}f(\tau)\frac{d}{dt}\delta(t-\tau)\,d\tau}$$

as was to be shown!

Answer 2

The "surface term" is 0 due to the delta. That is, the term from the integration by parts $$ \left [ f(\tau) \delta(t - \tau) \right ]_{\tau'}^{\tau''} = f(\tau'')\delta(t-\tau'') - f(\tau')\delta(t-\tau') = 0 $$ and you recover the expression obtained through the other method. In the case where $t=\tau''$ or $t=\tau'$, it depends on how you define the Dirac-delta, in a physics sense or in the sense of distributions. Often in physics one defines $$ \int_0^a f(t) \delta(t) \mathrm{d}t = \frac{1}{2} f(0) $$ for example, but then the concept of $\frac{\mathrm{d} \delta}{\mathrm{d} t}$ is another matter altogether.