I Dirac delta function integrated on a finite interval

[Hill]

TL;DR

How to account for integration limits with Dirac delta function?

Which is correct:
$$\int_{-1}^1 \delta (x-x_0) \, dx =\begin{cases} 1, -1\leq x_0 \leq 1 \\ 0, \text { otherwise} \end{cases}$$
or
$$\int_{-1}^1 \delta (x-x_0) \, dx =\begin{cases} 1, -1< x_0 < 1 \\ 0, \text { otherwise} \end{cases}$$
?

 

[Hill]

OK, I know. The first one.

 

[Hill]

Sorry, not so sure, again. Because, e.g., $$\int_{-1}^2 \delta (x-1) \, dx = 1$$ but if the first case above is correct, then $$\int_{-1}^2 \delta (x-1) \, dx = \int_{-1}^1 \delta (x-1) \, dx +\int_1^2 \delta (x-1) \, dx =2$$ and if the second case is correct, then $$\int_{-1}^2 \delta (x-1) \, dx = \int_{-1}^1 \delta (x-1) \, dx +\int_1^2 \delta (x-1) \, dx =0$$ So, how does it work?

 

[PeroK]

By symmetry, we must have:
$$\int_0^1 \delta(x) dx = \frac 1 2$$I use a rule of thumb that what holds for $\delta (x)$ must hold for the limit of appropriate Gaussians.

 

[Hill]

By symmetry, we must have:
$$\int_0^1 \delta(x) dx = \frac 1 2$$I use a rule of thumb that what holds for $\delta (x)$ must hold for the limit of appropriate Gaussians.

I see. Then you disagree with the following example from the QFTftGA:

By your definition, it should rather be $$...=\begin{cases} 1, -1< x_0 < 1 \\ \frac 1 2, x_0=-1\\ \frac 1 2, x_0=1\\ 0, \text { otherwise} \end{cases}$$ Is it so?

 

[PeroK]

I'd say so. I can check my notes, but it seems a minor detail. I told you the maths was wild and woolly.

As I recall, Tobias Osborne gives a good definition of the functional derivative somewhere in his lectures.

 

[Hill]

I told you the maths was wild and woolly.

Yes, you did, and I keep it in mind.

 

[Hill]

I'd say so. I can check my notes, but it seems a minor detail. I told you the maths was wild and woolly.

As I recall, Tobias Osborne gives a good definition of the functional derivative somewhere in his lectures.

This other example from the book is perhaps more "wild and woolly," but I might be wrong.

They use $$\frac {\partial} {\partial y}[f(y)+\epsilon\delta(y-x)]=f'+\epsilon\delta'(y-x)$$ Can we do this? Derivative of $\delta()$?

A bit later, they integrate by parts:

taking $\delta$ outside. Is it OK?

P.S. There should be a typo: $dy$ is missing in the last integral.

 

[Hill]

Finally, I'll go ahead and trust that these manipulations with Dirac delta are fine. I like this derivation:

They do not explain why the term in square brackets on the third line vanishes, but I understand that it is so because $t_i<t<t_f$, and this is because we vary the trajectory $x(t)$ everywhere except the end points.

 

[PeroK]

This is Chapter 1.3. I studied from this book in 2020 so I can't remember everything. I have, however, found my notes.

Interesting, I have no notes on this section! This suggests there was something I didn't like about it! I guess I must have decided that L&B had produced the E-L equations by some exotic mathematics that I didn't understand or didn't like or both. But, the net result was just the E-L equations. Which I was already completey familair with. So, I moved on. I guess I was hoping that their particular derivation of E-L wouldn't turn out to be important. In the worst case, I'd have to come back to this section and try to figure it out.

If I'm studying advanced material like this, with no tutor, then I have to be pragmatic.

Also, when I later watched Tobias Osborne's videos, his approach to functional derivatives made much more sense.

 

[romsofia]

Here are two lecture notes (physics motivated, both under 6 pages) that helped me:
This one does go into the splitting bounds apart on page 5: http://jacobi.luc.edu/DiracDelta.pdf
This one is more pertaining to your HW thread (Higher order functional derivatives), and it may be more useful for you: https://www.physics.usu.edu/Wheeler/QFT2016/Notes/QFT09FunctionalDerivatives.pdf

However, you'll most likely need to go review a more mathematical oriented text on distributions if you want more generalizations and rigor.