Shankar (in his book Principle of Quantum Mechanics book,page 64) mentions that instead of integrating with respect to dx' in $$\int \delta '(x-x') f(x')dx'=\frac{df(x)}{dx},$$ where $$ \delta '(x-x') = \frac{d}{dx}(\delta(x-x')),\tag{1.10.20}$$ we can do it with respect to $dx$ and get the following: $$\int \delta '(x-x') f(x')dx=-\frac{df(x')}{dx'}$$ I am not sure how this is true since I think $f(x')$ acts as a constant in this integral.

  • $\begingroup$ The paragraph below equation 1.10.28. $\endgroup$ – NegativeTension Nov 29 '17 at 6:32
  • $\begingroup$ I think, you are missing a minus in your first equation, the last one is correct. edit: No, sorry, x and x' are mixed up in the last one. The correct relation is $$\int \delta '(x-x') f(x')dx'=-\frac{df(x)}{dx},$$ $\endgroup$ – Photon Nov 29 '17 at 9:42
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    $\begingroup$ You are misquoting the book. Quote (1.10.20) in its entirety, as well as his conclusion. He explains the dual derivation impeccably, but first it must be read correctly. You should be spooked by the different variables in the r.h.sides of your first and 3rd equations. $\endgroup$ – Cosmas Zachos Nov 29 '17 at 16:20
  • $\begingroup$ I see where the misquoted him. He says the following "Some people prefer tointegrate $\delta'(x-x')$ over the first index, in which case it pulls out -#\frac{df}{dx}#". $\endgroup$ – NegativeTension Nov 29 '17 at 18:29
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    $\begingroup$ I'm voting to close this question as off-topic because misqoutes and typos are not interesting/useful to anybody. To reopen this post (v3), consider to provide exact block quotes. Or if it is not a quote, then make clear it is your own deductions. $\endgroup$ – Qmechanic Nov 29 '17 at 18:36

Your last equation is incorrect, it should have $f(x)$ instead of $f(x')$ . Let's first see why. It's incorrect because if it were true: $$\int_{\mathcal R}\delta'(x-x')f(x')dx = -f'(x')$$ The integration is over x, so $f(x')$ can be taken out of the integral: $$f(x')\int_{\mathcal R}\delta'(x-x')dx = -f'(x')$$ But $\delta'$ is odd, which means that the integral is zero; resulting in: $$f'(x')=0$$ Which is clearly a contradiction since this is not necessarily true for an arbitrary $f$.

So now that we know that's wrong, let's see what the correct expression is. we know that for any open interval $\mathcal R$ containing $x'=x$, we have : $$\int_{\mathcal R}f(x')'\delta'(x-x')dx' = f'(x)$$ (Note the order of $x$ and $x'$ in the argument). Now $x$ and $x'$ are just labels for some arbitrary numbers, so the statement remains true when we exchange these labels, i.e. $$\int_{\mathcal R}f(x)\delta'(x'-x)dx = f'(x')$$ Now since $\delta'$ is an odd "function", we get a minus sign by flipping the argument: $$\int_{\mathcal R}f(x)\delta'(x-x')dx = - f'(x')$$ Which is the desired result.


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