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So I've been doing some QM and I keep coming across the following type of integrals:

$$ \int f(x) \frac{\partial}{\partial x} \delta(x-x') dx. $$

I know that I should integrate by parts but then I always get $0$, since:

$$ \int \frac{\partial}{\partial x} \delta(x-x') dx = 0. $$

One such example that I encountered is the following:

enter image description here

So my question is: How do I evaluate such integrals?

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  • $\begingroup$ Why would partial integration give $0$?! Whats wrong with the derivation you cite? Also, please do not use images, use MathJax for math symbols and equations. And give the exact reference. $\endgroup$
    – Tobias Fünke
    Commented Oct 17, 2023 at 12:50
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    $\begingroup$ Well, when you integrate by parts, you have two terms. Your vanishing term is only one of those. $\endgroup$
    – Jeanbaptiste Roux
    Commented Oct 17, 2023 at 12:51
  • $\begingroup$ So using integration by parts we have $$ \int udv = uv - \int vdu $$ but taking dv to be $$ \frac{\partial}{\partial x} \delta(x-x') dx $$ gives v=0, so uv = 0 and vdu =0 But here it seems that vdu is not 0... So why's that? $\endgroup$
    – Gytis Vejelis
    Commented Oct 17, 2023 at 13:01
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    $\begingroup$ v is not zero, uv is not zero, only [uv] happens to be zero by other parts of our assumptions, and vdu is also not zero. $\endgroup$
    – naturallyInconsistent
    Commented Oct 17, 2023 at 13:19
  • $\begingroup$ Sorry, I'm still very confused, why is v non-zero? but uv is zero? $\endgroup$
    – Gytis Vejelis
    Commented Oct 17, 2023 at 13:25

1 Answer 1

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As pointed out in the comments, you have to use integration by parts. The first term, $f(x)\delta(x-x')$ has to be evaluated at $-\infty$ and $+\infty$. Those 2 vanish because the Dirac delta is zero everywhere except for $x=x'$. The remaining part is simply $$ -\int \mathrm{d}x \frac{\mathrm{d}f(x)}{\mathrm{d}x}\delta(x-x') = -\frac{\mathrm{d}f(x)}{\mathrm{d}x}\big|_{x'}$$

You can get an intuitive picture by approximating the Delta function by a nascent version, and thinking about its derivative. It would look like a function with two sharp peaks, the first positive and the second negative, very close to each other. This effectively extracts from the integral the values of $f(x-\Delta x/2)$ and $f(x+\Delta x/2)$, where $\Delta x$ is the width of the nascent delta. It is now clear where the minus derivative of $f$ comes from, except for the missing $1/\Delta x$. This factor actually comes from the fact that the derivative of the nascent delta function is roughly scaled by a factor of $1/\Delta x$ with respect to the nascent delta itself.

In the figure, the orange line represents $dv$, which is the derivative of $v$. As you can see, $v$ is not zero, but instead becomes a delta function in the appropriate limit.

enter image description here

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  • $\begingroup$ I still can't work this out. So using integration by parts: $$ \int f(x)\frac{\partial}{\partial x} \delta (x-x') dx \\ u=f(x), \\ dv=\frac{\partial}{\partial x} \delta (x-x') dx \\ du = df, \\ v = \int \frac{\partial}{\partial x} \delta (x-x') dx = 0 $$ hence uv = $0$ and vdu should also be $0$, no? $\endgroup$
    – Gytis Vejelis
    Commented Oct 17, 2023 at 13:49
  • $\begingroup$ Your problem is in $v$. You need a $v$ to be a "function" whose derivative gives you the derivative of a delta function. I'm being a bit lax here with the formalism of distributions, but in plain english, a function whose derivative is the derivative of a delta function, is a delta function. So the correct function for $v$ is not 0, but a delta function, which is 0 everywhere except at $x=x'$. You can use a nascent delta to reproduce the steps of integration by parts with more intuition. I will add some figures to illustrate this in the answer. $\endgroup$
    – Diego
    Commented Oct 17, 2023 at 14:04
  • $\begingroup$ @GytisVejelis The problem is that you set $du=df$, but then all of a sudden $dx$ appears in the integral. You should really have $du = \frac{df}{dx}dx$, and in place of $dx$ in your final expression, it should be $\frac{df}{dx}dx$, which is what Diego has and is what is written in the excerpt in your OP. You're just using integration by parts wrong. $\endgroup$
    – march
    Commented Oct 17, 2023 at 15:21

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