Consider the following integrals in variables $x,y$ over the whole $\mathbb{R}$, where $a,b\in\mathbb{R}/0$ are constants:

$$\int dx \int dy ~\delta(x-a)\delta(y-b\,x)=\int dy ~\delta(y-b\,a)=1$$

In the following we will evaluate the above integrals in a slightly different way and obtain a completely different result. First, we note the following general identity for delta distributions, where $f(x)$ is an arbitrary differentiable function and $C$ is a constant in respect to $x$:


This identity is easy to prove since it is a direct consequence of the well known identity $\delta(f(x))=\sum_{x_i\in f(x)=0}\frac{1}{f'(x_i)}\delta(x-x_i)$. Explicitly we see:

$$\int dx F(x) \delta(f(x))=\int \frac{df(x)}{f'(x)} F(x) \delta(f(x))=\\=\int \frac{df(x)}{f'(x)} F(x) C\delta(Cf(x))=\int dx F(x) C\delta(Cf(x))$$ Where $F(x)$ is some test function.

Now, let us set $F(x)=\delta(x-a)$ in the above example, and rescale the remaining delta function $\delta(y-b\,x)$ by $C=\frac{y}{y-b\,a}$. (We can use $y$ in $C$ because it is held fixed during $x$ integration):

$$\int dx \int dy ~\delta(x-a)\delta(y-b\,x)=\int dx \int dy ~F(x)\delta(y-b\,x) \\=\int dx \int dy ~F(x)C\delta\Big(C(y-b\,x)\Big) =\int dx\int dy ~\delta(x-a)\frac{y}{y-b\,a}\delta\Big(y\frac{y-b\,x}{y-b\,a} \Big) \\ =\int dy \frac{y}{y-b\,a} \delta(y)=0$$

Clearly, the same integral cannot be equal to 1 and 0 simultaneously. Therefore, I must have done something wrong in the above. Please, point out my mistake.

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    $\begingroup$ you say $F(x)$ is a test function, but then you set $F(x)=\delta(x-a)$. test functions, though, must be smooth and compactly supported. $\endgroup$ – symplectomorphic Apr 15 '14 at 5:35
  • $\begingroup$ Isn't any distribution "smooth" in a distributional sense? $\endgroup$ – Kagaratsch Apr 15 '14 at 5:37
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    $\begingroup$ test functions are functions, in the strict sense; distributions aren't. $\endgroup$ – symplectomorphic Apr 15 '14 at 5:38
  • $\begingroup$ I think the only subtlety to watch out for is that the peak of $\delta(x-a)$ should be away from the peak of $\delta(y-b\,x)$. If this is true, then $\delta(x-a)$ is definitely smooth on the support of $\delta(y-b\,x)$, namely constant zero. $\endgroup$ – Kagaratsch Apr 15 '14 at 5:39
  • $\begingroup$ For all the guys who come here and actually think that $\delta(x-a)$ is not supposed to be treated as a test function: If you look at the very first line of computation above, there is no other way than to treat one of the delta distributions as a test function on the support of the other if you want to evaluate the double integral at all. Think about it on your own, before you jump to conclusions of another persons comment. $\endgroup$ – Kagaratsch Apr 15 '14 at 22:01

The distribution which is written here as $\delta(x-a) \delta(y-bx )$is just the two dimensional Dirac function $\delta(x)\delta(y)$ (more pedantically $ \delta(x)\otimes\delta(y)$, a notation preferred by mathematicians in order not to confuse it with a pointwise product which can be problematic for distributions) composed with the mapping $(x,y)\mapsto(X,Y)$ of the plane where $X=x-a$, $Y=y-bx$. The concept of the composition of a distribution with a smooth diffeomorphism is well-defined (a formal definition is in the treatise of Schwartz). In the case which occurs most frequently in theoretical physics---composing the Dirac function with a diffeomorphism--- this gives the usual formula as in the Wikipedia article on distributions which is often used as an ad hoc definition. In this situation (composition with an affine mapping with Jacobian determinant $1$) the result is very simple and natural---it is the Dirac function with singularity at $(-a,0)$ and so the value of the integral is $1$.

  • $\begingroup$ Neither are they for a mathematician. Could you explain why this is relevant to the above answer? $\endgroup$ – jena Apr 27 '14 at 9:58
  • $\begingroup$ It is not, I read your answer too quickly and my comment was more a reflex than anything else. Sorry about the noise, I will delete it. $\endgroup$ – Did Apr 27 '14 at 11:27
  • $\begingroup$ Hi jena! Thanks for your reply. I thought the singularity (support) of the two dimensional delta function was at $(a,ab)$. How do you arrive at $(-a,0)$? Also, the above example is oversimplified (missing a test function and involving simple arguments), but my actual interest is in the class of generally valid transformations (and their proper jacobians) in a multi dimensional case. Could you say a couple of words about that or direct me to some literature? Thank you! $\endgroup$ – Kagaratsch May 11 '14 at 22:13

I thought about it for a while and here is my best attempt to explain this.

Even though $y$ is fixed during $x$ integration, its fixed value still can be selected dynamically. Therefore, the constant $C$ should be finite for any $y\in\mathbb{R}$. Clearly, $C=\frac{y}{y-b\,a}$ is identically zero at $y=0$, so that in this particular point $C$ does not define a meaningful scale. Since $y=0$ is a regular point in the initial expression for the integrals, it should stay regular after a rescaling. Therefore, $C=\frac{y}{y-b\,a}$ is simply not a valid choice.


Having thought about it more, it became clear to me that the point $y=0$ cannot possibly give a contribution. In the initial integral both delta functions give support at $x=a$ and $y=ab$. This cannot change when a proper choice of $C$ is applied. However, in $C=\frac{y}{y-b\,a}$ the denominator is singular on the support of the delta functions, therefore this is the actual reason why this choice is not allowed. I feel that this explanation is now sufficient.

Please let me know if this makes sense, or if something deeper is involved here.


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