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I am trying to get a better understanding on how we can manipulate the infinitesimal dx in an integral $$\int f(x) dx$$

I have come across the following $$ d(\cos (x)) = -\sin(x) dx$$

Therefore

$$\int^{x=2\pi}_{x=0} dx \sin(x) \cos(x) = - \int^{x = 2\pi}_{x=0} d(\cos(x)) \cos(x) = - \dfrac{1}{2} [ \cos^{2}(x)]^{x=2\pi}_{x=0} = -\dfrac{1}{2}[1-1] = 0$$

This looks to me like the chain rule can be applied to infinitesimals in analogy to differentiation.

However, today I'm trying to solve the following problem : prove $$\delta(ax) = \dfrac{\delta(x)}{|a|}$$

Following the hint I looked at $$\int d(ax)\delta(ax) = 1 = \int d(ax)\delta(-ax)$$ Since $$\int d(ax)\delta(ax) = 1 \quad \text{and} \quad \delta(x) = \delta(-x)$$

From this it would seem $$d(ax) = |a|dx$$ giving $$\int d(ax)\delta(ax) = |a|\int dx \delta(ax) = |a|\int dx \delta(-ax) = \int dx \delta(x) = 1$$ as expected.

I would have naively assumed $d(ax) = a \space dx$

In summary, I have no idea how to treat d(f(x)), and I'm not sure where to look for information. Could someone help me gain a better understanding ? Unfortunately I have only taken a few undergraduate maths courses so far, so I couldn't understand anything too complex.

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    $\begingroup$ The discrepancy is due to mixing the theory of differential forms, which formalizes calculations like $d(\cos \theta) = -\sin \theta d\theta$ for taking the differential of orientation-reversing (or even non-injective) functions, and the measure theory used by Lebesgue integration where measures cannot be negative and have no orientation. Someone more conversant in measure theory can hopefully give you a complete answer. $\endgroup$
    – user7530
    Commented Aug 24, 2020 at 17:46
  • $\begingroup$ If you include the limits on the integrals and transform them, then you can use $d(ax) = a \, dx$: $$ \int_{-\infty}^{\infty} d(-x) = \int_{\infty}^{-\infty} (-dx). $$ $\endgroup$
    – md2perpe
    Commented Aug 24, 2020 at 18:21

2 Answers 2

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The answer of md2perpe is the good way to prove what you want to prove. Another way to solve your problem, is to remark that defining the Heaviside function $H = \mathbb{1}_{\mathbb{R}_+}$, one has $H' = \delta_0$ and $H(ax) = \mathrm{sign}(a)\,H(x)$. Therefore $$ \begin{align*} \delta_0(a\,x) &= H'(a\,x) = \frac{1}{a} \frac{\mathrm d}{\mathrm d x} (H(a\,x)) \\ &= \frac{1}{a} \frac{\mathrm d}{\mathrm d x} (\mathrm{sign}(a)\,H(x)) = \frac{1}{|a|} H'(x) \\ &= \frac{1}{|a|} \delta_0(x) \end{align*} $$


I will here add some comment about the notation $\mathrm d(f(x))$. One of the problems with this notation is that $\mathrm d x$ denotes the Lebesgue measure, while $\delta$ (which I prefer to write $\delta_0$) is not a Lebesgue measurable function but also a measure. So one should not use the expression $$ ∫ \delta_0(x) \,\mathrm{d} x $$ but either $∫ f(x) \,\mathrm{d} x$ if $f$ is a Lebesgue measurable function, and $∫ f\,\delta_0 = f(0)$ if $f$ is a $\delta_0$ measurable function (e.g. a function continuous in $0$). In some sense, a measure is only defined on sets and not on points, so if we identify $\mathrm d x$ with the indication of a local volume, then we should rather write $$ ∫ f(x) \,\delta_0(\mathrm{d}x) $$

An other good formalism is the one of the Stieltjes integral (see e.g. https://en.wikipedia.org/wiki/Lebesgue%E2%80%93Stieltjes_integration). In this formalism, if $g$ is a function of bounded variations, then one can define $$ ∫ f\,\mathrm{d}g = \int f(x)\,\mathrm{d}g(x) $$ and actually, since $g$ is of bounded variations if and only if its derivative in the sense of distributions $g'$ is a measure. So, as a distribution, we have $$ \langle g',f\rangle = ∫ f(x) \,\mathrm{d}g(x) $$ (or if you do not know distributions, let say that if $g'$ is integrable then we have $\int f\,g' = ∫ f \,\mathrm{d}g$). So, to have coherent notations, one should write $∫ f\,\mathrm d g$ to indicate that one integrate with respect to the measure $g'$, and not $g$. For example, for the Dirac delta, this gives $$ ∫ f(x)\,\mathrm{d}H(x) = ∫ f(x)\,\delta_0(\mathrm{d}x) = \langle \delta_0,f\rangle = f(0) $$ Here the first integral is well defined as a Lebesgue-Stieltjes integral, the second as an integral with respect to a measure and the third as a distribution.

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  • $\begingroup$ (+1) This is a very well-written and compact synopsis that addresses the OP's issues, in my opinion, at an appropriate level fit for purpose. Well done! $\endgroup$
    – Mark Viola
    Commented Sep 18, 2020 at 16:35
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Let $\varphi$ be a test function.

If $a>0$ then $$ \int_{-\infty}^{\infty} \delta(ax) \, \varphi(x) \, dx = \{ y=ax \} = \int_{-\infty}^{\infty} \delta(y) \, \varphi(y/a) \, \frac{1}{a} dy \\ = \frac{1}{a} \varphi(0) = \int_{-\infty}^{\infty} \frac{1}{a} \delta(x) \, \varphi(x) \, dx . $$

If $a<0$ then $$ \int_{-\infty}^{\infty} \delta(ax) \, \varphi(x) \, dx = \{ y=ax \} = \int_{\infty}^{-\infty} \delta(y) \, \varphi(y/a) \, \frac{1}{a} dy = - \int_{-\infty}^{\infty} \delta(y) \, \varphi(y/a) \, \frac{1}{a} dy \\ = -\frac{1}{a} \varphi(0) = \int_{-\infty}^{\infty} \frac{1}{-a} \delta(x) \, \varphi(x) \, dx . $$ Thus, for any $a\neq 0,$ $$ \int_{-\infty}^{\infty} \delta(ax) \, \varphi(x) \, dx = \int_{-\infty}^{\infty} \frac{1}{|a|} \delta(x) \, \varphi(x) \, dx . $$

Since this is valid for all test functions $\varphi$ we have $$ \delta(ax) = \frac{1}{|a|} \delta(x). $$

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  • $\begingroup$ Thank you very much, this really clears things up for the delta function identity. Your comment also helped me understand I didn't realise I was using different variables for the limits $1=\int^{x=\infty}_{x=-\infty}dx\delta(x)=\int^{ax=\infty}_{ax=-\infty}dax \space \delta(ax)=\dfrac{1}{|a|}\int^{ax=\infty}_{ax=-\infty}dax \space \delta(x)$ $\endgroup$
    – Mr Lolo
    Commented Aug 29, 2020 at 12:10

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