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I'm a physicist, not a mathematician, so I ask here:

We know, that a fundamental property of Dirac's Delta "function" is

$$\int_{-\infty}^{+\infty} f(\tau) \delta(t-\tau) d \tau = f(t)$$

But what about the following:

$$\int_{-\infty}^{t} f(\tau) \delta(t-\tau) d \tau = f(t) \tag{1}$$

If that would be true, then

$$\int_{-\infty}^{+\infty} f(\tau) \delta(t-\tau) d \tau = \int_{-\infty}^{t} f(\tau) \delta(t-\tau) d \tau + \int_{t}^{\infty} f(\tau) \delta(t-\tau) d \tau $$

From the right side we would get $2f(t)$ while the left side gives $f(t)$

I found (1) in a book, but I'm not really happy with that conclusion...Is there something wrong in the book?

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  • $\begingroup$ OK - I see this topic is discussed somehow controversial and the integral is not well defined. $\endgroup$ – MichaelW Feb 18 '17 at 21:10
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    $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$ – Qmechanic Feb 18 '17 at 21:14
  • $\begingroup$ -1. Not useful. It is irrelevant whether you are a physicist or a mathematician or a taxicab driver. The question is about mathematics and has no physics content. $\endgroup$ – sammy gerbil Feb 19 '17 at 1:12
  • $\begingroup$ This is a bit of a late comment but in physics there are two answers. One is “It’s physics so there must be something which is physically correct happening so there’s another way to solve the problem/it might not matter”. The other is that I’ve seen some physics calculations (note: not my specialty at all) which did depend on this value. In particular $\int_{-\infty}^t\delta(x)f(x)\mathrm d x=H(t)f(0)$ and $\int_t^\infty\delta(x)f(x)\mathrm d x=(1-H(t))f(0)$ where $H$ is the Heaviside step function and to get “correct” answers in physics one needed $H(0)=\frac12$ (normally it shouldn’t matter) $\endgroup$ – Dan Robertson Jan 1 '18 at 15:54
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Note that $$ \int_{-\infty}^{t} f(\tau) \delta(t-\tau) d \tau, \tag{A} $$ under the change of variables $\tau\to\tau+t$, becomes $$ \int_{-\infty}^{0} g(\tau) \delta(\tau) d \tau \tag{B} $$ where I have defined $g(\tau)\equiv f(\tau+t)$. The purpose of this post is to investigate to what extent is the integral $(\mathrm B)$ meaningful.

If you change the upper limit to something slightly greater than zero, then the integral above becomes $$ \int_{-\infty}^{\epsilon} g(\tau) \delta(\tau) d \tau=g(0)\tag{B.1} $$ while if you replace the upper integral with a small negative number, then the integral vanishes, $$ \int_{-\infty}^{-\epsilon} g(\tau) \delta(\tau) d \tau=0\tag{B.2} $$

But with an upper limit that is exactly zero, there is no clear meaning we could ascribe to the integral $(\mathrm B)$. There are different conventions one could follow, and it seems that the book OP is reading chose $(\mathrm B.1)$ to define what the integral $(\mathrm B)$ means.

Nevertheless, there is another possible meaning we could ascribe to $(\mathrm B)$, one which is perhaps much more common to find in standard books. The trick is to think of $\delta$ in terms of a nascent delta function, $$ \delta(\tau)\to\lim_{\epsilon\to0}\delta_\epsilon(\tau) $$ with $\delta_\epsilon(\tau)$ a well-behaved function of $\tau$.

If you take, as it is usually done, a symmetric function $\delta_\epsilon(+\tau)=\delta_\epsilon(-\tau)$, then it is clear that $$ \lim_{\epsilon\to0}\int_{-\infty}^{0} g(\tau) \delta_\epsilon(\tau) d \tau=\frac12 g(0) \tag{B.3} $$ which is in fact the most common convention to define the integral $(\mathrm B)$. It is the diplomatic choice between $(\mathrm B.1)$ and $(\mathrm B.2)$: it is half their sum.

You have the three possibilities, $(\mathrm B.1,\mathrm B.2,\mathrm B.3)$, to define the integral $(\mathrm B)$, all of which are pretty natural. Conventions vary from book to book, and which of these is more convenient depends on the context. In some extreme cases, only one of them is consistent with the rest of the formalism.

In a more formal setting, one could attempt to give a proper meaning to the integral $(\mathrm B)$ in the context of distribution theory. The naïve approach $$ (\delta\cdot 1_{\mathbb R^-})[g]\overset?=\int_{-\infty}^{+\infty} \delta(\tau)1_{\mathbb R^-}(\tau)g(\tau) $$ is meaningless because the product of the distributions $\delta$ and $1_{\mathbb R^-}$ is undefined (their singular support, $\{0\}$, coincides). The second approach $$ \delta[g\cdot 1_{\mathbb R^-}]\overset?=\int_{-\infty}^{+\infty} \delta(\tau)1_{\mathbb R^-}(\tau)g(\tau) $$ is meaningless too because $g\cdot1_{\mathbb R^-}$ is not in general a continuous function, and therefore it is not in the domain of $\delta$. The only case where this is a well-defined approach is when $g(0)=0$, in which case $g\cdot1_{\mathbb R^-}$ is continuous, and the three integrals $(\mathrm B.1,\mathrm B.2,\mathrm B.3)$ agree: they all vanish.

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