Edit: $\delta$ is the Dirac distribution and $\Theta$ is the unit step function.

In my theoretical physics studies I came across the following identity, that is supposed to be valid for all $t \in \mathbb{R}$

\begin{equation} \int \limits_{-\infty}^t \exp(t'-t) \delta(t - t') \mathrm{d}t'= \frac{1}{2} \end{equation}

which I could (contextually) make sense of as something I had been taught in the first semester

\begin{equation} \int \limits_{-\infty}^{\infty} \Theta(t'-t) \delta(t - t') \mathrm{d}t'= \frac{1}{2} \end{equation}

which is seriously riddled with inconsistency issues.

This sounds just like normal theoretical physics reasoning to me, but is there a way to formalize it?
Can we see the Dirac distribution as an element of a dual space of discontinuous functions in a proper sense? - I suspect not.
Or is it perhaps possible to introduce a kind of measure space where the Dirac measure can be seen to average at jump discontinuities? - I believe this should be possible in one dimension, but I am not sure.

  • $\begingroup$ Is your $\Theta$ the unit step function? $\endgroup$ Commented Jan 20, 2017 at 12:41
  • $\begingroup$ Yes it is. I also added that to the question. $\endgroup$
    – iolo
    Commented Jan 20, 2017 at 12:46
  • $\begingroup$ Well, you could make it a functional on the space of semicontinuous functions by saying that it takes a function to the mean of the left and right limits. That is the extension of delta as a functional on $C({\bf R})$, and a pretty reasonable one (although certainly not unique: you could very well put any weighted mean of the two limits, although it may be the unique time-symmetrical extension, which is something a physicist might appreciate). How one would justify the first equality, I have no idea. $\endgroup$
    – tomasz
    Commented Jan 20, 2017 at 13:00
  • $\begingroup$ I just realized I had made a typo: The upper limit in the second integral is supposed to be $\infty$. Hence going from the first integral to the second is just inserting th $\Theta$ function. $\endgroup$
    – iolo
    Commented Jan 20, 2017 at 13:05
  • 1
    $\begingroup$ $\delta$ as average of the left and right limits may be realized as continuous functional on the space of jump-discontinuous functions equipped with the locally uniform convergence, but this topology seems not desirable in the sense that it is inseparable. Moreover, under this topology, any sufficiently small open set will fix the location of most jumps and that does not seem to qualify as useful. Skorokhod topologies may be a possible substitute, but then $\delta$ is no longer a continuous functional since Skorokhod topologies allows jumps to be wiggled in a neighborhood. $\endgroup$ Commented Jan 20, 2017 at 13:52

2 Answers 2


We know that $$\int_{-\infty}^\infty \exp(t'-t)\delta(t-t')\,dt'=\exp(0)=1$$ for any $t$, since $\int_{-\infty}^\infty f(x)\delta(x)\,dx=\delta(f)=f(0)$ for $f\in\mathscr{D}(\Bbb{R})$ and, even though $\exp\notin \mathscr{D}(\Bbb{R})$, we can pretend it is. What is more, $\delta(t-t')=0$ for $t'\ne t$ (or, to use the more official lingo, the support of $\delta$ is $\{0\}$), so really the integrand in the integral above is concentrated on $\{t\}$. It seems plausible that half the integral's value is in the half-integral below $t$, $\int_{-\infty}^t$, and the other half is in the half-integral above $t$, $\int_t^{\infty}$. That gives us the desired ${1\over2}$.

I don't quite know how to justify the integral with $\Theta$ in it.

Comment A. The official story on $\delta$ is conveniently found in Rudin Functional Analysis ch. 6.

Comment B. A closed question recently asked "Is there another Mathematics?" For me, the calcuations in Dirac's Principles of Quantum Mechanics and books of that ilk constitute another mathematics, which can be compared to the stuff in Rudin and in Reed and Simon, which is the true mathematics.


A very simple way to think about it, which is not too far from how the Dirac delta is formally defined, is that any integral in dx of a product in which the delta(x-c) is one of the terms, is the value of the remaining terms of the product at c. So, in your case, your equation becomes $e(t)=\frac{1}{2}$ or $t=-\ln(2)$.

  • $\begingroup$ You misinterpreted the question. The above equations are supposed to be identities for all $t \in \mathbb{R}$ $\endgroup$
    – iolo
    Commented Jan 20, 2017 at 13:01

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