While calculating an inverse Laplace transform Wolfram Alpha returned to me the following output:

7 + 2 DiracDelta[-1 + t] + 14 DiracDelta[t] + HeavisideTheta[-1 + t] + 16 DiracDelta'[t]

What does DiracDelta'[t] mean? A derivative of Dirac Delta function? Wouldn't that be infinite at $0$ and zero everywhere else? That is, basically the Dirac Delta function itself?


"Infinite at zero and zero everywhere else" is a woefully inadequate description of the dirac delta.

The best (and usually literal) definition of the dirac delta is basically that the notation resembling an integral containing a dirac delta is defined to mean evaluation:

$$ \int_{-\infty}^{\infty} f(x) \delta(x-a) \, \mathrm{d}x := f(a) $$

whenever $f$ is continuous at $a$. Notation involving the derivative is defined by a similar formula:

$$ \int_{-\infty}^{\infty} f(x) \delta'(x-a) \, \mathrm{d}x := -f'(a) $$

where $f$ is continuously differentiable at $a$. The idea behind the definition is that it is is meant to invoke partial integration; to imagine a hypothetical calculation

$$ \int_{-\infty}^{\infty} \left( f(x) \delta'(x-a) + f'(x) \delta(x-a) \right) \, \mathrm{d}x = (f(x) \delta(x-a))\big|_{x=-\infty}^{x=\infty} = 0 $$

There is a systematic approach to this sort of stuff: they're called distributions. On a suitable space of test functions, this partial integration formula is the definition of the derivative.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy