# How to represent DiracDelta?

I need to represent δ(t-t0).

The hint is: In cases involving Dirac delta functions, onemay use the regularized delta function δ(t) = ε/[π(t^2 + ε^2)] approaching δ(t) in the limit ε → 0^+.

But I don't know how to insert that limit in Mathematica or represent the function in a different way.

• This limit is not a usual limit, but the limit in the weak topology. Such limits are not imlemented in current CASes. All that is not a simple matter. See that Wiki article and/or Encyclopedia of Mathematics as a first reading. Something similar: an atom cannot be represented as a very small ball. – user64494 Feb 28 at 17:13
• What will you be using DiracDelta[] for? – J. M.'s ennui Feb 28 at 17:34
• I'm not using DiracDelta for anything, I just want to represent it. – Marina Nebot Feb 28 at 17:44
• @MarinaNebot The limit you mentioned should be taken already after the delta-function is integrated over t in convolution with some other function. It only makes sense when you think of a delta function as a kernel of an integral operator. If you simply need to represent it somehow, why using the regularization and not the standard abstract delta-function notation (or, DiracDelta[] in WL)? Or, if you want to use the regularized version, simply denote it somehow and use that notation. Either way, you don't need to take the limit just for representation purposes, and it would make no sense. – Leonid Shifrin Feb 28 at 17:55
• Comments are not for extended discussion; this conversation has been moved to chat. – Kuba Feb 28 at 23:59

## 1 Answer

The delta_function(t-t0) is used as an operator multiplied by some other function g(t) inside a definite integral over all t from plus to minus infinity and maps the function g(t) to a specific value g(t0) determined by the zero argument of the delta_function. To illustrate this look for example with a function t Cos[t] at

Integrate[(e/(Pi*((t - t0)^2 + e^2)))*(t*Cos[t]), {t, -Infinity, Infinity}]


giving some lengthy output containing a few expressions containing Floor functions . Determine their value for typical parameter values like

Floor[(Pi + 2*Arg[e + I*t0])/(4*Pi)] /. {e -> 10^(-12), t0 -> 5}


and enter their values into the lengthy integration result . Then simplify to get in this case :

(t0*Cos[t0] - e*Sin[t0])/E^e


Take the limit e -> 0 and the test function evaluated at t = t0 remains .

• @user64494 That may or may not be true in a strict mathematical sense, but physicists use it all the time.. – Andreas Feb 28 at 20:48
• @user64494 Simple-minded approaches generally give the right answer in physical situations where the width of the impulse represented by the delta is irrelevant to the answer. That mathematicians can come up with un-physical problems where delta is problematic is irrelevant to science and engineering. The way you get into trouble in the real world with delta is when the time scale of the impulse actually matters. The theory of distributions cannot repair this. – John Doty Feb 28 at 21:59
• @user64494 Emotional? Nah, just pragmatic. Delta functions are tools that get a certain kind of job done: I'm no more emotionally attached to them than I am to my collection of screwdrivers. – John Doty Feb 28 at 22:31
• @user64494 as long as e is finite there is no problem. The smaller it gets the better the approximation of the Lorentz curve to the delta needle will be and the result will come ever closer to t cos(t) – Andreas Feb 28 at 22:34
• Why is the integral over R of delta(t-t0) times t cos(t) not sensible? It seems like the sort of thing the delta functional was made to handle. – Daniel Lichtblau Mar 1 at 3:39