Integrating using one sided limits

Questions
1

I am finding my way around Mathematica but am stuck.

As part of the learning process I am using know functions and known outcomes. In this particular case calculating the Laplace transform of a Dirac Delta function from the integral formula.

Integrate[DiracDelta[t] E^(-s*t), {t, 0, \[Infinity]}]

what I actually want is the lower bound to be lim->0 from below. The above integral produces:

1 - HeavisideTheta[0]

rather than 1.

If I use LaplaceTransform I get 1.

Using Mathematica V10. I have tried using Lim as the lower bound for the integration, however, it returns the same result. Setting the value to something <0 yields 1.

How do I set the integration to the lim->0 to produce 1?

2

Maybe:

 Limit[Integrate[DiracDelta[t] E^(-s*t), {t, a, \[Infinity]}, Assumptions -> a >= 0], a -> 0, Direction -> "FromBelow"]
 (* 1 *)
3

Thanks. I had tried using a limit but obviously not correctly.

Note that it works if I replace “FromBelow” with 1 (as per documentation on limits. If I use “FromBelow” it does not evaluate. It also works without the assumption a>=0

Further if I use a → -1 i.e. the limit from above it correctly returns 0.

4

Hi Ted! The problem is that DiracDelta is not really a function, but a positive measure (or, more generally a Schwartz’s distribution). So, we must use a sequence of functions converging towards DiracDelta to obtain the Laplace Transform.
Try for instance this sequence:

f[t_, r_] := If[t > r, 0, 1/r]
Plot[Table[f[t, r], {r, 1/10, 2, 1/10}], {t, 0, 3}, PlotRange -> All]

Then, compute the Laplace transform, and its limit when the width converge to zero:

Integrate[f[t, r] E^(-s*t), {t, 0, \[Infinity]}, Assumptions -> r > 0]
Limit[%, r -> 0]

You will obtain 1.

5

Thanks Claude. It makes sense, I was hoping Mathematica would do the dirty work for me :>

6

I would argue that this is risky since it involves a “hidden” HeavisideTheta[0] (which is left undefined).

In[211]:= Integrate[DiracDelta[t] E^(-s*t), {t, a, \[Infinity]}, 
 Assumptions -> a >= 0]

(* Out[211]= HeavisideTheta[-a] *)

Might be safer to go the opposite direction to get the limit-from-negative-side effect.

In[213]:= Integrate[DiracDelta[t] E^(-s*t), {t, a, [Infinity]},
Assumptions → a < 0]

(* Out[213]= 1 *)

The limit as a->0 from below is now trivially 1.

7

It turns out that Mathematica is a big fan of Steely Dan.

I mean, who knew?

8

Have always liked Steely Dan :slight_smile:

9

Is that because the Heaviside function is undefined at 0?

Adding the assumption that a<0 makes perfect sense, in fact it answers my underlying lack of knowledge.

10

Yes. It is undefined really for the reason you might suspect: you don’t know if you arrived zero from above, below or neither (e.g. an integral that has the singular point, 0, as a bound of integration).