# Use of the delta function

• I
Gold Member
I've come across the equation $$\int_0^1 dx \frac{dA(x)}{dx} + B = C = \text{finite}$$ in my readings on a certain topic in physics and, in both articles i have read, the following step is taken $$\int_0^1 dx \left( \frac{dA(x)}{dx} + (B-C)\delta(1-x) \right) = \text{finite}$$

For the purposes of my question, I don't think the explicit form of the functions A,B,C are of concern. The reason ##\delta(1-x)## is chosen is associated with the underlying physics which I also think is not important. It seems to me that the replacement $$1 = \int_0^1 \delta(1-x) dx$$ has been made to go from line 1 to line 2 in the above. My questions are:

1) Why is this replacement even correct? The zero of the delta function occurs at the edge of the interval so I don't see why the integral should evaluate to one.
2) I put this integral into wolfram alpha. It returns ##\int_0^1 dx \delta(1-x) = \theta(0)##, where ##\theta(0)## is the step function evaluated at zero. Upon subsequently putting ##\theta(0)## into wolfram alpha it returns 1. So a) how is ##\theta(0)## obtained analytically and b) why is ##\theta(0) = 1##?

Thanks!

Svein
1. Integration a delta function is equivalent to an evaluation. Thus $\int_{a}^{b}\delta(x_{0}) dx$ = 1 if a ≤ x0 ≤ b and 0 otherwise.
2. In the same way $\int_{a}^{b}f(x)\delta(x_{0})dx$ =f(x0) if a ≤ x0 ≤ b and 0 otherwise.

Gold Member
1. Integration a delta function is equivalent to an evaluation. Thus $\int_{a}^{b}\delta(x_{0}) dx$ = 1 if a ≤ x0 ≤ b and 0 otherwise.
2. In the same way $\int_{a}^{b}f(x)\delta(x_{0})dx$ =f(x0) if a ≤ x0 ≤ b and 0 otherwise.
Ok, so then what is wrong with the following calculation: $$\int_0^1 \delta(1-x) dx = \int_0^1 \theta(x) \delta(1-x) dx = \theta(1) = 1?$$
(In the first equality I insert a theta function since this equals one for x >0 so I inserted one). The reason I think it is incorrect is because wolfram alpha gives me ##\theta(0)## as the result.

stevendaryl
Staff Emeritus
1. Integration a delta function is equivalent to an evaluation. Thus $\int_{a}^{b}\delta(x_{0}) dx$ = 1 if a ≤ x0 ≤ b and 0 otherwise.
2. In the same way $\int_{a}^{b}f(x)\delta(x_{0})dx$ =f(x0) if a ≤ x0 ≤ b and 0 otherwise.

I think your rules aren't quite right. I think it should be something like (assuming $a < b$):

$\int_a^b \delta(x-x_0) dx$
• ... $=1$ if $a < x_0 < b$
• ...$=0$ if $a < b < x_0$ or $x_0 < a < b$
• ...$=\frac{1}{2}$ if $x_0 = a$ or $x_0 = b$
You have to have the last case if you want $\int_a^b \delta(x-x_0) dx + \int_b^c \delta(x-x_0) dx = \int_a^c \delta(x-x_0) dx$

stevendaryl
Staff Emeritus
Ok, so then what is wrong with the following calculation: $$\int_0^1 \delta(1-x) dx = \int_0^1 \theta(x) \delta(1-x) dx = \theta(1) = 1?$$
(In the first equality I insert a theta function since this equals one for x >0 so I inserted one). The reason I think it is incorrect is because wolfram alpha gives me ##\theta(0)## as the result.
It's tricky working with discontinuous functions (and the delta function isn't actually even a function). But I would say that:

$\int_0^1 \delta(1-x) dx = \int_0^1 \delta(x) = \frac{1}{2}$

That's the most sensible answer, since
$\int_a^1 \delta(x) dx = 1$ if $a < 0$
$\int_a^1 \delta(x) dx = 0$ if $a > 0$

So when $a=0$, we should have:
$\int_0^1 \delta(x) dx = \frac{1}{2}$

As I said, you need this if you want it to be the case that $\int_{-1}^1 \delta(x) dx =\int_{-1}^0 \delta(x) dx + \int_{0}^1 \delta(x) dx$

Gold Member
I agree it seems like a plausible definition as you write it but then how to explain

a) the validity of the insertion of the delta from line 1 to 2 in my OP?

b) the fact wolfram alpha gives ##\int_0^1 \delta (1-x) dx = \theta(0) ## which it then, after inputting ##\theta(0)##, says is 1?

Svein Sorry, an error crept into my definitions. The correct definitions should be
1. Integration a delta function is equivalent to an evaluation. Thus $\int_{a}^{b}\delta(x - x_{0}) dx$ = 1 if a ≤ x0 ≤ b and 0 otherwise.
2. In the same way $\int_{a}^{b}f(x)\delta(x - x_{0})dx$ =f(x0) if a ≤ x0 ≤ b and 0 otherwise.

Gold Member
Unfortunately I am more confused now than before I posted my question - @Svein and @stevendaryl have given different (as far as I can see) opinions on how to use the delta 'function' when the zero of its argument coincides with one of the endpoints.

Applying what @Svein says to @stevendaryl suggestion of rewriting ##\int_{-1}^{1} \delta(x) dx = \int_{-1}^0 \delta(x) dx + \int_0^1 \delta(x) dx## gives me 1 on the lhs and 2 on the rhs (!) so either: a) we give up the assertion that ##\int_{-1}^{1} \delta(x) dx = \int_{-1}^0 \delta(x) dx + \int_0^1 \delta(x) dx## (perhaps ok if the delta function is not considered to be continuous at 0) or b) the equation ##\int_a^b \delta(x-x_0) dx = 1## if ##a \leq x_0 \leq b##, with ##x_o## strictly being included in the endpoints of the integration interval is wrong. (i.e I would agree that ##\int_a^b \delta(x-x_0) dx = 1## if ##a < x_0 < b## is certainly true).

I understand all these 'fallacies' probably come about from sloppiness in the use of the delta 'function' as a function but hopefully there can be a reconcilation about what has been said.

All my questions in the OP still remain.

Thanks!

stevendaryl
Staff Emeritus
I agree it seems like a plausible definition as you write it but then how to explain

a) the validity of the insertion of the delta from line 1 to 2 in my OP?

b) the fact wolfram alpha gives ##\int_0^1 \delta (1-x) dx = \theta(0) ## which it then, after inputting ##\theta(0)##, says is 1?
In my opinion, it's more convenient to use
$\theta(x) = 0$ when $x < 0$
$\theta(x) =\frac{1}{2}$ when $x = 0$
$\theta(x) = 1$ when $x > 0$

On the other hand, delta functions and step functions are introduced to make certain calculations easier. The edge cases should be analyzed separately. I think you have to go back to the original problem to see what makes sense.

• jasonRF
Stephen Tashi
2) I put this integral into wolfram alpha. It returns ##\int_0^1 dx \delta(1-x) = \theta(0)##, where ##\theta(0)## is the step function evaluated at zero.
Which Wolfram step function did you use?
The Wolfram Language represents the Heaviside generalized function as HeavisideTheta, while using UnitStep to represent the piecewise function Piecewise[[PLAIN]http://mathworld.wolfram.com/images/equations/HeavisideStepFunction/Inline7.gif[PLAIN]http://mathworld.wolfram.com/images/equations/HeavisideStepFunction/Inline8.gif1, [Broken] x >= 0[PLAIN]http://mathworld.wolfram.com/images/equations/HeavisideStepFunction/Inline9.gif[PLAIN]http://mathworld.wolfram.com/images/equations/HeavisideStepFunction/Inline10.gif] [Broken] (which, it should be noted, adopts the convention http://mathworld.wolfram.com/images/equations/HeavisideStepFunction/Inline11.gif instead of the conventional definition [PLAIN]http://mathworld.wolfram.com/images/equations/HeavisideStepFunction/Inline12.gif). [Broken]

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Gold Member
@Stephen Tashi Oh I see thanks, I didn't know of the difference - I wrote in 'integral from 0 to 1 of delta(1-x)' into wolfram alpha and it gives me theta(0). Next to the output it tells me '##\theta(x)## is the Heaviside step function' but when I click on that to view its properties it takes me to unitstep so I think it is indeed using the Piecewise ##[\left\{ \left\{ 1, x \geq 0\right\} \right\}]## definition, so that ##\theta(0) = 1##. But, if I understand correctly, I could take ##\theta(0) = 1/2## in accordance with what @stevendaryl was saying - so it is just a matter of choice and wolfram just happens to choose one convention?

In any case, how did wolfram derive that ##\int_0^1 dx \delta(1-x) = \theta(0)##?

Stephen Tashi
In any case, how did wolfram derive that ##\int_0^1 dx \delta(1-x) = \theta(0)##?
I don't use Wolfram alpha, so I don't know. We can also ask why it bothered to express the answer as "##\theta(0)##" instead of just giving a numerical result. Does Wolfram alpha have a feature like "Show me the steps you used"?

A humorous speculation is that it did a change of variable: ##\int_0^1 1 \delta(1-x) dx = \int_1^0 - 1\delta(y) dy = \int_0^1 1 \delta(y)dy ##

FactChecker
Gold Member
If the delta function is interpreted as a measure at 0, then I think that any integral range which includes 0 should give the full integral value of 1.

stevendaryl
Staff Emeritus
If the delta function is interpreted as a measure at 0, then I think that any integral range which includes 0 should give the full integral value of 1.
As I said, that would imply that $\int_{-1}^{0} \delta(x) dx + \int_{0}^{1} \delta(x) dx = 2$, while the usual rules for integration would say $\int_{-1}^{0} \delta(x) dx + \int_{0}^{1} \delta(x) dx = \int_{-1}^{+1} \delta(x) dx = 1$

• FactChecker
FactChecker
Gold Member
As I said, that would imply that $\int_{-1}^{0} \delta(x) dx + \int_{0}^{1} \delta(x) dx = 2$, while the usual rules for integration would say $\int_{-1}^{0} \delta(x) dx + \int_{0}^{1} \delta(x) dx = \int_{-1}^{+1} \delta(x) dx = 1$
Good point. I missed that in your earlier post. But if the measure definition of delta function and Lebesgue integrals are used, including 0 in both integrals should sum to 2.

Gold Member
Hi FactChecker,
Good point. I missed that in your earlier post. But if the measure definition of delta function and Lebesgue integrals are used, including 0 in both integrals should sum to 2.
Could you elaborate on what you mean here?

FactChecker
Gold Member
Hi FactChecker,

Could you elaborate on what you mean here?
Using general measure theory and Lebesgue integration, you can make a measure that integrates to 1 at the single point x=0. But then if you include 0 in both integrals ∫-10 and ∫0+1, then you have 2 instead of 1. You would need to only include 0 in one integral and not in the other ( like ∫-1-0 + ∫0+1 where -0 means the limit as the integral interval upper value approaches 0 from below.)

• CAF123
Gold Member
I see, so do you mean if we can make a measure so that ##\int_0^1 dx \delta(x) = 1## then we should define ##\int_{-1}^1 dx \delta(x) = 1 = \int_{-1}^{-0} dx \delta(x) + \int_0^1 dx \delta(x) = 0 + 1?## (I haven't done any measure theory so I just want to see if this is a correct take on things)

Also, going back to the claim in the OP where they go from
$$\int_0^1 dx \frac{dA(x)}{dx} + B = C = \text{finite}$$
to
$$\int_0^1 dx \left( \frac{dA(x)}{dx} + (B-C)\delta(1-x) \right) = \text{finite}$$
I guess all the author has done is defined the measure ##dx## so that ##\int_0^1 dx \delta(1-x)=1##?

Thanks!

pwsnafu
I see, so do you mean if we can make a measure so that ##\int_0^1 dx \delta(x) = 1## then we should define ##\int_{-1}^1 dx \delta(x) = 1 = \int_{-1}^{-0} dx \delta(x) + \int_0^1 dx \delta(x) = 0 + 1?## (I haven't done any measure theory so I just want to see if this is a correct take on things)
The Dirac measure is defined so that if a subset ##A \subset \mathbb{R}## contains the point 0 then it has measure 1. Otherwise the measure is 0. I'm going to write this as ##\delta(A)##. Now the Lebesgue integral in our case is (without going into details) ##\int_A f \, d\delta = f(0)\delta(A)##.

Let ##A = [-1,1]##. Lebesgue integration allows us to spit up ##A## but only if the union is disjoint. We can split this into either ##[-1,0] \cup (0,1]## or ##[-1,0) \cup [0,1]## but not ##[-1,0]\cup[0,1]##.

This is an elegant way to define the Dirac, but comes with the downside of the derivatives of the Dirac delta not compadible with "derivatives" as understood in measure theory (known as Radon-Nikodym derivatives).

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• FactChecker
Gold Member
Hi stevendaryl, (I'm still dabbling over this :P)
As I said, that would imply that $\int_{-1}^{0} \delta(x) dx + \int_{0}^{1} \delta(x) dx = 2$, while the usual rules for integration would say $\int_{-1}^{0} \delta(x) dx + \int_{0}^{1} \delta(x) dx = \int_{-1}^{+1} \delta(x) dx = 1$
Why do we want to have that ##\int_{-1}^1 \delta(x) dx = \int_{-1}^0 \delta(x)dx + \int_0^1 \delta(x) dx## in the first place? The split at zero is where the delta function has its infinite spike so should we expect to be able to write this equation a priori? (it seems like a double counting in some sense)

stevendaryl
Staff Emeritus
• 