How Does the Dirac Delta Function Simplify to f(0) in Integrals?

[LLT71]

I somehow understand "why" that works, but when it comes to prove it I fail... I managed to "half-prove" it by
saying:
∫f(x)δ(x)dx=f(a)*δ(a)+...+f(0)*δ(0)+...+f(b)*δ(b)=f(a)*0+...f(0)*1+...+f(b)*0=f(0)
note: integral goes from "a" to b", a<0, b>0 and δ(x) is Dirac-delta function
but I figured out I've missed to put Δx in every term and now it confuses me even more...
help!

 

[PeroK]

I somehow understand "why" that works, but when it comes to prove it I fail... I managed to "half-prove" it by
saying:
∫f(x)δ(x)dx=f(a)*δ(a)+...+f(0)*δ(0)+...+f(b)*δ(b)=f(a)*0+...f(0)*1+...+f(b)*0=f(0)
note: integral goes from "a" to b", a<0, b>0 and δ(x) is Dirac-delta function
but I figured out I've missed to put Δx in every term and now it confuses me even more...
help!

What's your definition of $\delta$?

 

[LLT71]

What's your definition of $\delta$?

it equals 1 when x=0, and it equals zero when x is different than zero (Dirac function)

 

[PeroK]

it equals 1 when x=0, and it equals zero when x is different than zero (Dirac function)

That's not the Dirac function. That's a function that is equivalent to the zero function when integrating over an interval.

 

[LLT71]

That's not the Dirac function. That's a function that is equivalent to the zero function when integrating over an interval.

ahhhhh yeah you are right... at zero dirac function equals +inf and integral from -inf to +inf of Dirac function equals one by definition. how I'm going to attack this problem now or better to say how I'm going to attack it anyway?

 

[PeroK]

ahhhhh yeah you are right... at zero dirac function equals +inf and integral from -inf to +inf of Dirac function equals one by definition. how I'm going to attack this problem now or better to say how I'm going to attack it anyway?

It really depends on your starting point in defining the Dirac delta function.

 

[LLT71]

It really depends on your starting point in defining the Dirac delta function.

if I say I want to prove it in a manner similar from the first post (by using definition of definite integral) then what "type of definition" of Dirac function should I use?

 

[PeroK]

if I say I want to prove it in a manner similar from the first post (by using definition of definite integral) then what "type of definition" of Dirac function should I use?

You could assume that it's a function whose integral about the point 0 is 1, no matter how small the interval containing 0.

You could assume $f$ ia continuous at 0 as well.

 

[LLT71]

You could assume that it's a function whose integral about the point 0 is 1, no matter how small the interval containing 0.

You could assume $f$ ia continuous at 0 as well.

well, now I think my high-school math skills aren't sufficient to actually prove it... let me ask if you could prove it in this definite integral fashion:
∫f(x)δ(x)dx=f(a)*δ(a)+...+f(0)*δ(0)+...+f(b)*δ(b)=f(a)*0+...f(0)*1+...+f(b)*0=f(0) cause that's the furtherst I "understand". thanks!

 

[PeroK]

well, now I think my high-school math skills aren't sufficient to actually prove it... let me ask if you could prove it in this definite integral fashion:
∫f(x)δ(x)dx=f(a)*δ(a)+...+f(0)*δ(0)+...+f(b)*δ(b)=f(a)*0+...f(0)*1+...+f(b)*0=f(0) cause that's the furtherst I "understand". thanks!

If you're at high school level I wouldn't worry about this sort of thing. Instead take this result as essentially a postulated property of the delta function: that when integrated with another function it picks out the function value at a specific point.

 

[LLT71]

If you're at high school level I wouldn't worry about this sort of thing. Instead take this result as essentially a postulated property of the delta function: that when integrated with another function it picks out the function value at a specific point.

actually I'm at college where math is frozen to math from high school (don't ask me why, I hate it). no matter what, as always, I wanted to go step further in my understanding. we have a DSP (Digital Signal Processing) subject and actually I have to self-study it... so tell me what kind of math I need in order to decompose this problem and actually "understand" it. if it is a calculus based proof (at any levels) you can post it or give me names/external links to any other non-calculus based proofs. I want to give it a try. and yes, I will be very thankful!

 

[PeroK]

actually I'm at college where math is frozen to math from high school (don't ask me why, I hate it). no matter what, as always, I wanted to go step further in my understanding. we have a DSP (Digital Signal Processing) subject and actually I have to self-study it... so tell me what kind of math I need in order to decompose this problem and actually "understand" it. if it is a calculus based proof (at any levels) you can post it or give me names/external links to any other non-calculus based proofs. I want to give it a try. and yes, I will be very thankful!

The Dirac delta function is an exception in that it's not a function and it is difficult to use rigorously. Neither straight calculus nor real analysis will help. It's use in physics as an infinite spike function is invaluable and I would accept it as a unique creature for this reason.

I think you'd gain relatively little from the time and effort needed to study it formally. Time that would be better spent on calculus, linear algebra or differential equations.

 

[LLT71]

The Dirac delta function is an exception in that it's not a function and it is difficult to use rigorously. Neither straight calculus nor real analysis will help. It's use in physics as an infinite spike function is invaluable and I would accept it as a unique creature for this reason.

I think you'd gain relatively little from the time and effort needed to study it formally. Time that would be better spent on calculus, linear algebra or differential equations.

thank you for your guidance! I always wondered why they put any of these subjects before you gain deep understanding of every math concept behind it. I hope that's not the case with universities around the world.

 

[micromass]

The property in the OP is actually the definition of the Dirac delta function.

 

[FactChecker]

Since the delta function is 0 except at x=0, you can simplify the integrand to get rid of all f(x) except f(0). Work from there.

 

[micromass]

I don't get why everybody is trying to prove this. It cannot be proven. It's a definition.

 

[FactChecker]

I don't get why everybody is trying to prove this. It cannot be proven. It's a definition.

His definition has no function f(x). That is very common.

 

[micromass]

His definition has no function f(x). That is very common.

Common but wrong. The property cannot be proven from his definition.

 

[FactChecker]

Common but wrong. The property cannot be proven from his definition.

I could prove it as long as integration retains its linear properties.

 

[micromass]

I could prove it.

Then I'm sorry to say that your proof is wrong.

 

[FactChecker]

Then I'm sorry to say that your proof is wrong.

I stand corrected. The formal definition does say exactly what you are saying (that f(x) is included in the definition).

That being said, I suspect that the real analysis formal definition is beyond the scope of the OP. I base that on the definition of the delta function that the OP gave. In that case, a reasonable heuristic "proof" from his definition can be done just by assuming that the integral retains its linear properties.

 

[LLT71]

I stand corrected. The formal definition does say exactly what you are saying (that f(x) is included in the definition).

That being said, I suspect that the real analysis formal definition is beyond the scope of the OP. I base that on the definition of the delta function that the OP gave. In that case, a reasonable heuristic "proof" from his definition can be done just by assuming that the integral retains its linear properties.

you mean something like this:
http://math.stackexchange.com/questions/73010/proof-of-dirac-deltas-sifting-property?rq=1
ok, now I know it's a definition, but I wanted to share this with you to see if by this you meant "real analysis formal definition" (using infinitesimally small epsilon/delta interval around point).

 

[PeroK]

you mean something like this:
http://math.stackexchange.com/questions/73010/proof-of-dirac-deltas-sifting-property?rq=1
ok, now I know it's a definition, but I wanted to share this with you to see if by this you meant "real analysis formal definition" (using infinitesimally small epsilon/delta interval around point).

That's the sort of thing I had in mind, BUT the delta function is definitely not the place to start with formal analysis.

Also, epsilon & delta as used in analysis are NOT infinitesimals. They are simply positive real numbers. But that's another story.

 

[micromass]

One cool way to see/define the delta function is as the hypothetical limit of a sequence of functions. For example, you can look at the following functions and see the delta function as their limit (what it means for these functions to have a limit is the tricky bit of course).

 

[FactChecker]

you mean something like this:
http://math.stackexchange.com/questions/73010/proof-of-dirac-deltas-sifting-property?rq=1
ok, now I know it's a definition, but I wanted to share this with you to see if by this you meant "real analysis formal definition" (using infinitesimally small epsilon/delta interval around point).

As far as I know, the best formal development of the Dirac delta function involves measure theory and Lebesgue integration. Those are important concepts in a real analysis class. (see the section "As a measure" in https://en.wikipedia.org/wiki/Dirac_delta_function )

 

[Svein]

As far as I know, the best formal development of the Dirac delta function involves measure theory and Lebesgue integration. Those are important concepts in a real analysis class. (see the section "As a measure" in https://en.wikipedia.org/wiki/Dirac_delta_function )

In fact, the Stieltjes integral defines the integral against a "step measure" which also turns out to be an evaluation at the step point times the size of the step.

 

[Stephen Tashi]

you mean something like this:
http://math.stackexchange.com/questions/73010/proof-of-dirac-deltas-sifting-property?rq=1

The argument given there is not a proof because it assumes at the outset that the integrand $f(x) \delta (x -t)$ is a function that has an integral. If we use the definition of "integral" given in introductory calculus ( e.g. as defined terms of the limit of Riemann sums) together with the definition that $\delta(x-t) = 0$ when $x \ne t$, we can prove $\int_{-\infty}^{\infty} f(x)\delta(x-t) dx = 0$. That contradicts what we want to be true.

The fact that the dirac delta "function" is commonly defined with the condition $\int_{0 -a}^{0+a} \delta(x) dx = 1$ shows the delta function is not actually a function. For, example, what would you say to a calculus student who defined a function by saying: " I define $f(x)$ as $f(x) = x^2$ and $\int_{-1}^{1} f(x) = 13.7 \$"?

Once you define a function, you don't have any choice about what its integrals are going to be - except that you can choose different definitions of the general concept of integration.

To approach the dirac delta function coherently, we must revise the definition of integration - or at least the notation for integration.

One way to do this is to define the notation $\int_{a}^{b} f(x) \delta(x) dx$ to mean something besides the usual kind of integral. This has the drawback of making the notation "$\int$" ambiguous, but that is a frequent transgression in writing mathematics.

Let $\{g_n\}$ be the sequence of functions given by $g(n) = \frac{1}{\sqrt{2 \pi} \sigma} e^{-\frac{x^2}{2 \sigma^2}}$ where $\sigma_n = 1/n$.

(This defines $g_n$ to be the probability density for a normal distribution with mean zero and standard deviation $\frac{1}{n}$. Like any probability density, $g_n$ has the property that $\int_{-\infty}^{\infty} g_n(x) dx = 1$ .)

The graph of these functions is illustrated by the curves that @micromass showed. As $n$ increases they become more and more peaked over the value $x = 0$.

Then we define the notation $\int_{a}^{b} f(x) \delta(x) dx$ to mean $\lim_{n \rightarrow \infty} \big{[} \int_{a}^{b} f(x) g_n(x) dx \big{]}$ if that limit exists.

Notice that this approach defines an integral with the symbol "$\delta(x)$ in the integrand without defining the symbol "$\delta(x)$" to have any meaning by itself.

There are probably more sophisticated was to define $\delta(x)$ than the method I illustrated, but the method I illustrated indicates the sort of thing you must do to have a logically consistent structure.

 

[pwsnafu]

Let $\{g_n\}$ be the sequence of functions given by $g(n) = \frac{1}{\sqrt{2 \pi} \sigma} e^{-\frac{x^2}{2 \sigma^2}}$ where $\sigma_n = 1/n$.

(This defines $g_n$ to be the probability density for a normal distribution with mean zero and standard deviation $\frac{1}{n}$. Like any probability density, $g_n$ has the property that $\int_{-\infty}^{\infty} g_n(x) dx = 1$ .)

It worth pointing out that sequential theories of generlized functions don't specify a specific $\{g_n\}$, rather they define the properties they want delta sequences to satisfy, then define Dirac as the common limit across all such delta sequences. If you don't do it this way you end up with $\int f(x) \delta^2(x) \, dx$ to be whatever you want.