# I Prove that ∫f(x)δ(x)dx=f(0)

1. Jan 22, 2017

### 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!

2. Jan 22, 2017

### PeroK

What's your definition of $\delta$?

3. Jan 22, 2017

### LLT71

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

4. Jan 22, 2017

### PeroK

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

5. Jan 22, 2017

### LLT71

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 gonna attack this problem now or better to say how I'm gonna attack it anyway?

6. Jan 22, 2017

### PeroK

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

7. Jan 22, 2017

### LLT71

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?

8. Jan 22, 2017

### PeroK

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.

9. Jan 22, 2017

### LLT71

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!

10. Jan 22, 2017

### PeroK

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.

11. Jan 22, 2017

### LLT71

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 wanna give it a try. and yes, I will be very thankful!

12. Jan 22, 2017

### PeroK

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.

13. Jan 22, 2017

### LLT71

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.

14. Jan 22, 2017

### micromass

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

15. Jan 22, 2017

### 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.

16. Jan 22, 2017

### micromass

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

17. Jan 22, 2017

### FactChecker

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

18. Jan 22, 2017

### micromass

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

19. Jan 22, 2017

### FactChecker

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

20. Jan 22, 2017

### micromass

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

21. Jan 22, 2017

### FactChecker

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.

Last edited: Jan 22, 2017
22. Jan 23, 2017

### LLT71

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).

23. Jan 23, 2017

### PeroK

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.

24. Jan 23, 2017

### 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).

25. Jan 23, 2017

### FactChecker

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 )