   Dirac delta squared
  So I was reading about the delta function on wikipedia, and as most textbooks it gives examples with the derivative, and integration with substitution and such. However I cant remember having seen any explanations with powers of it.For example integral of (f(x) δ(x))2 dxUsing integration by parts it looks like it must be zero. Allthough that doesnt make much sense to me. But since the dirac delta isnt really a function I'm not even sure if the IBP rules apply. Wolfram alpha doesnt seem to know either. Clues anyone?(putting this in the lounge as it's not game related)   
  Taking the dirac delta function as the limit of the gaussian function as the variance approaches zero, some quick mathematica work suggests that the square of the function has infinite area.   ;  I've only ever seen the dirac delta used to represent an impulse in a dynamic systems class I took.  It's an entity with zero width and infinite height, so I don't know if you can just square it and expect anything sensible.  Perhaps you might need to look at approaching the problem from the frequency domain?And that is where my knowledge on the subject ends :(   ;  Here is what I did:∫ (f(x) δ(x))2 dxU = f(x)δ(x), U' = f'(x)δ(x) + f(x)δ'(x)V' = f(x)δ(x),  V = f(0)so we getf(x)δ(x)f(0) - ∫ f'(x)δ(x) dx - ∫ f(x)δ'(x) dx (from -∞ to ∞)By the definition of the dirac delta funtion we havef(x)δ(x)f(0)|∞∞ = 0 - 0 = 0 (since δ(x) is zero anywhere except at x = 0)∫ f'(x)δ(x) dx = f'(0)∫ f(x)δ'(x) dx = -f'(0)So the whole thing cancels out. Any flaws here?   ;  The formula(*) ∫∞∞ f(x) δ(x - x0) dx = f(x0)would seem to imply on substituting δ for f that∫∞∞ δ(x) δ(x - x0) dx = δ(x0).However, since δ is a generalized function (distribution) and f is supposed to be an ordinary function in (*), I suspect that this manipulation is not valid, especially considering that if one were to set x0 = 0, then one would get the answer δ(0), which is "infinite." As a heuristic, though, this suggests that the integral of δ2 should be infinite.Honestly, though, I don't know. [smile]   ;  Well the thing is, I started looking into this because I was trying to minimize the second derivative of functions that may have discontinous derivatives at certain places, eg:The second derivative of this function can be defined as a delta function (multiplied by the difference in f'(x) between the two sides) where f'(x) is discontinous.Now it's easy to minimize |f''(x)| this way (btw I mean minimize with respect to other parameters, not find the minimum or anything), but it looks likely that it should be possible to minimize (f''(x))2 too. It doesnt make sense to me that it should be zero or infinite.Edit: I also tried using δ(x) = lim (b -> 0) sqrt(pi/2)b-1 exp(-(x/b)2) like Sneftel did and arrived at infinity [Edited by - Jesper T on August 4, 2009 2:29:35 PM];  Pass the dutchie 'pon the left hand side...   