The discussion centers on the definition of the delta function, specifically \(\delta(x-a-ib)\), where \(a\) and \(b\) are real numbers and \(i\) represents the square root of -1. It is established that a generalized function, or distribution, does not possess a value at a specific point, but rather acts as a functional that assigns values based on integrals over a set. The standard delta function \(\delta(x)\) assigns \(f(0)\) to any function \(f\), while the shifted delta function \(\delta(x-a)\) assigns \(f(a)\). For complex arguments, \(\delta(x-a-bi)\) assigns \(f(a+bi)\) to functions of complex numbers.
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No, you don't. A generalized function (distribution) does not HAVE a value at a specific point. A generalized function is a "functional" that assigns a number to every function. We can think of the functions as a subset of the generalized functions by saying that the function f(x) is the functional that to any function g(x) assigns the number \int_A f(x)g(x)dx where "A" is some given set we use to define our generalized functions.zetafunction said:can we really give a definition of \delta (x-a-ib) a,b real and 'i' means the square root of -1
if i try it in the sense of generalized function for any x a and b i get the result oo unless b is zero