Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Zeta function for complex argument

  1. Jul 23, 2009 #1
    can we really give a definition of [tex] \delta (x-a-ib) [/tex] 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
  2. jcsd
  3. Jul 23, 2009 #2


    User Avatar
    Science Advisor

    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 [itex]\int_A f(x)g(x)dx[/itex] where "A" is some given set we use to define our generalized functions.

    The standard delta function, [itex]\delta(x)[/itex] is the functional that assigns f(0) to every function f. The "shifted" delta function, [itex]\delta(x- a)[/itex] is the functional that assigns f(a) to every function f. For a complex number, a+ bi, [itex]\delta(x-a-bi)[/itex] is the functional that assigns the value f(a+bi) to every function f. Of course, for that to make sense we must be talking about functions of complex numbers, not functions of real numbers only.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook