Function collection with specific properties

I'm looking for at least one answer to each number. If I know some function that holds I will put the answer, but, if there others, I would like to know too.

If is there's a similar function with the property described, I would like to know too.

These functions could be very useful to join formulas, describe completely some kind of properties and so many applications. Several times I need something similar to this, so I decided search and collect.

PS.: is prefereable to me functions defined in the complex domain or $f_{x}:\mathbb{C}\implies\mathbb{C}$, where x is the function number. But, probably will be interesting to the functions 1 to 6, answers defined on reals ($f_{x}:\mathbb{R}\implies\mathbb{R}$).

Is there a function that:

Is import to say, below we have some function definitions, but this isn't the kind of definition what I'm looking for. I'm searching for definitions who didn't use comparision to find its value. For example, $f_1(k)$ is equal to 1, as defined, but this value was found using comparision. But using the Kronecker Delta, we find this, without comparision. And this is the strong power of the Kronecer Delta. The full power of these functions can be viewed when applicating $f_6$ to some problems, for example.

1)$f_1(x)=\begin{cases} 0 & x\ne k \\ 1 & x= k \end{cases}$

2)$f_2(x)=\begin{cases} 0 & x< k \\ 1 & x\geq k \end{cases}$

3)$f_3(x)=\begin{cases} 0 & x< k \\ 1 & x> k \end{cases}$

4)$f_4(x)=\begin{cases} 0 & x< k \\ i & x= k \\ 1 & x> k \end{cases}$

5)$f_5(x)=\begin{cases} 0 & x\in(\mathbb{Q}-\mathbb{Z}) \\ 1 & x\in \mathbb{Z} \end{cases}$

6)$f_6(x)=\begin{cases} 0 & x\in \mathbb{Q}\\ 1 & x \in\mathbb{I} \quad(\mathbb{I}=\mathbb{R}-\mathbb{Q}) \end{cases}$

7)$f_7(x)=\begin{cases} 0 & x \in \mathbb{R}\\ 1 & x \in (\mathbb{C}-\mathbb{R}) \end{cases}$

8)$f_8(x)=\begin{cases} 0 & x \in \mathbb{R}\\ i & x=a+bi \quad (a,b\in \mathbb{R}-\{0\})\\ 1 & x=ai \quad (a\in \mathbb{R}-\{0\}) \end{cases}$

I'm not looking for pieceswise answers like $f(x)=\begin{cases} 0 & x< 0 \\ 1 & x\geq 0 \end{cases}$ that we need comput using comparision, but maybe special functions like Headvise step function($H(x) = \int_{-\infty}^x { \delta(t)} \, \mathrm{d}t$).

I will write the answers given or collected in the next section. Appreciate the help, thanks.

1)

Kronecker Delta

$\delta_{x,k} = \frac1{2\pi i} \oint_{|z|=1} z^{x-k-1} dz=\frac1{2\pi} \int_0^{2\pi} e^{i(x-k)\varphi} d\varphi$

2)

Maybe a interpretation of Heaviside Step Function, not sure:

$H(x-k) = \int_{-\infty}^{x+k} { \delta(t)} \, \mathrm{d}t$

3)

4)

5)

6)

7)

8)

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In each case (once you've specified your domain and codomain), you have defined a function. Are you possibly asking for formulas (of a certain type) for these functions? –  Bill Cook Apr 14 '12 at 1:29
So you're asking how to define these functions as integrals of other functions? –  Alex R. Apr 14 '12 at 1:32
@Sam, I would say yes. –  GarouDan Apr 14 '12 at 1:39
@GarouDan: you are probably receiving downvotes because you ask whether a function exists even though the answer is obvious. Yes a function exists, given by the definitions you state. You should clarify the question to narrow down the types of formulas you consider acceptable. –  Grumpy Parsnip Apr 15 '12 at 0:15
@GarouDan I find this answer pointless. As Bill says, you have already defined the functions you are "looking" for. –  Pedro Tamaroff Apr 15 '12 at 0:41

$$f_2(x) = \frac{x-k}{\sqrt{(x-k)^2 + \epsilon^2}}+1$$
This will get sharper as $\epsilon \to 0$.
Eg. Let $k = 3$, $\epsilon = 0$.
PatrickMcLaren,$\lim_{e\to 0}\sqrt{-\frac{x-k}{\sqrt{e^2+(x-k)^2}}}$ is very similar to $f_4(x)$. A bit different you propose. I had implement your function and it didnt't hold to $f_2(x)$, I think we can use it to define $f_4(x)$. Am I correct? (Very interesting function, thx.) –  GarouDan Apr 14 '12 at 15:32
Sorry, there was an error in the sign. Your example should work for $f_4(x)$, yes. –  Patrick McLaren Apr 14 '12 at 16:46
PatrickMcLaren. Looks almost works. $f_2(3)$, when $k=3$ and $\epsilon$ tends to $0$ is equal 1. So $f_2(2)=0$,$f_2(3)=1$,$f_2(4)=2$. This what the Mathematica says. You confirm? –  GarouDan Apr 15 '12 at 1:12