The closed form of of the Dirac Delta Function I know that $\delta(x)$ is the Dirac Delta function - satisfying -$$\int^{\infty}_{- \infty}\delta(x)dx=1$$Can anyone tell me a closed form of  $\delta(x)$ . I guess it might be in a form of $e^{a(x)}$ for some $a(x)$
 A: The Dirac delta function is not a function. Every function which is zero everywhere and "not zero" at zero has integral zero - changing the value of the function at one point shouldn't change the integral (since you can make the partition of the domain of integration arbitrarily small around that bad point).
Think about the other part of the definition $\int \delta \phi = \phi(0)$. This is closer to how the Dirac Delta "function" is defined - it is actually a linear functional (a linear function going to the real numbers) on a space of "test functions," which is the vector space of all functions (typically $R^n \to C$) that are infinitely differentiable and which are zero outside of some bounded set (i.e. compactly supported). In this case, the dirac delta functional is the linear map that takes a test function and sends it to its value at zero. It is easy to check that this is linear. 
It is also true that this is a continuous linear functional when you put the appropriate topology on the space of test functions, but communicating the construction of that topology few words is exceedingly difficult (without using phrases like inductive limit of Frechet spaces, at least). If you take the space of all of these continuous linear functionals on the test functions you get what are called distributions. Intuitively, you can think of distributions as the family of functions that associate each test function to some number in a way that is linear and also continuous with respect to the maximum values of the partial derivatives of each test function. (So that if two test functions have similar partial derivatives of all orders, then they are sent to nearby numbers.)
Many of these distributions have no closed form (in the sense that you want), including the dirac delta function. (You can prove that there is no function $f$ that satisfies $\int f \phi = \phi(0)$ for all test functions $\phi$.) However, many familiar functions can be thought of as distributions, just by taking sending $f$ to the functional $\phi \to \int f \phi$ - at least if you are willing to ignore differences between functions that occur on "measure zero sets". (Which is to say, {the constant function 1} and {the function which is constantly 1 except when it is zero when it is 12} are the same as distributions. A point is a measure zero set. Any countable set is a measure zero set. The middle thirds cantor set is also a measure zero set. Fat cantor sets are not measure zero sets. Functions which are identical except on measure zero sets give rise to the same distribution because integration ignores measure zero sets.)
What is cool about this is that if you think about a function as a distribution instead of as a "function", then placing it inside the vector space of distributions actually allows you to assign meaningful partial derivatives of every order. These are called its distributional derivatives. For example, the function $H$, which is zero when $x \leq 0$ and 1 when $x > 0$, has a distributional derivative equal to the dirac delta functional. The way this works is that we define the distributional derivative to be a formal differentiation by parts (this is not nonsense - it is the transposition of the differentiation operator from the space of test functions to itself up to the space of distributions, the idea being that you can take a differentiation of a distribution and pull it down to a differentiation on the test functions it is acting on. This is like taking the transpose of a matrix, but in an infinite dimensional vector space.)
$\int H' \phi = - \int H \phi' = - \phi (\infty) + \phi(0) = \phi(0)$, where the $\phi(\infty)$ term disappears because each of the  functions $\phi$ are zero outside of some bounded set (which may vary for each $\phi$.)
If you draw pictures of these functions it makes some intuitive sense that the derivative of $H$ is the dirac delta function - there is a spike at zero, and it is zero everywhere else.
Maybe this gives you some of the intuition you are looking for. 
A: As others have said, it is not a function at all.  Sometimes you can do something that gets close.  Though there is no simple formula, but you can sometimes use any highly peaked function of unit area.  So $f(x)=\lim_{a \to 0} \begin {cases} 0& |x| \gt a\\\frac 1{2a} & |x| \le a \end {cases}$ or a Gaussian with width going to zero are two possibilities.  Your answer should not depend on which you use. But you will always have some sort of limit to get the $f(x)=0$ if $x \neq 0$
A: The Dirac Delta is a distribution, that is, a linear functional
$$\delta: \bigg\{\mbox{functions}\bigg\}\longrightarrow \mathbb{C}.$$
It is an evaluation map that can be expressed as
$$\delta(f) = f(0).$$
One often thinks of distributions as represented by functions --- in this case, as a function whose graph is a spike at the origin --- and writes them as functions, e.g. $\int \delta(x)f(x)\ dx = f(0)$, but they are not, strictly speaking, functions.
A: I've created the following function which will approach dirac delta function outputs
$$\lim_{c \to \infty} \left [ (\frac{y}{y-x})\cdot((y-c)^2+x^2)=0 \right ]$$
The function was formed by taking the function for a horizontal line at 0, divided by a 45º line (y=x) through the origin to create a hole in the function at x=0, then multiplied by a circle with radius 0 whose origin approaches (0,∞) to create a point which approaches (0,∞). As x reaches 0 y while approach infinity to cancel out the c value (y-origin of the circle point) thus bringing the evaluation back to zero. Y can’t reach infinity, only approach it (at the same rate c approaches infinity) otherwise you will be left with the undefined form $\frac{\infty}{\infty}$. For all other x-values, y will be 0 thanks to $\frac{y}{y-x}$. 
Because of a limit evaluation at x=0 wolfram alpha will fail since it doesn’t support this, but you can evaluate by hand and see for yourself that these terms will approach these values.
