What makes you think that it is not the definition?
What makes you think that it is?
The Dira delta "function" is defined using a non-standard Lebesgue integral of a continuous compactly supported function f(x), as a measure [imath]\delta (dx)[/imath] with the property
[imath]\displaystyle \int _a^b f(x) \, \delta (dx) = \begin{cases} f(0) & , \, a \leq 0 \leq b \\ 0 & , \, \text{else} \end{cases}[/imath]
which reduces to the expression I gave you above when we consider a Riemann integral over an interval on the real line.
Sometimes it is more loosely defined as a "function" with the property
[imath]\displaystyle \int _a^b f(x) \, \delta (x) \, dx = \begin{cases} f(0) & , \, a \leq 0 \leq b \\ 0 & , \, \text{else} \end{cases}[/imath]
for all smooth functions f(x), with the constraint
[imath]\displaystyle \int _{-\infty} ^{\infty} \delta (x) \, dx = 1[/imath]
though this is not strictly correct.
The statement
[imath]\delta (x) = \begin{cases} {\infty} & , x= 0 \\ 0 & , \, \text{else} \end{cases}[/imath]
is somewhat intuitive, but is not a valid definition of any function on the real numbers, and is not precise enough to define it properly anyway. There is too much wiggle room. As blamocur asks, how would you tell the difference between [imath]\delta (x)[/imath] and [imath]2 \delta (x)[/imath]?
Please check your source.
-Dan