Definition of dirac delta function why is it $1 / \epsilon$ and not some constant / epsilon? The definition that $$\int_{-\infty}^{\infty} \delta(t) = 1$$
is based off of: $$\lim_{\epsilon\to 0} 1/\epsilon * \epsilon$$
So the limit of the value would approach one, but what if you change the function to $2/\epsilon$ or any other arbitrary constant. 
In that case, the dirac delta function would become: $$\int_{-\infty}^{\infty} \delta(t) = c$$ 
Which would lead to different answers when solving differential equations that include the delta function. 
It's different, but wouldn't it still be a correct model for the derivative of the step function?
 A: The Dirac delta is, in its strictest definition, a linear function on the space of continuous functions defined as
$$
\langle δ,f\rangle=f(0).
$$
Formally, but not with any integration theory behind it, this is also written as
$$
\int δ(x)·f(x)\,dx = f(0)
$$
Inserting the constant function $f(x)=1$ leads to
$$
\int δ(x)\, dx = 1
$$
Again, this is not an integral, this is the evaluation of the constant $1$ at $x=0$ written as if it were an integral.
There exist several approximation of the Dirac delta of the form $\phi_ϵ(x)=\frac1ϵ\phi(\frac xϵ)$ where $\phi$ is non-negative, integrable over $\Bbb R$ with integral value $1$. For these the integral
$$
\int_{\Bbb R}\phi_ϵ(x)·f(x)\,dx=\int_{\Bbb R}\phi(u)·f(ϵu)\,du
$$
is sensible as Riemann or Lebesgue integral, if the function classes of $f$ and $\phi$ match ($f$ continuous and $\phi$ with compact support or $f$ tempered and $\phi$ fast falling or ...)
With the given assumptions on $\phi$ one then gets
$$
\lim_{ϵ\to0}\int_{\Bbb R}\phi_ϵ(x)·f(x)\,dx=f(0),
$$
but in this the limes and the integral can not be exchanged in any analytical sense, it is only formally or symbolically that the limes is written as if one could exchange them.
