The Dirac-delta function $$\delta(x-a)$$ goes to $\infty$ at $a$. I'm having trouble to understand what $$k\delta(x) \qquad k \in \mathbb{R}$$ then represents. How can I scale up (or down) infinity? Or does it just mean that $$\int_{-\infty}^{\infty} k\delta(x-a)dx=k $$
And if that's right, what's the difference in interpretation between $\delta(x-a)$ and $k\delta(x-a)$? Can someone please shed some light on the intuition here?
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You can think of the function like a limit of a parametrized family of functions.
With a growing value of a parameter the function's value at $x=0$ grows and its values at bigger $|x|$ vanish.
As a result we get functions, whose graphs become thinner and thinner while getting taller and taller.
Please see the animated example at Wikipedia:
https://en.wikipedia.org/wiki/File:Dirac_function_approximation.gif
The important thing is, we define those functions so that the integral from $-\infty$ to $+\infty$ of those functions in the family remains constant, equal $1$.
So using $\delta(x)$ in an integral-like expression works more or less like selecting the value of the other term at a specific argument.
The Dirac delta is not actually a function, but rather a distribution. However, the interpretation of it being a function helps to build some intuition as you say. The idea is that, because it is "infinite at some point $x=a$ and zero otherwise", when you integrate a function times the delta $\delta(x-a)$ you only get the contribution of the function at $x=a$, that is: $$\int_{\mathbb{R}}f(x)\delta(x-a)dx = f(a)$$ If $f$ is a constant function, say $f(x) = k$ for every $x \in \mathbb{R}$, then the above equality becomes: $$\int_{\mathbb{R}}k\delta(x-a)dx = k$$ because $f(a) = k$ trivially. Note that this last relation implies: $$\int_{\mathbb{R}}\delta(x-a)dx = 1$$ by the linearity of the integral.