Evaluating Dirac delta expression $\int_{-\infty}^\infty f(x, n) \delta(x - n) dx$ where $f(x,n) = (2x + 3n)^2$?

Question

The dirac delta function is usually defined as:

$$\int_{-\infty}^\infty f(x) \delta(x - n) dx = f(n).$$

But what if the function $f$ is also dependent on $n$? For example what if we had: $$f(x, n) = (2x + 3n)^2.$$

In this case can we say: $$\int_{-\infty}^\infty f(x, n) \delta(x - n) dx = f(n, n) = (2n + 3n)^2 = 25n^2.$$

Is this valid?

Answer 1

Yes, it is correct. the function $f(x,n)$ is really only a function of $x$ in this integral. You can think $n$ as a parameter.

Answer 2

Yes. Regard $n$ as a fixed parameter, and evaluate the rest of the integrand at the value of $x$ which makes the argument of $\delta$ vanish (in this case, at $x=n$), so you get $f(n,n)$ as you showed.