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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?

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    $n$ is a constant and not a variable. So yeah, what you wrote the second time is indeed true. – 2017-02-15

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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.

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    I was also wondering about the more complicated case involving vectors in $\mathbb{R}^3$? - http://math.stackexchange.com/questions/2145872/evaluating-the-dirac-delta-expression-int-mathbbr3-sum-n-in-mathbbz – 2017-02-15
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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.