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$\delta(x)$ is the Dirac delta function and $\epsilon>0$

$\int_{-\epsilon}^\epsilon\delta(ax)dx$=$\int_{-\epsilon}^\epsilon\delta(u)d(\frac ua)=\frac1a$

My question is does this hold more generally? For a smooth function $f(x)$ where there is a zero at the origin and there is a radius $\epsilon$ about the origin(let this be the region $R$) that is non-zero, and a smooth function $g(x)$ that is smooth on $R$, does $\int_{-\epsilon}^\epsilon\delta(f(x))g(x)dx=\frac{g(0)}{f'(0)}$?

So does $\int_{-.005}^{.005}\delta(7x+x^2)(14x+37\cos(x))dx=\frac{37}{7}$?

Jacob Claassen
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2 Answers2

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In general, no - your calculations end up with you getting some term with $\frac{g(0)}{f'(0)}$, but actually this term should be $\frac{g(x)}{f'(x)}$. In your first example, you got lucky that $g$ and $f'$ were independent of $x$, so they looked like they were evaluated at $0$.

What it would simplify to is $$\int_{-\epsilon}^{\epsilon}\delta(f(x))g(x)\,dx=\int_{f(-\epsilon)}^{f(\epsilon)}\delta(u)\cdot\frac{g(x)}{f'(x)}\,du$$ We then evaluate this at $u=0$, assuming that $f(-\epsilon)$ and $f(\epsilon)$ retain their sign, allowing $0$ to remain within this range. We then evaluate $f'$ and $g$ at the points when $u=f(x)=0$. In your example, this does indeed happen to be at $x=0$, but this would not be the case for arbitrary $f$. This is the reason why your second integral worked out to be $\frac{37}7$. As long as $f$ satisfies these conditions, then your rule holds.

In general, it is a struggle to convert that $x$ term into something in terms of $u$.

John Doe
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The identity you are looking for is $\delta(f(x))=\sum_i \frac{\delta(x-x_i)}{|f'(x_i)|} $, where $x_i$ are the roots of $f$.

Cf. http://mathworld.wolfram.com/DeltaFunction.html

Hyperplane
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