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Sorry to add to all the delta function questions already here but I couldn't find a related one.

Can anyone explain why $ \int_{-\infty}^{\infty} f(x) \delta(x^2-9)=\frac 16(f(-3)+f(3)) $?

I can have tried rewriting the delta function as $\delta((x+3)(x-3))$ but am unsure where to proceed from here.

Thanks in advance

user1887919
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1 Answers1

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We will split the integral in two and make a substitution. $$\int_0^\infty f(x)\delta(x^2-9)\,dx=\int_0^\infty f(\sqrt x )\delta (x-9)\dfrac{1}{2x} \,dx=\dfrac{f(3)}{6}$$ Similarly, evaluating the other part of the integral and adding we get the answer.

gebruiker
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Ishan Banerjee
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