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I need to evaluate $\int_0^\infty \phi(x)\delta(x^2-1)\, dx$ but have very little context for knowing which steps are valid when working with $\delta$. A $u$-substitution seems appropriate, so instead finding

$$\int_0^\infty \phi(x)\delta(u)\frac{du}{2x}$$

however, I'm not sure of how to deal with the $\phi(x)/x$. Should I next infer $x = \pm\sqrt{u+1}$ and then infer that this integral is

$$\frac{\phi(-1)}{-2} + \frac{\phi(1)}{2}$$

But that seems to rely on $\phi$ being defined for negative values, which seems like it shouldn't be necessary.

For context it may help to know that I only have by definition that

$$\int_{-\infty}^\infty \phi(x)\delta(x-x_0)\ dx = \phi(x_0)$$

Addem
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  • For $f\in C^1$ with finitely many zeros where $f' \ne 0$ and $\phi \in C^0$ then $\int_{-\infty}^\infty \phi(x)\delta(f(x)) dx \overset{def}= \lim_{\epsilon \to 0^+}\int_{-\infty}^\infty \phi(x)\frac{1_{|f(x)| < \epsilon}}{2\epsilon} dx $ is well-defined. – reuns Nov 28 '17 at 21:31
  • @reuns I had a mistake in my question, I've edited. – Addem Nov 28 '17 at 21:48
  • @reuns If I understand the comment, this would suggest that I look at $\lim_{e\rightarrow 0^+}\int_{1-\epsilon}^{1+\epsilon}\frac{\phi(x)}{2\epsilon}\ dx$, yes? – Addem Nov 28 '17 at 21:50
  • Note that since the integral is over $x$ from zero to infinity, we use only the positive square roots. – Joel Nov 28 '17 at 21:54
  • Since $f(1) = 0$ and $f(x) = (x-1) f'(1)+o(|x-1|)$ $\quad\lim_{\epsilon\rightarrow 0^+}\int_0^2\phi(x)\frac{1_{|f(x)| < \epsilon}}{2\epsilon}\ dx=\lim_{\epsilon\rightarrow 0^+}\int_{1-\frac{\epsilon}{|f'(1)|}}^{1+\frac{\epsilon}{|f'(1)|}}\frac{\phi(x)}{2\epsilon}\ dx= \ldots$ – reuns Nov 28 '17 at 22:07
  • @reuns I have no clue how to make sense of what is going on there. It may help to know that this is in the context of an introductory ODE course where the only prereqs are Linear Algebra and Calculus, and I just happen to have a decent grasp of introductory Analysis so I can follow this material better than the average student in the class ... but still I don't get why you would seem to linearize at 1, nor why you'd divide $\epsilon$ by $f'(1)$. – Addem Nov 28 '17 at 22:16

1 Answers1

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Consider the integral $\int_0^\infty \phi(x) \delta(x^2-1)dx$. Proceed with a $u$ substitution setting $u = x^2 -1$, which yields $du = 2xdx$.

We select $x = \sqrt{u+1}$. We only need the positive square root, since $x \in [0,\infty)$ according to the integration bounds. This gives $dx = \frac{du}{2\sqrt{u+1}}$

Making the substitutions: $$\int_0^\infty \phi(x) \delta(x^2-1)dx = \int_{-1}^\infty \frac{\phi(\sqrt{u+1})}{2\sqrt{u+1}} \delta(u) du = \frac{\phi(1)}{2}$$

I should mention that $\int_{a}^b \phi(x) \delta(x-x_0) dx = \phi(x_0)$ for any interval $[a,b]$ that contains $x_0$. The integral is zero when $x_0 \not \in [a,b]$. We don't require the knowledge of $\phi$ over all of $(-\infty, \infty)$, since the delta function is zero everywhere but at the origin (in the undergrad ODE sense, that is. the delta function isn't technically a "function").

Joel
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