I'm trying to evaluate

$$ \int_{-\infty}^\infty{\delta(t^2)} dt $$

If it were just $ \delta (t) $ the area would be $1$, but I don't think this is the case here because the area of the function should be smaller since it approaches $0$ faster.

I attempted a change of variables with $\tau = t^2$, but that ends up with

$$ \int_{-\infty}^\infty\ \frac{\delta(\tau)}{2\sqrt{\tau}} d\tau $$

However, if I do this, it should evaluate to $ \frac{1}{2\sqrt{0}}$ which would be undefined.

Is there a different change of variables I should attempt?

  • $\begingroup$ I have removed my answer since it was not correct: thank to @reuns for pointing out the main error. However, $\delta(t^2)$ is not a distribution, as I clearly stated. $\endgroup$ – Daniele Tampieri Sep 28 '18 at 13:15
  • $\begingroup$ Could you give some context? From where have you got that integral? $\endgroup$ – md2perpe Sep 28 '18 at 14:28

It is not well-defined. Consider $\delta_{\epsilon}(t) = \frac{1}{2\epsilon}\mathbf{1}_{[-\epsilon,\epsilon]}(t)$ which we know to converge to $\delta$ as distribution. Then

$$ \int_{-\infty}^{\infty} \delta_{\epsilon}(t^2) \, dt = \frac{1}{\sqrt{\epsilon}} \to \infty \quad \text{as} \quad \epsilon \to 0^+. $$

On the other hand, we can give a reasonable interpretation of $\delta(t^2)$ on a suitable space of test functions. Indeed, define $\varphi_+'(0) := \lim_{h \to 0^+} \frac{\varphi(h)-\varphi(0)}{h}$ (resp. $\varphi_-'(0) := \lim_{h \to 0^-} \frac{\varphi(h)-\varphi(0)}{h}$) if the limit exists. Then consider the space

$$\mathcal{A} = \{ \varphi \in C_c(\mathbb{R}) : \text{$\varphi(0) = 0$ and $\varphi_{\pm}'(0)$ exist.} \}$$


\begin{align*} \int_{-\infty}^{\infty} \varphi(t) \delta_{\epsilon}(t^2) \, dt &= \frac{1}{2\epsilon} \int_{-\sqrt{\epsilon}}^{\sqrt{\epsilon}} \varphi(t) \, dt \\ &= \frac{1}{2} \int_{-1}^{1} \frac{\varphi(\sqrt{\epsilon}u) - \varphi(0)}{\sqrt{\epsilon}} \, du \qquad (t = \sqrt{\epsilon}u) \\ &\xrightarrow[\epsilon \to 0^+]{} \frac{1}{2} \left( \int_{0}^{1} \varphi_+'(0) u \, du + \int_{-1}^{0} \varphi_-'(0) u \, du \right) \\ &= \frac{\varphi_+'(0) - \varphi_-'(0)}{4}. \end{align*}

In view of this computation, $\delta(t^2)$ can be realized as what captures the 'jump discontinuity' of $\varphi'$ at $0$, provided $\varphi(0) = 0$.


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