# Integral of Dirac-delta function $t^2$.

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?

• 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. – Daniele Tampieri Sep 28 '18 at 13:15
• Could you give some context? From where have you got that integral? – 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

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

Then

\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 $$\varphi(0) = 0$$.