I need to prove that the order of the distribution given by $$\langle T,\varphi\rangle=\int_\mathbb{R}(\varphi(t^{-2},\sin t)-\varphi(0,\sin t)) dt,$$ where $\varphi\in C^\infty_0(\mathbb{R^2})$ is exactly 1. I have already shown that it is less than or equal to 1 but I haven't been able to show that it can't be 0. I am allowed to use the fact that the distribution $$\langle T',\varphi\rangle=\int_0^{+\infty}(\varphi(x)-\varphi(0))\frac{dx}{x^k}$$ has order 1 whenever $1<k<2$.
Could somebody give me a hint?
Using the variable substitution $t=1/s$ we get $$ \langle T,\varphi\rangle %= \int_\mathbb{R}(\varphi(t^{-2},\sin t)-\varphi(0,\sin t)) \, dt %= \left(\int_{-\infty}^0 + \int_0^\infty \right) (\varphi(t^{-2},\sin t)-\varphi(0,\sin t)) \, dt %= \{ s:=1/t \} %= \left(\int_{-\infty}^{0} + \int_0^{\infty} \right) (\varphi(s^2,\sin \frac{1}{s})-\varphi(\infty,\sin \frac{1}{s})) \, \frac{ds}{s^2} = \int_{\mathbb{R}} \left( \varphi(s^2,\sin \frac{1}{s}) - \varphi(0,\sin \frac{1}{s}) \right) \, \frac{ds}{s^2} $$
Now, $$ \varphi(s^2,\sin \frac{1}{s}) = \varphi(0,\sin \frac{1}{s}) + s^2 \varphi'_1(0,\sin \frac{1}{s}) + s^4 o(s) $$ so close to $s=0$ the integrand is $\varphi'_1(0,\sin \frac{1}{s})$ suggesting that $T$ has order 1.
Perhaps you can make this more precise?