Prove order of distribution $\Lambda_{1/x}$ is 1 The distribution is defined as:
$$\Lambda_{1/x}(\varphi)=\lim_{\varepsilon\rightarrow0+}\int_{\mathbb{R}\backslash(-\varepsilon,\varepsilon)}\frac{\varphi(x)}{x}\ \mathrm{d}x$$
I tried integrating per partes. In my result $\varphi_{1,2}$ denotes the first/second primitive function.
$$\Lambda_{1/x}(\varphi)=\lim_{\varepsilon\rightarrow0+}\frac{\varphi_1(\varepsilon)-\varphi_1(-\varepsilon)}{\varepsilon}+\int_{\mathbb{R}\backslash(-\varepsilon,\varepsilon)}\frac{\varphi_1(x)}{x^2}\ \mathrm{d}x=$$
$$=\varphi(0)+\lim_{\varepsilon\rightarrow0+}\frac{\varphi_2(\varepsilon)-\varphi_2(-\varepsilon)}{\varepsilon^2}-2\int_{\mathbb{R}\backslash(-\varepsilon,\varepsilon)}\frac{\varphi_2(x)}{x^3}\ \mathrm{d}x=$$
Now because the Maclaurin series of $\varphi_2(\varepsilon)-\varphi_2(-\varepsilon)=2\varphi_2'(0)\varepsilon+O(\varepsilon^3)=2\varphi_1(0)\varepsilon+O(\varepsilon^3)$ so if I evaluate the limit of the fraction right away it diverges.
I'm stuck here. Is my way correct? How do I continue? Or is there a better way?
 A: I found answer to my question here
The solution is using the easily obtained identity
$$\lim_{\varepsilon\rightarrow0+}\int_{\mathbb{R}\backslash(-\varepsilon,\varepsilon)}\frac{\varphi(x)}{x}\ \mathrm{d}x=\lim_{\varepsilon\rightarrow0+}\int_{\varepsilon}^{+\infty}\frac{\varphi(x)-\varphi(-x)}{x}\ \mathrm{d}x$$
As $\varphi\in\mathbb{C}^\infty$ the mean value theorem tells me that $\exists\xi\in(-x,x):\frac{\varphi(x)-\varphi(-x)}{2x}=\varphi'(\xi)$
$$\Rightarrow\frac{\varphi(x)-\varphi(-x)}{x}\leq2\sup\{|\varphi'(\eta)|;\ \eta\in(-x,x)\}$$
$\varphi$ has compact support $K\subset(-a,a)$. This implies:
$$\lim_{\varepsilon\rightarrow0+}\int_{\mathbb{R}\backslash(-\varepsilon,\varepsilon)}\frac{\varphi(x)}{x}\ \mathrm{d}x\leq2a\sup\{|\varphi'(\eta)|;\ \eta\in(-x,x)\}$$
From this we know that the order is $\leq1$. Now the order cannot be 0. Let's define $\varphi_e:\ 0<\varphi_e<1,\ \mathrm{supp}\varphi_e\subset[e,4e]$, $\varphi_e=1$ on $[2e,3e]$. Then
$$\lim_{\varepsilon\rightarrow0+}\int_{\varepsilon}^{+\infty}\frac{\varphi_e(x)-\varphi(-x)}{x}\ \mathrm{d}x\geq\frac{1}{4e}\sup\{|\varphi(\eta)|;\ \eta\in(-x,x)\}$$
