I'm trying to solve this exercise
If $T \cdot x=1$ in the sense of distributions then $T=\textrm{p.v.}\left(\frac{1}{x}\right)+c \delta_{0}$
Where $\delta_{0}$ is the Dirac delta distribution centered at the origin and $$\left\langle\textrm{p.v.}\left(\frac{1}{x}\right), \varphi\right\rangle=\lim _{\varepsilon \rightarrow 0} \int_{\mathbb{R} \backslash(-\varepsilon, \varepsilon)} \frac{\varphi(x)}{x} d x$$
For any test function $\varphi \in C_c^\infty(\mathbb{R})$
Here's my attempt, equality in the distribution sense must mean equality for any test function, so
$$\int_{\mathbb{R}}T\cdot x \varphi dx = \int_\mathbb{R} \varphi dx \implies \int_\mathbb{R} \varphi (T \cdot x - 1)dx = 0 \implies \int_\mathbb{R} x\varphi (T - \frac{1}{x})dx = 0$$
Since $\varphi$ is any arbitrary test function $\psi = x \varphi$
$$\implies \int_{\mathbb{R} \setminus (-\varepsilon, \varepsilon)} \psi (T - \frac{1}{x})dx + \int_{-\varepsilon}^\varepsilon \psi (T - \frac{1}{x})dx = 0$$
Taking $\varepsilon \rightarrow 0$ gives
$$\int_{\mathbb{R}}T\psi = \lim _{\varepsilon \rightarrow 0} \int_{\mathbb{R} \backslash(-\varepsilon, \varepsilon)} \frac{\psi(x)}{x} d x$$
So $T$ is the Cauchy principal value. But where does the dirac delta appear? If I know nothing about $T$ is it justified to assume that $\langle T, \varphi \rangle = \int_\mathbb{R} T\varphi$ i.e. "integral of something" times the test function? The dirac delta does not seem to really follow example as its just defined as $\langle \delta, \varphi \rangle = \varphi(0)$, nor does the principal value.
