Let $\varphi$ be a test function (belonging to the set of smooth functions with compact support) We define the distribution principal value of $1/x$:
$$\left\langle\operatorname{p.\!v.}\left(\frac{1}{x}\right),\varphi\right\rangle: =\lim_{\varepsilon\to 0^+}\left[\int_{-\infty}^{-\varepsilon} \frac{\varphi(x)}{x} \, \mathrm{d}x + \int_{\varepsilon}^{+\infty} \frac{\varphi(x)}{x} \, \mathrm{d}x\right]$$
How do I show it is an odd distribution?
An odd distribution $T$, is such that $T(-t)=-T(t)$
My try:
\begin{align}\left\langle\operatorname{p.\!v.}\left(\frac{1}{x}\right),\varphi\right\rangle&=\lim_{\varepsilon\to 0^+}\left[\int_{-\infty}^{-\varepsilon} \frac{\varphi(-x)}{-x} \, \mathrm{d}(-x) + \int_{\varepsilon}^{+\infty} \frac{\varphi(-x)}{-x} \, \mathrm{d}(-x)\right]\\\left\langle\operatorname{p.\!v.}\left(\frac{1}{-x}\right),\varphi\right\rangle&=\lim_{\varepsilon\to 0^+}\left[\int_{-\infty}^{-\varepsilon} \frac{\varphi(-x)}{x} \, \mathrm{d}(x) + \int_{\varepsilon}^{+\infty} \frac{\varphi(-x)}{x} \, \mathrm{d}(x)\right]\end{align}
But I don't know what to do with the $\varphi(-x)$, since it is an arbitrary test function, I can't say of it is odd or even. Besides I am not sure if it is ok to do $\left\langle\operatorname{p.\!v.}\left(\frac{1}{-x}\right),\varphi\right\rangle$ in the first place, because of the angular brackets.
How do I go about it?
\begin{align}\left\langle\operatorname{p.\!v.}\left(\frac{1}{x}\right),\varphi\right\rangle&=\lim_{\varepsilon\to 0^+}\left[\int_{-\infty}^{-\varepsilon} \frac{\varphi(x)}{x}\,dx+ \int_{\varepsilon}^{+\infty} \frac{\varphi(x)}{x}\,dx\right]\\&=\lim_{\varepsilon\to 0^+}\int_{\varepsilon}^{+\infty} \frac{\varphi(x)-\varphi(-x)}{x}\,dx\\\left\langle\operatorname{p.\!v.}\left(-\frac{1}{x}\right),\varphi\right\rangle&=\lim_{\varepsilon\to 0^+}\int_{\varepsilon}^{+\infty} \frac{\varphi(-x)-\varphi(x)}{-x}\,d(-x)\\&=-\left\langle\operatorname{p.\!v.}\left(\frac{1}{x}\right),\varphi\right\rangle\end{align}