Define $$T: C_{0}^{\infty}(\mathbb{R})\to \mathbb{R}$$ by $$T(\phi)=\int_{-\infty}^{\epsilon}\frac{\phi(x)}{x}+\int_{-\epsilon}^{\epsilon}\frac{\phi(x)-\phi(0)}{x}+\int_{\epsilon}^{\infty}\frac{\phi(x)}{x}$$
I need to show that $T$ is independent of $\epsilon$.
For this I take $\epsilon $ and $\delta$ different from each other and both $\gt 0$ . Without loss of generality I assume that $\epsilon \gt \delta$. Then I consider the difference $$|\int_{-\infty}^{\epsilon}\frac{\phi(x)}{x}+\int_{-\epsilon}^{\epsilon}\frac{\phi(x)-\phi(0)}{x}+\int_{\epsilon}^{\infty}\frac{\phi(x)}{x}-\int_{-\infty}^{\delta}\frac{\phi(x)}{x}-\int_{-\delta}^{\delta}\frac{\phi(x)-\phi(0)}{x}-\int_{\delta}^{\infty}\frac{\phi(x)}{x}|$$ $$=|\int_{-\epsilon}^{-\delta}\frac{\phi(x)}{x}+\int_{-\epsilon}^{-\delta}\frac{\phi(x)-\phi(0)}{x}+\int_{\delta}^{\epsilon}\frac{\phi(x)}{x}+\int_{\delta}^{\epsilon}\frac{\phi(x)-\phi(0)}{x}|$$
Then by taking the maximum of upper bounds of $\phi$ and $\phi'$ as $M$, I came to $2M((\epsilon-\delta)+\log(\frac{\epsilon}{\delta}))$. This is no good for me as I cannot make it less than an arbitrary $\epsilon_1$.
Then I thought of another way:
I defined
$$T_n(\phi)=\int_{-n}^{\epsilon}\frac{\phi(x)}{x}+\int_{-\epsilon}^{\epsilon}\frac{\phi(x)-\phi(0)}{x}+\int_{\epsilon}^{n}\frac{\phi(x)}{x}$$
Then if I differentiate $T_n(\phi)$ with respect to $\epsilon$, it is easy to see that the result will come out to be $0$. Since this is true for all $n$ , I think it can be concluded from here for the general case as well.
Is this line of thinking correct??
Is there anyother way??
Thanks for the help!!
Since $\phi$ has compact support, there is an $N$ such that $n\ge N\implies T_n=T$. This makes your last argument correct.
Another aay too do it is to observe that if the support of $ºphi$ is contained in $[_N,N]$, then $$\begin{align} \int_{-\infty}^{-\epsilon}\frac{\phi(x)}{x}\,dx+\int_{\epsilon}^{\infty}\frac{\phi(x)}{x}\,dx&=\int_{-N}^{-\epsilon}\frac{\phi(x)}{x}\,dx+\int_{\epsilon}^{N}\frac{\phi(x)}{x}\,dx\\ \int_{-N}^{-\epsilon}\frac{\phi(x)-\phi(0)}{x}\,dx+\int_{\epsilon}^{N}\frac{\phi(x)-\phi(0)}{x}\,dx. \end{align}$$