I'm reading the original article on distance covariance (link), and throughout the article the author uses the following lemma:
Can someone please explain what he actually means by "principal value sense". If we use the limit of integrals over increasing annulus, then with $A_\varepsilon= \mathbb{R}^d \setminus (B(0,\varepsilon)\cup B(0,1/\varepsilon)^c)$ we easily see that $$ 0\leq 1_{A_\varepsilon}(t) \frac{1-\cos(t^\intercal x)}{\|t\|^{d+\alpha}} \uparrow 1_{\mathbb{R}^d \setminus \{0\}}(t) \frac{1-\cos(t^\intercal x)}{\|t\|^{d+\alpha}} , $$ when $\varepsilon \downarrow 0$ (does not make sense for negative $\varepsilon$), for any $t\in\mathbb{R}^d$. Thus Lebesgue's monotone convergence theorem yields that $$ \lim_{\varepsilon \to 0} \int_{\mathbb{R}^d \setminus A_\varepsilon} \frac{1-\cos(t^\intercal x)}{\|t\|^{d+\alpha}} \, d\lambda^d(t) = \int_{\mathbb{R}^d \setminus \{0\}} \frac{1-\cos(t^\intercal x)}{\|t\|^{d+\alpha}} \, d\lambda^d(t), $$ so I'm wondering why they haven't just stated the lemma in terms of this integral.
I obviously haven't understood the term “principal value sense” correctly.
Can anyone please provide me with the correct interpretation of the integral in the lemma?
Your interpretation is correct. This integral converges, so the principle value is unnecessary. However, if you want to consider
\begin{equation*} \int_{\mathbb{R}^d} \frac{1-\exp(i\left<t,x\right>)}{\|t\|^{d+\alpha}}dt \end{equation*}
for $\alpha>1$, then you need the principal value. Notice that
\begin{equation*} \int_{\mathbb{R}^d\backslash A_{\epsilon}} \frac{\sin(\left<t,x\right>)}{\|t\|^{d+\alpha}}dt \end{equation*}
is zero for every $\epsilon>0$ since the integrand is odd. Therefore the limit as $\epsilon\rightarrow 0$ exists, even though the function is not Lebesgue integrable.