This is definitely wrong. The two first integrals are defined as principal values, allowing to cancel the sine parts, and conclude with the lemma (actually, as OP already knows, the lemma itself do not need to be expressed as principal value). But for the third one, it is not even clear how a principal value could be defined. Since one needs to bound away both $t$ and $s$ from zero, I would go for integrating over $D_\epsilon = \left(\mathbb{R}^p \setminus (\epsilon B_p \cup \epsilon^{- 1}B_p^{c}) \right) \times \left(\mathbb{R}^q \setminus (\epsilon B_q \cup \epsilon^{- 1}B_q^{c}) \right)$, where $B_p$ and $B_q$ are the appropriate unit balls. But now set $Y = 0$, so that for $X \in \mathbb{R}^p$ and $\epsilon > 0$,
$$\int_{D_\epsilon}
\frac{1 - \exp(\mathrm{i}\left<t,X\right>)}
{\Vert t \Vert^{p+1} \Vert s \Vert^{q+1}} \mathrm{d}t \mathrm{d}s =
\int_{\mathbb{R}^p \setminus (\epsilon B_p \cup \epsilon^{- 1}B_p^{c})}
\frac{1}{\Vert s \Vert^{q+1}}
\int_{\mathbb{R}^p \setminus (\epsilon B_p \cup \epsilon^{- 1}B_p^{c})}
\frac{1 - \cos(\left<t,X\right>)}
{\Vert t \Vert^{p+1}} \mathrm{d}t \mathrm{d}s.$$
But then by virtue of the lemma, with $X \neq 0$, the inner integral (over $t$) is bounded away from zero for $\epsilon$ small enough, and thus the above quantity diverges. This is far from being equal to zero, as the third equality would imply.
