Principal value integral of complex exponential I'm reading the article Brownian distance covariance and stumbled upon a equality I can't seem to derive myself. We are first presented with the following lemma:

and after stating this lemma, the authors follow up with a remark in which my problem lies:

Now the first two integrals are rather simple to show using the above lemma but the third iterated principal value integral is where my troubles begin. 
I have tried multiple approaches so far, two of them being to rewrite the nominator:
$$
1-\exp(it^\intercal x + i s^\intercal y) = (1-\exp(it^\intercal x)) + (1-\exp(is^\intercal y)) -(1-\exp(it^\intercal x))(1-\exp(is^\intercal y))
$$
but this approach yields nothing since the second term yields a the inner principal value integral of $\infty$ and vice versa for the first term when evaluating the outer principal value integral. Another approach I tried was to write
\begin{align*}
1-\exp(it^\intercal x + i s^\intercal y) =& 1- \cos(t^\intercal x + s^t y) -i\sin (t^\intercal x + s^\intercal y) \\
 =& 1- \cos t^\intercal x \cos s^\intercal y + \sin t^\intercal x \sin s^\intercal y -i (\sin t^\intercal x \cos s^\intercal y + \cos t^\intercal x \sin s^\intercal y) \\
 =&(1-\cos t^\intercal x) + (1- \cos s^\intercal y)-(1-\cos t^\intercal x)(1-\cos s^\intercal y) \\ 
&+ \sin t^\intercal x \sin s^\intercal y -i (\sin t^\intercal x \cos s^\intercal y + \cos t^\intercal x \sin s^\intercal y) \\
\end{align*}
In this expression the third term integrates to the negative value of the wanted solution, so its close but not the wanted expression (even if I ignore the sign, I can't get the other terms to yield a zero integral).
Am approaching this is the wrong way or is there something I have missed? Any solutions/hints would be appreciated.
 A: There is a problem with the last equality. Consider a set $A\subset\mathbb{R^q}$ where  $\left<s,Y \right>$ is bounded away from integer multiples of $2\pi$. For $t$ in a  small enough neighborhood of the origin, call it $B$, the numerator is bounded away from $0$. So the singularity is not canceled out. Hence, for $s\in A$ the integral
\begin{equation*}
\int_B 
\frac{1-\exp\{i\left<t,X \right>+ i\left<s,Y \right>   \}}
{|t|_p^{1+p} |s|_q^{1+q}}
dt
\end{equation*}
is not defined (or infinite). This implies
\begin{equation*}
\int_{\mathbb{R}^p}
\frac{1-\exp\{i\left<t,X \right>+ i\left<s,Y \right>   \}}
{|t|_p^{1+p} |s|_q^{1+q}}
dt
\end{equation*}
is not defined (or infinite).
The inner integral must be defined a.e. for convergence, and this is not the case.
A: Just to clarify the answer of Dunham, take the easy example with $Y = 0$, which I develop on MathOverflow. Both reassuring and puzzling in the same time, this results is given but actually never used in the "Brownian distance covariance" paper. Even more puzzling, the same equality was already given, without being used either, in the earlier paper on distance covariance.
