Derivations in paper "Brownian Distance Covariance" and their intuition I am reading a paper and I am stuck at following points. I tried to analyze but do not understand where to start.
The paper "Brownian distance covariance" contains following lemma
Notations: $<t,s>$ is scalar product of vectors t and s. $|x|_p$ is the euclidean norm of $x$ in $\mathbb{R}^d$. $\Gamma(.)$ is the complete gamma function$
Lemma:
If $0<\alpha<2$ then for all $x$ in $\mathbb{R}^d$
$\int_{\mathbb{R}^d} \frac{1-cos<t,x>}{|t|_d^{d+\alpha}} dt=C(d,a)|x|_d^\alpha$, where $C(d,\alpha)=\frac{2 \pi^{d/2} \Gamma(1-\alpha/2)}{\alpha 2^\alpha\Gamma((d+\alpha/2)/2)}$
The integrals at 0 and $\infty$ are meant in principal value sense: $\lim_{\epsilon \to0} \int_{\mathbb{R}^d\\ \{\epsilon B+\epsilon^{-1}B^c\}}$, where $B$ is a unit ball in $\mathbb{R}^d$ centered at 0 and $B^c$ is complement of B
The proof of the lemma is given in another paper to which i dont have access. I dont need detailed proof but just the intuition of this.
Further the paper says:
Taking $\alpha=1$ consider $c_d=C(d,1)$
Above mentioned lemma implies there exist constants $c_p$ and $c_q$ such that for X in $\mathbb{R}^p$ and Y in $\mathbb{R}^q$,
$\int_{\mathbb{R}^p}\frac{1-exp\{i<t,X>\}}{|t|_p^{1+p}}dt=c_p|X|_p$
$\int_{\mathbb{R}^q}\frac{1-exp\{i<s,Y>\}}{|s|_q^{1+q}}ds=c_q|Y|_q$
$\int_{\mathbb{R}^p}\int_{\mathbb{R}^q}\frac{1-exp\{i<s,Y>+i<t,X>\}}{|s|_q^{1+q} |t|_p^{1+p}}ds dt=c_p|X|_p c_q|Y|_q$
The above matter is used for explaining the use of $w=c_p|X|_p c_q|Y|_q$ as weight function in finding $\int_{\mathbb{R}^{p+q}} |f_{X,Y}(t,s)-f_X(t) f_Y(s)|^2 w(t,s) dt ds$
I know that $1-cos<t,X>$ is related to $\exp \{i<t,X>\}$but I am unable to get a clear picture.
 A: For any $x \in \mathbb{R}^d$ there exists an orthogonal matrix $R \in \mathbb{R}^{d \times d}$ (i.e. $R^{-1} = R^T$, $|\det R|=1$) such that $R \cdot x = |x| e_1$ where $e_1 = (1,0,\ldots,0)$ denotes the 1st unit vector. In particular,
$$\langle R^T s, x \rangle = \langle s, R x \rangle = |x| s_1$$
for any $s=(s_1,\ldots,s_d) \in \mathbb{R}^d$. Hence, if we substitute $t := R^T s$, we obtain
$$\begin{align*} \int_{\mathbb{R}^d} \frac{1-\cos \langle t,x \rangle}{|t|^{d+\alpha}} \, dt &= \int_{\mathbb{R}^d} \frac{1-\cos (s_1 |x|)}{|s|^{d+\alpha}} \, ds. \end{align*}$$
Finally, we can substitute $r := |x| s$ to conclude
$$ \int_{\mathbb{R}^d} \frac{1-\cos \langle t,x \rangle}{|t|^{d+\alpha}} \, dt = |x|^{\alpha} \int_{\mathbb{R}^d} \frac{1-\cos(r_1)}{|r|^{d+\alpha}} \, dr. \tag{1}$$
It requires much more work to calculate the constant $$C(d,\alpha) = \int_{\mathbb{R}^d} \frac{1-\cos(r_1)}{|r|^{d+\alpha}} \, dr,$$ but usually the constant is not of interest.
To prove the other equalities, note that
$$\int_{\mathbb{R}^d} \frac{\sin \langle t, X \rangle}{|t|^{d+\alpha}} \, dt = 0$$
by symmetry. Combining this with the identity
$$1- \exp(i \, \langle t,X \rangle) = (1-\cos(\langle t,X \rangle)) - i \sin (\langle t,X \rangle)$$
and $(1)$ proves the claimed identites.
Remark: The lemma actually shows how to calculate the characteristic exponent of an $\alpha$-stable symmetric Lévy process.
