Prove $ \int_{\mathbb{R}^d} \frac{|e^{i\langle \xi, y \rangle} + e^{- i\langle \xi, y \rangle} - 2|^2 }{|y|^{d+2}}dy = c_d |\xi|^2 $ My question is: Prove that there exists some constant $c_d$ such that for any $\xi \in \mathbb{R}^d$:
$$\displaystyle \int_{\mathbb{R}^d} \dfrac{|e^{i\langle \xi, y \rangle} + e^{- i\langle \xi, y \rangle} - 2|^2 }{|y|^{d+2}}dy = c_d |\xi|^2 $$
where $\langle \xi, x \rangle = \displaystyle \sum_{j=1}^{d} \xi_j y_j$. I have
$$ \dfrac{|e^{i\langle \xi, y \rangle} + e^{- i\langle \xi, y \rangle} - 2|^2 }{|y|^{d+2}} = \dfrac{|2\cos\langle \xi, y \rangle - 2|^2 }{|y|^{d+2}} = \dfrac{16|\sin^2\dfrac{\langle \xi, y \rangle}{2} |^2 }{|y|^{d+2}} \leq \dfrac{16}{|y|^{d+2}}$$
integrable since $d+2 > d$ so the mapping $y \mapsto \dfrac{|e^{i\langle \xi, y \rangle} + e^{- i\langle \xi, y \rangle} - 2|^2 }{|y|^{d+2}}$ is integrable in $\mathbb{R}^d$. But I don't know how to find prove the equality above. Are there any ideas for this problem?
Thank you so much.

Requested Form of the Integral
Let $T$ be an orthogonal linear transformation where $T(\xi)=|\xi|(1,0,0,\dots,0)$.
$$
\begin{align}
\int_{\mathbb{R}^d}\frac{\left|\,e^{i\langle\xi,y\rangle}+e^{-i\langle\xi,y\rangle}-2\,\right|^2}{|y|^{d+2}}\,\mathrm{d}y
&=\int_{\mathbb{R}^d}\frac{\left|\,e^{i|\xi|y_1}+e^{-i|\xi|y_1}-2\,\right|^2}{|y|^{d+2}}\,\mathrm{d}y\tag{1a}\\
&=|\xi|^2\underbrace{\int_{\mathbb{R}^d}\frac{\left|\,e^{iy_1}+e^{-iy_1}-2\,\right|^2}{|y|^{d+2}}\,\mathrm{d}y}_{c_d}\tag{1b}
\end{align}
$$
Explanation:
$\text{(1a)}$: substitute $y\mapsto T^{-1}(y)$, noting that
$\phantom{\text{(1a):}}$ the Jacobian of $T$ is $1$, so $\mathrm{d}T^{-1}(y)=\mathrm{d}y$
$\phantom{\text{(1a):}}$ $T$ is an isometry, so $\left|T^{-1}(y)\right|=|y|$
$\phantom{\text{(1a):}}$ $T$ is orthogonal, so $\left\langle\xi,T^{-1}(y)\right\rangle=\left\langle T(\xi),y\right\rangle=|\xi|y_1$
$\text{(1b)}$: substitute $y\mapsto y/|\xi|$

Computing the Constant
$$
\begin{align}
\int_{x_n=t}\frac{\mathrm{d}x}{\left(t^2+|x|^2\right)^{\frac{d+2}2}}
&=\int_0^\infty\frac{\omega_{d-2}r^{d-2}\,\mathrm{d}r}{\left(t^2+r^2\right)^{\frac{d+2}2}}\tag{2a}\\[3pt]
&=\frac{\omega_{d-2}}{|t|^3}\int_0^\infty\frac{r^{d-2}\,\mathrm{d}r}{\left(1+r^2\right)^{\frac{d+2}2}}\tag{2b}\\
&=\frac{\omega_{d-2}}{|t|^3}\frac{\sqrt\pi}4\frac{\Gamma\!\left(\frac{d-1}2\right)}{\Gamma\!\left(\frac{d+2}2\right)}\tag{2c}\\[6pt]
&=\frac1{|t|^3}\frac{\pi^{d/2}}{d\,\Gamma\!\left(\frac{d}2\right)}\tag{2d}
\end{align}
$$
Explanation:
$\text{(2a)}$: convert from rectangular to polar in $x_n=t$
$\text{(2b)}$: substitute $r\mapsto rt$
$\text{(2c)}$: Beta integral
$\text{(2d)}$: $\omega_{d-1}=\frac{2\pi^{d/2}}{\Gamma\left(\frac{d}2\right)}$
Since $e^{i\langle\xi,y\rangle}+e^{-i\langle\xi,y\rangle}\in\mathbb{R}$,
$$
\begin{align}
\left|\,e^{i\langle\xi,y\rangle}+e^{-i\langle\xi,y\rangle}-2\,\right|^2
&=\left(\,e^{i\langle\xi,y\rangle}+e^{-i\langle\xi,y\rangle}-2\,\right)^2\\
&=16\sin^4(\langle\xi,y\rangle/2)\tag3
\end{align}
$$
Therefore,
$$
\begin{align}
\int_{\mathbb{R}^d}\frac{\left|\,e^{i\langle\xi,y\rangle}+e^{-i\langle\xi,y\rangle}-2\,\right|^2}{|y|^{d+2}}\,\mathrm{d}y
&=\frac{16\pi^{d/2}}{d\,\Gamma\!\left(\frac{d}2\right)}\int_{-\infty}^\infty\frac{\sin^4(|\xi|t/2)}{|t|^3}\,\mathrm{d}t\tag{4a}\\
&=\frac{8\pi^{d/2}|\xi|^2}{d\,\Gamma\!\left(\frac{d}2\right)}\int_0^\infty\frac{\sin^4(t)}{t^3}\,\mathrm{d}t\tag{4b}\\
&=\frac{8\pi^{d/2}|\xi|^2}{d\,\Gamma\!\left(\frac{d}2\right)}\int_0^\infty\frac{\cos(2t)-\cos(4t)}{t}\,\mathrm{d}t\tag{4c}\\
&=\frac{8\pi^{d/2}|\xi|^2}{d\,\Gamma\!\left(\frac{d}2\right)}\,\log(2)\tag{4d}
\end{align}
$$
Explanation:
$\text{(4a)}$: apply $(2)$ and $(3)$
$\text{(4b)}$: substitute $t\mapsto2t/|\xi|$ and apply symmetry
$\text{(4c)}$: integrate by parts twice
$\text{(4d)}$: Frullani integral
Thus,
$$
c_d=\frac{8\pi^{d/2}\log(2)}{d\,\Gamma\!\left(\frac{d}2\right)}\tag5
$$
