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 $

Question

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.

Answer 1

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 $y\mapsto y/|\xi|$

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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, $$ c_d=\frac{8\pi^{d/2}\log(2)}{d\,\Gamma\!\left(\frac{d}2\right)}\tag5 $$

Answer 2

Define $F(\xi) = \int_{\mathbb{R}^d} \dfrac{|e^{i\langle \xi, y \rangle} + e^{- i\langle \xi, y \rangle} - 2|^2 }{|y|^{d+2}}dy$. Show that $F(\kappa \xi) = |\kappa|^{\alpha} F(\xi)$ for some $\alpha \in \mathbb{R}$ (independent of $\xi \in \mathbb{R}^{d} \setminus \{0\}$ and $\kappa \in \mathbb{R}$), and then show that $F(O(\xi)) = F(\xi)$ for each orthogonal transformation $O$. Conclude that $F(\xi) = c \|\xi\|^{\alpha}$ for some $c > 0$.