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Prove $ int_{mathbb{R}^d} frac{|e^{ilangle xi, y rangle} + e^{- ilangle xi, y rangle} - 2|^2 }{|y|^{d+2}}dy = c_d |xi|^2 $

Mathematics Asked by Nga NTQ on November 28, 2020

My question is: Prove that there exists some constant $c_d$ such that for any $xi in mathbb{R}^d$:
$$displaystyle int_{mathbb{R}^d} dfrac{|e^{ilangle xi, y rangle} + e^{- ilangle xi, y rangle} – 2|^2 }{|y|^{d+2}}dy = c_d |xi|^2 $$
where $langle xi, x rangle = displaystyle sum_{j=1}^{d} xi_j y_j$. I have
$$ dfrac{|e^{ilangle xi, y rangle} + e^{- ilangle xi, y rangle} – 2|^2 }{|y|^{d+2}} = dfrac{|2coslangle xi, y rangle – 2|^2 }{|y|^{d+2}} = dfrac{16|sin^2dfrac{langle xi, y rangle}{2} |^2 }{|y|^{d+2}} leq dfrac{16}{|y|^{d+2}}$$
integrable since $d+2 > d$ so the mapping $y mapsto dfrac{|e^{ilangle xi, y rangle} + e^{- ilangle xi, y rangle} – 2|^2 }{|y|^{d+2}}$ 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.

2 Answers

Requested Form of the Integral

Let $T$ be an orthogonal linear transformation where $T(xi)=|xi|(1,0,0,dots,0)$. $$ begin{align} int_{mathbb{R}^d}frac{left|,e^{ilanglexi,yrangle}+e^{-ilanglexi,yrangle}-2,right|^2}{|y|^{d+2}},mathrm{d}y &=int_{mathbb{R}^d}frac{left|,e^{i|xi|y_1}+e^{-i|xi|y_1}-2,right|^2}{|y|^{d+2}},mathrm{d}ytag{1a}\ &=|xi|^2underbrace{int_{mathbb{R}^d}frac{left|,e^{iy_1}+e^{-iy_1}-2,right|^2}{|y|^{d+2}},mathrm{d}y}_{c_d}tag{1b} end{align} $$ Explanation:
$text{(1a)}$: substitute $ymapsto T^{-1}(y)$, 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 $leftlanglexi,T^{-1}(y)rightrangle=leftlangle T(xi),yrightrangle=|xi|y_1$
$text{(1b)}$: substitute $ymapsto y/|xi|$


Computing the Constant $$ begin{align} int_{x_n=t}frac{mathrm{d}x}{left(t^2+|x|^2right)^{frac{d+2}2}} &=int_0^inftyfrac{omega_{d-2}r^{d-2},mathrm{d}r}{left(t^2+r^2right)^{frac{d+2}2}}tag{2a}\[3pt] &=frac{omega_{d-2}}{|t|^3}int_0^inftyfrac{r^{d-2},mathrm{d}r}{left(1+r^2right)^{frac{d+2}2}}tag{2b}\ &=frac{omega_{d-2}}{|t|^3}frac{sqrtpi}4frac{Gamma!left(frac{d-1}2right)}{Gamma!left(frac{d+2}2right)}tag{2c}\[6pt] &=frac1{|t|^3}frac{pi^{d/2}}{d,Gamma!left(frac{d}2right)}tag{2d} end{align} $$ Explanation:
$text{(2a)}$: convert from rectangular to polar in $x_n=t$
$text{(2b)}$: substitute $rmapsto rt$
$text{(2c)}$: Beta integral
$text{(2d)}$: $omega_{d-1}=frac{2pi^{d/2}}{Gammaleft(frac{d}2right)}$

Since $e^{ilanglexi,yrangle}+e^{-ilanglexi,yrangle}inmathbb{R}$, $$ begin{align} left|,e^{ilanglexi,yrangle}+e^{-ilanglexi,yrangle}-2,right|^2 &=left(,e^{ilanglexi,yrangle}+e^{-ilanglexi,yrangle}-2,right)^2\ &=16sin^4(langlexi,yrangle/2)tag3 end{align} $$ Therefore, $$ begin{align} int_{mathbb{R}^d}frac{left|,e^{ilanglexi,yrangle}+e^{-ilanglexi,yrangle}-2,right|^2}{|y|^{d+2}},mathrm{d}y &=frac{16pi^{d/2}}{d,Gamma!left(frac{d}2right)}int_{-infty}^inftyfrac{sin^4(|xi|t/2)}{|t|^3},mathrm{d}ttag{4a}\ &=frac{8pi^{d/2}|xi|^2}{d,Gamma!left(frac{d}2right)}int_0^inftyfrac{sin^4(t)}{t^3},mathrm{d}ttag{4b}\ &=frac{8pi^{d/2}|xi|^2}{d,Gamma!left(frac{d}2right)}int_0^inftyfrac{cos(2t)-cos(4t)}{t},mathrm{d}ttag{4c}\ &=frac{8pi^{d/2}|xi|^2}{d,Gamma!left(frac{d}2right)},log(2)tag{4d} end{align} $$ Explanation:
$text{(4a)}$: apply $(2)$ and $(3)$
$text{(4b)}$: substitute $tmapsto2t/|xi|$ and apply symmetry
$text{(4c)}$: integrate by parts twice
$text{(4d)}$: Frullani integral

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

Correct answer by robjohn on November 28, 2020

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

Answered by Peter Morfe on November 28, 2020

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