# 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.

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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