# integral with hyperspherical coordinates?

I need to prove that $$\int_{\mathbb{R}^d} \frac{{\lvert e^{i\langle\xi,y\rangle}+e^{-i\langle\xi, y\rangle}-2\rvert}^2}{\lVert y \rVert^{d+2}}dy$$ < $$\infty$$ for every linear form $$\xi$$ on $$\mathbb{R}^d$$.

I thought it was going to be pretty straightforward using the fact that $$\frac{{\lvert e^{i\langle\xi,y\rangle}+e^{-i\langle\xi, y\rangle}-2\rvert}^2}{\lVert y \rVert^{d+2}} \leq \frac{16}{\lVert y \rVert^{d+2}}$$ and hyperspherical coordinates.

But as I wrote$$\int_{\mathbb{R}^d} \frac{1}{\lVert y \rVert^{d+2}}dy = \int_{S^{d-1}}\left(\int_{0}^{\infty} \frac{r^{d-1}}{r^{d+2}}dr \right)dy,$$I realized it diverges.

So now I'm stuck and I don't really see how to prove the first statement.

• My bad the x should be y
– DLM
Commented Oct 2, 2022 at 15:01
• Using $e^{i\langle\xi,y\rangle}+e^{-i\langle\xi, y\rangle}-2=\big(e^{\frac i 2\langle\xi,y\rangle}-e^{-\frac i 2\langle\xi, y\rangle}\big)^2=-4\sin^2\frac{\langle\xi, y\rangle}{2}$ $$\int_{\mathbb{R}^d} \frac{{\lvert e^{i\langle\xi,y\rangle}+e^{-i\langle\xi, y\rangle}-2\rvert}^2}{\lVert y \rVert^{d+2}}dy=2^4\int_{\mathbb{R}^d} \frac{\sin^4\frac{\langle\xi, y\rangle}{2}}{\lVert y \rVert^{d+2}}dy=2^4\int_{S^{d-1}}d\Omega \int_0^\infty\frac{\sin^4\frac{\xi \,r\cos\phi_1}{2}}{r^{d+2}}r^{d-1}dr$$ Commented Oct 2, 2022 at 16:48
• where $\int d\Omega$ denotes the integration over all angles in the spheric system of coordinates, and $\phi_1$ is the angle between $\vec\xi$ and the polar axis of the system (this is our choice how to choose the axis' direction). The integral $\displaystyle\int_0^\infty\frac{\sin^4\frac{\xi \,r\cos\phi_1}{2}}{r^{d+2}}r^{d-1}dr=\int_0^\infty\frac{\sin^4\frac{\xi \,r\cos\phi_1}{2}}{r^3}dr$ converges at $r\to0$ and $r\to\infty$. Commented Oct 2, 2022 at 16:48
• Thanks, I got there but how do you prove that it converges in 0 and more generally that $\int_0^1\frac{\sin^n x}{x^n}dx$ converges in 0
– DLM
Commented Oct 2, 2022 at 17:53
• oh yeah I had forgotten that haha
– DLM
Commented Oct 2, 2022 at 19:00

Though it does not have the direct relation to the formulated task, it would be interesting to get a closed form of $$\displaystyle I(\xi,n)=\int_{\mathbb{R}^n} \frac{{\lvert e^{i\langle\vec\xi,\vec y\rangle}+e^{-i\langle\vec\xi, \vec y\rangle}-2\rvert}^2}{\lVert \vec y \rVert^{n+2}}dy$$. Denoting $$\displaystyle\xi=|\vec\xi|$$ and $$\displaystyle r=|\vec y|$$ $$I(\xi,n)=2^4\int_{S^{n-1}}d\Omega \int_0^\infty\frac{\sin^4\frac{\xi \,r\cos\phi_1}{2}}{r^3}dr=4\xi^2\int_{S^{n-1}}\cos^2\phi_1d\Omega\int_0^\infty\frac{\sin^4x}{x^3}dx\tag{1}$$ For n-sphere $$d\Omega$$ is defined as $$\displaystyle d\Omega=\sin^{n-2}\phi_1\sin^{n-3}\phi_2...\sin\phi_{n-2}\,d\phi_1...d\phi_{n-2}$$ ( here)

The interval of integration over $$d\phi_1$$ is $$[0;\pi]$$, therefore $$\int_0^\pi\cos^2\phi_1\sin^{n-2}\phi_1d\phi_1=\int_0^\pi\sin^{n-2}\phi_1d\phi_1-\int_0^\pi\sin^n\phi_1d\phi_1$$ Integrating the second term by part, $$\int_0^\pi\sin^n\phi_1d\phi_1=(n-1)\int_0^\pi\sin^{n-2}\phi_1\cos^2\phi_1d\phi_1=(n-1)\int_0^\pi\sin^{n-2}\phi_1(1-\sin^2\phi_1)d\phi_1$$ $$\int_0^\pi\sin^n\phi_1d\phi_1=\frac{n-1}{n}\int_0^\pi\sin^{n-2}\phi_1d\phi_1$$ $$\int_0^\pi\cos^2\phi_1\sin^{n-2}\phi_1d\phi_1=\frac{1}{n}\int_0^\pi\sin^{n-2}\phi_1d\phi_1$$ and, therefore, $$\int\cos^2\phi_1d\Omega=\frac{1}{n}\int d\Omega\tag{2}$$ To evaluate the last integral, we consider $$\int_{-\infty}^\infty...\int_{-\infty}^\infty dx_1...dx_n e^{-x_1^2-...-x_n^2}=\pi^\frac{n}{2}=\int d\Omega\int_0^\infty e^{-r^2}r^{n-1}dr$$ $$=\frac{1}{2}\int d\Omega\int_0^\infty e^{-t}t^{n/2-1}dt=\frac{1}{2}\Gamma\Big(\frac{n}{2}\Big)\int d\Omega\,\,\Rightarrow\,\,\int d\Omega=\frac{2\pi^\frac{n}{2}}{\Gamma\big(\frac{n}{2}\big)}$$ and $$\int_{S^{n-1}}\cos^2\phi_1d\Omega=\frac{1}{n}\int_{S^{n-1}} d\Omega=\frac{2\pi^\frac{n}{2}}{n\,\Gamma\big(\frac{n}{2}\big)}\tag{3}$$ Now we only have to evaluate $$\displaystyle J=\int_0^\infty\frac{\sin^4x}{x^3}dx$$. Integrating a couple of times by part, $$J=\int_0^\infty\frac{\sin2x\sin^2x}{x^2}dx=\int_0^\infty\big(2\cos2x\sin^2x+\sin^22x\big)\frac{dx}{x}$$ $$=\int_0^\infty\frac{\cos2x-\cos4x}{x}dx=\Re\,\int_0^\infty\frac{e^{2ix}-e^{4ix}}{x}dx$$ Integrating along a closed contour the complex plane (formally, it means the substitution $$x=it$$), $$J=\int_0^\infty\frac{e^{-2t}-e^{-4t}}{t}dt=\ln2\tag{4}$$ (in the last step we used the Frullani theorem)

Putting (3) and (4) into (1) $$\boxed{\,\,I(\xi,n)=\frac{8\ln2}{n}\frac{\pi^\frac{n}{2}}{\Gamma\big(\frac{n}{2}\big)}\xi^2\,\,}$$ It is easy to check independantly that, for example, at $$n=3$$ it gives the correct result: $$I(\xi, n=3)=\frac{16\pi\ln2}{3}\xi^2$$

• Interesting indeed !
– DLM
Commented Oct 4, 2022 at 6:38