Louis Meunier

Remarks

A Special Spherical Integral

I recently had to deal with the following surface integral (in computing the Fourier transform of the surface measure of a sphere): I=B(x,t)eizydSy, where B(x,t)Rd the surface of the d-ball centered at xRd with radius t>0, and zy the usual Euclidean “dot” product; namely x,t, and z are all fixed. Finding a “nice” closed form for the expression proceeds as follows. First, we translate so the integral is about the origin, by sending yyx, I=B(0,t)eizyeizxdSy=eizxB(0,t)eizydSy=:eizxI(z). Next, notice that I(z) is rotationally symmetric. Namely, if A:RdRd is a rotation matrix, then I(Az)=B(0,t)exp(i(Az)y)dSy=B(0,t)exp(izAy)dSy, where A the hermitian adjoint of A; rotation matrices are unitary, so A=A1. Changing variables by letting u:=A1y, then, the new domain of integration remains B(0,t) since A a rotation, and since det(A1)=det(A)1=1, the resulting Jacobian of the transformation is unity. Thus, we find I(Az)=I(z) indeed.

From this observation, then, we may assume without loss of generality that z is a scalar of the usual basis vector, e1, namely say z=|z|e1, with || the Euclidean norm. With this, I(z)=B(0,t)exp(i|z|y1)dSy, with y=(y1,,yd). Finally, rewriting this exponential using Euler’s formula and writing the integral using d-dim spherical coordinates, we find (identifying y1=tcosϕ1) I(z)=0π02π[cos(|z|tcos(ϕ1))+isin(|z|tcos(ϕ1))]×td1sind2(ϕ1)sin(ϕd2)dϕ1dϕd1=2πtd1[j=1d30πsinj(ϕ)dϕ]×[0πcos(|z|tcosϕ)sind2(ϕ)dϕ+0πisin(|z|tcosϕ)sind2(ϕ)dϕ]. The iterated product term simplifies using a well-known Gamma-function identity j=1d30πsinj(ϕ)dϕ=π(d3)/2j=1d3Γ(j+12)Γ(j+22)=π(d3)/2Γ(d12), and the right-most integral is identically zero, since the sin terms are symmetric about π/2. Finally, the first integral term is, up to a constant depending only on d, t, and |z|, equal to the Bessel function Jd22(t|z|) (see here, section 3.3). All together, and explicitly writing out this constant, we find I(z)=2πtd1π(d3)/2Γ(d12)Γ(12)Γ(d12)(12t|z|)(d2)/2Jd22(t|z|)=(2π)d/2td/2|z|d22Jd22(t|z|), from which we arrive finally at a relatively-nice formula I=(2π)d/2td/2|z|(d2)/2eizxJd22(t|z|). Throughout, we tacitly assumed d2. Indeed, if d=1, the “unit ball” is just the open interval (xt,x+t) and so the surface of such a ball is the set xt,x+t, so I=eizy|y=xtx+t=eiz(x+t)eiz(xt)=eizx[eizteizt]=2ieizxsin(x)=2icos(xz)sin(x)2sin(x)sin(xz).

[*A similar computation can be made for the same integral but now over the ball B(x,t). I’ll omit the details but describe some necessary computations. If we call the surface integral above I(τ) (over the sphere B(x,τ)), the analogous integral can be written as 0tI(τ)dτ, which eventually leads to having to compute the integral (up to some constants related to x,t,ξ, and the dimension d)* 0tτd2Jd22(τ|ξ|)dτ, *which citing again this reference, equals td2|ξ|1Jd2(t).*]