A Special Spherical Integral

I recently had to deal with the following integral (in computing the Fourier transform of the surface measure of a sphere): $$I={\int }_{\partial B(x,t)}{e}^{iz\cdot y}dy,$$ where $\partial B(x,t)\subset R^d$ the surface of the $d$-ball centered at $x\in R^d$ with radius $t>0$, and $z\cdot y$ 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 $y\mapsto y-x$, $$I={\int }_{\partial B(0,t)}{e}^{iz\cdot y}{e}^{iz\cdot x}dy=e^{iz\cdot x}{\int }_{\partial B(0,t)}{e}^{iz\cdot y}dy=:e^{iz\cdot x}\cdot I\left(z\right).$$ Next, notice that $I\left(z\right)$ is rotationally symmetric. Namely, if $A:R^d\to R^d$ is a rotation matrix, then $$I\left(Az\right)={\int }_{\partial B(0,t)}\exp \left(i\right(Az)\cdot y)dy={\int }_{\partial B(0,t)}\exp (iz\cdot A^{\ast }y)dy,$$ where $A^{\ast }$ the hermitian adjoint of $A$; rotation matrices are unitary, so $A^{\ast }=A^{-1}$. Changing variables by letting $u:=A^{-1}y$, then, the new domain of integration remains $\partial B(0,t)$ since $A$ a rotation, and since $det\left(A^{-1}\right)=det(A{)}^{-1}=1$, the resulting Jacobian of the transformation is unity. Thus, we find $I\left(Az\right)=I\left(z\right)$ indeed.

From this observation, then, we may assume without loss of generality that $z$ is a scalar of the usual basis vector, $e_1$, namely say $z=|z|\cdot e_1$, with $|\cdot |$ the Euclidean norm. With this, $$I\left(z\right)={\int }_{\partial B(0,t)}\exp \left(i|z|y_1\right)dy,$$ with $y=(y_1,\dots ,y_d)$. Finally, rewriting this exponential using Euler’s formula and writing the integral using $d$-dim spherical coordinates, we find (identifying $y_1=t\cos {\varphi }_1$) $$I\left(z\right)={\int }_0^{\pi }\cdots {\int }_0^{2\pi }\left[\cos \right(|z|t\cos \left({\varphi }_1\right))+i\sin (|z|t\cos \left({\varphi }_1\right)\left)\right]\times t^{d-1}{\sin }^{d-2}\left({\varphi }_1\right)\cdots \sin \left({\varphi }_{d-2}\right)d{\varphi }_1\cdots d{\varphi }_{d-1}=2\pi t^{d-1}\left[\prod _{j=1}^{d-3}{\int }_0^{\pi }{\sin }^j\right(\varphi \left)d\varphi \right]\times [{\int }_0^{\pi }\cos \left(|z|t\cos \varphi \right){\sin }^{d-2}\left(\varphi \right)d\varphi +{\int }_0^{\pi }i\sin \left(|z|t\cos \varphi \right){\sin }^{d-2}\left(\varphi \right)d\varphi ].$$ The iterated product term simplifies using a well-known Gamma-function identity $$\prod _{j=1}^{d-3}{\int }_0^{\pi }{\sin }^j\left(\varphi \right)d\varphi ={\pi }^{(d-3)/2}\prod _{j=1}^{d-3}\frac{\Gamma \left(\frac{j+1}{2}\right)}{\Gamma \left(\frac{j+2}{2}\right)}=\frac{{\pi }^{(d-3)/2}}{\Gamma \left(\frac{d-1}{2}\right)},$$ and the right-most integral is identically zero, since the sin terms are symmetric about $\pi /2$. Finally, the first integral term is, up to a constant depending only on $d$, $t$, and $|z|$, equal to the Bessel function $J_{\frac{d-2}{2}}\left(t|z|\right)$ (see here, section 3.3). All together, and explicitly writing out this constant, we find $$I\left(z\right)=2\pi t^{d-1}\cdot \frac{{\pi }^{(d-3)/2}}{\Gamma \left(\frac{d-1}{2}\right)}\cdot \frac{\Gamma \left(\frac{1}{2}\right)\Gamma \left(\frac{d-1}{2}\right)}{(\frac{1}{2}t|z|{)}^{(d-2)/2}}{J}_{\frac{d-2}{2}}\left(t|z|\right)=(2\pi {)}^{d/2}\frac{t^{d/2}}{|z{|}^{\frac{d-2}{2}}}\cdot J_{\frac{d-2}{2}}(t|z|),$$ from which we arrive finally at a relatively-nice formula $$I=(2\pi {)}^{d/2}\frac{t^{d/2}}{|z{|}^{(d-2)/2}}{e}^{iz\cdot x}\cdot J_{\frac{d-2}{2}}(t|z|).$$ Throughout, we tacitly assumed $d>2$. Indeed, if $d=1$, the “unit ball” is just the open interval $(x-t,x+t)$ and so the surface of such a ball is the set $x-t,x+t$, so $$I=e^{iz\cdot y}{|}_{y=x-t}^{x+t}=e^{iz(x+t)}-e^{iz(x-t)}=e^{izx}[e^{izt}-e^{-izt}]=2i{e}^{izx}\sin \left(x\right)=2i\cos \left(xz\right)\sin \left(x\right)-2\sin \left(x\right)\sin \left(xz\right).$$

% If $d=2$, integrating over (“around”) the circle centered at $x$ of radius $t$