Some trigonometric integrals and the Fourier transform of a spherically symmetric exponential function

TL;DR

This paper presents a simplified method for computing the Fourier transform of a spherically symmetric exponential function, leveraging symmetry and homogeneity to reduce complexity.

Contribution

It introduces a more straightforward approach to evaluate the Fourier transform of spherically symmetric exponential functions using polar coordinates and symmetry properties.

Findings

Simplified Fourier transform calculation method

Reduction to rational trigonometric integral

Connection to harmonic function Fourier representations

Abstract

We calculate the Fourier transform of a spherically symmetric exponential function. Our evaluation is much simpler than the known one. We use the polar coordinates and reduce the Fourier transform to the integral of a rational function of trigonometric functions. Its evaluation turns out to be much easier than expected because of homogeneity and a hidden symmetry. Relationship with a Fourier integral representation formula for harmonic functions is explained.

Equations

$${\int }_{{\mathbb{R}}^n}{e}^{-2\pi a|x|}{e}^{-2\pi it\cdot x}dx=\frac{a\Gamma [(n+1)/2]}{{\pi }^{(n+1)/2}(a^2+|t{|}^2{)}^{(n+1)/2}}.$$

$$e^{-\beta }=\frac{1}{\sqrt{\pi }}{\int }_0^{\infty }\frac{e^{-u}}{\sqrt{u}}{e}^{-{\beta }^2/4u}du,\beta >0$$

$${\gamma }_n=2^{n}B\left(\frac{n+1}{2},\frac{n+1}{2}\right).$$

$${\int }_0^{\pi }\frac{{\sin }^{n}x}{(a+ip\cos x{)}^{n+1}}dx=\frac{{\gamma }_n}{(a^2+p^2{)}^{(n+1)/2}}(n=0,1,2,\dots ).$$

$$I$$

$$t\cdot \xi =|t|g{}^t(1,0,...,0))\cdot g\eta =|t|{}^t(1,0,...,0)\cdot \eta =|t|{\eta }_1.$$

$$I={\int }_{S^{n-1}}\frac{(n-1)!}{\{2\pi (a+2\pi i|t|{\eta }_1){\}}^n}d\eta .$$

$${\eta }_1=\cos {\theta }_1,{\eta }_2=\sin {\theta }_1\cos {\theta }_2,{\eta }_3=\sin {\theta }_1\sin {\theta }_2\cos {\theta }_3,\dots ,$$

$${\eta }_j=\sin {\theta }_1\sin {\theta }_2\cdots \sin {\theta }_{j-1}\cos {\theta }_j,\dots ,$$

$${\eta }_{n-1}=\sin {\theta }_1\sin {\theta }_2\cdots \sin {\theta }_{n-2}\cos {\theta }_{n-1},$$

$${\eta }_n=\sin {\theta }_1\sin {\theta }_2\cdots \sin {\theta }_{n-2}\sin {\theta }_{n-1},$$

$$d\eta =\prod _{j=1}^{n-1}{\sin }^{n-1-j}{\theta }_{j}d{\theta }_j.$$

$$I=\frac{(n-1)!}{(2\pi {)}^n}{\int }_0^{2\pi }d{\theta }_{n-1}\left(\prod _{j=2}^{n-2}{\int }_0^{\pi }{\sin }^{n-1-j}{\theta }_{j}d{\theta }_j\right){\int }_0^{\pi }\frac{{\sin }^{n-2}{\theta }_{1}d{\theta }_1}{(a+i|t|\cos {\theta }_1{)}^{n-1}}.$$

$$f(\lambda x)={\lambda }^{\delta }f(x)$$

$$x\frac{d}{dx}f(x)=\delta f(x),x>0.$$

$${\int }_0^{\pi }\frac{{\sin }^{n}x}{(a+b\cos x{)}^{n+1}}dx=\frac{{\gamma }_n}{(a^2-b^2{)}^{(n+1)/2}}(n=0,1,2,\dots ).$$

$${\int }_0^{\pi }\frac{{\sin }^{n}x}{(\sqrt{a}+\sqrt{b}\cos x{)}^{n+1}}dx=\frac{{\gamma }_n}{(a-b{)}^{(n+1)/2}}(n=0,1,2,\dots ).$$

$$\underset{a\to \infty }{\lim }{a}^{(n+1)/2}f(a,b)$$

$$={\int }_0^{\pi }{\sin }^{n}xdx={\gamma }_n.$$

$$2\frac{\partial f}{\partial q}$$

$$=\frac{-1}{2\sqrt{ab}}{\left[\frac{{\sin }^{n+1}x}{(\sqrt{a}+\sqrt{b}\cos x{)}^{n+1}}\right]}_0^{\pi }=0.$$

$$f(a,b)=f(p)=\frac{\text{const.}}{p^{(n+1)/2}}=\frac{\text{const.}}{(a-b{)}^{(n+1)/2}}.$$

$$I_n$$

$$=(n-1)\left\{{\int }_0^{\pi }\frac{{\sin }^{n-2}x}{(a+b\cos x{)}^{n+1}}dx-I_n\right\}$$

$$+(n+1)b{\int }_0^{\pi }\frac{{\sin }^{n}x\cos x}{(a+b\cos x{)}^{n+2}}dx.$$

$$\Bigl{(}b\frac{\partial}{\partial b}+n\Bigr{)}I_{n}=\frac{1}{n}\frac{\partial^{2}}{\partial a^{2}}I_{n-2}\quad(n\geq 2).$$

$${\int }_0^{\pi }\frac{{\sin }^{\mu -1}xdx}{(a+b\cos x{)}^{\mu }}=\frac{2^{\mu -1}}{\sqrt{(a^2-b^2{)}^{\mu }}}B\left(\frac{\mu }{2},\frac{\mu }{2}\right)$$

$$u(t)=\frac{1}{2(2\pi)^{n-1}}\int_{V}\Bigl{(}1-\frac{n-2}{2|y|}\Bigr{)}f(y/|y|)e^{-iz\cdot t}e^{-|y|}\Bigl{(}\frac{dx\wedge dy}{|y|}\Bigr{)}^{n-1}.$$

Full text

Department of Mathematical Sciences, Kwansei Gakuin University

Gakuen 2-1, Sanda, Hyogo, Japan 669-1337

yamane@kwansei.ac.jp

2000 Mathematics Subject Classification: Primary 33

1 Fourier transform

In this article we give a simple proof of the following theorem about the Fourier transform of a spherically symmetric function.

Theorem 1**.**

If $a>0$, we have

[TABLE]

The Fourier transform of $e^{-2\pi a|x|}$ is much harder to obtain than that of the Gaussian kernel $e^{-\pi\alpha|x|^{2}}$. Lower dimensional ($n\leq 3$) cases can be found in many books. For example, the one- and three-dimensional cases are [4, 12.23.9] and [4, 12.24] respectively. The general case can be found in the standard textbook [7]. The proof in [7, pp.6-7] based on the unfamiliar formula

[TABLE]

is technical and mysterious. We give a much simpler proof in this article. Since we are dealing with a spherically symmetric function $e^{-2\pi a|x|}$, it is natural to use the polar coordinates. Then we are led to integrals involving trigonometric functions. Their evaluation is interesting in its own right.

First we recall a result about the integral of the $n$-th power of $\sin x$ ([4, 3.621]).

Proposition 2**.**

Set $\gamma_{n}=\int_{0}^{\pi}\sin^{n}\theta\,d\theta$ for $n\in\mathbb{N}=\{0,1,2,\dots\}$. Then we have $\gamma_{n}=2(n-1)!!/n!!$ if $n$ is odd and $\gamma_{n}=\pi(n-1)!!/n!!$ if $n$ is even. (By convention, $(-1)!!=0!!=1$.) We also have

[TABLE]

The following proposition will be proved in Section 3.

Proposition 3**.**

For $a>0,p\in\mathbb{R},$ we have

[TABLE]

Now we prove Theorem 1. Set $x=r\xi,\xi\in S^{n-1}$. Then $dx=r^{n-1}d\xi$, where $d\xi$ is the surface-area element of $S^{n-1}$. Let us denote the left-hand side of (1) by $I$. Then we have

[TABLE]

Let $g\in\mathrm{SO}(n)$ be a special orthogonal matrix with the property $g^{t}(1,0,...,0)=|t|^{-1}t$. Set $\xi=g\eta$. Then $d\xi=d\eta$ and

[TABLE]

We have

[TABLE]

The variable $\eta$ and the surface-area element $d\eta$ can be written in terms of the angles $\theta_{1},\dots,\theta_{n-2}\in[0,\pi]$ and $\theta_{n-1}\in[0,2\pi]$ in the following way:

[TABLE]

We have

[TABLE]

By using Propositions 2 and 3, we obtain Theorem 1.

2 Positively homogeneous functions

The result of this section shall be used in the next section. A function $f(x)$ on $\mathbb{R}_{+}=\left\{x>0\right\}$ is said to be *positively homogeneous *of degree $\delta$ if

[TABLE]

holds for any $x>0,\lambda>0$. A typical example is $x^{\delta}$. It is the only positively homogeneous function of degree $\delta$ up to a constant factor. Indeed, we have $f(x)=x^{\delta}f(1)$.

By differentiation in $\lambda$, the equation (5) yields $xf^{\prime}(\lambda x)=\delta\lambda^{\delta-1}f(x)$. Plugging $\lambda=1$, we obtain

[TABLE]

Conversely, any solution of (6) is of the form $f(x)=Cx^{\delta}$ and is homogeneous of degree $\delta$. Summing up, positive homogeneity of degree $\delta$ is equivalent to (6).

Assume $\delta<0$. Then $f(x)=Cx^{\delta}$ diverges as $x\to+0$ if $C\neq 0$. In other words, the operator $xd/dx-\delta$ has a trivial kernel and is injective on the space of bounded differentiable functions on $0<x<L$, where $L$ is an arbitrary positive constant.

The differential operator $xd/dx$ is called the Euler operator and plays an important role in the theory of ordinary differential equations with regular singularities.

3 Integrating a rational function of trigonometric functions

In this section, we prove Proposition 3. It follows from Proposition 4 below by analytic continuation. Indeed, when $a>0$ is fixed, the integral $\int_{0}^{\pi}\sin^{n}x/(a+b\cos x)^{n+1}\,dx$ is a holomorphic function in the simply connected domain $\mathbb{C}\setminus\left\{b\leq-a,a\leq b\right\}.$

Proposition 4**.**

If $a>b>0$, we have

[TABLE]

Notice that the $n=0$ case is well-known as an example of residue calculus (e.g. [1]), but here we give a completely different proof. Proposition 4 is equivalent to Proposition 5 below.

Proposition 5**.**

If $a>b>0$, we have

[TABLE]

Now we prove Proposition 5. Let $f(a,b)$ be the left hand side of (8). It is positively homogeneous in the sense that $f(\lambda a,\lambda b)=\lambda^{-(n+1)/2}f(a,b),\lambda>0$. On the other hand, when $b$ is fixed and $a$ tends to $\infty$, we have by Lebesgue’s dominated convergence theorem

[TABLE]

Set $p=a-b,q=a+b$. Differentiation under the integral sign gives

[TABLE]

So we have discovered a hidden symmetry in $f$. It is independent of $q$ and is a function of $p$ alone. We denote $f(a,b)=f(p)$ by abuse of notation. The positive homogeneity means $f(p)$ is homogeneous of degree $-(n+1)/2$. Therefore

[TABLE]

This constant must be $\gamma_{n}$ because of (9). The proofs of Propositions 5, 4, 3 are now complete.

We give an alternative proof of Proposition 4. It appeared in [8] as a proof of Proposition 3. First, Proposition 4 holds for $n=0,1$. If $n=0$, we have $2\int_{0}^{\pi}=\int_{0}^{2\pi}$ and the right hand side is well-known in complex analysis (e.g. [1]). If $n=1$, the primitive can be found easily.

The remaining part can be proved by induction. Let $I_{n}$ be the left hand side of (7). If $n\geq 2$, integration by parts gives

[TABLE]

The integrals are derivatives of $I_{n-2}$ and $I_{n}$ with respect to $a$ and $b$ up to constant factors. We get

[TABLE]

If $I_{n-2}=\gamma_{n-2}(a^{2}-b^{2})^{-\frac{n-1}{2}}$, then $I_{n}=\gamma_{n}(a^{2}-b^{2})^{-\frac{n+1}{2}}$ satisfies (10). It is the unique solution of (10) because of the injectivity of $b\partial/\partial b+n$. Induction proceeds.

4 Generalization of Proposition 4

Proposition 4 can be found in [4, 3.665.1] in a more general form. It states

Proposition 6**.**

If $a>b>0,\mathrm{Re}\,\mu>0$, we have

[TABLE]

This proposition is a consequence of Propositions 2, 4 and Carlson’s theorem below.

Theorem 7**.**

(i) If $f(z)$ is analytic and bounded for $\mathrm{Re}\,z\geq 0$ and if $f(n)=0$ for $n=0,1,2,\dots$, then $f(z)$ is identically zero.

(ii) If $f(z)$ is analytic and bounded for $\mathrm{Re}\,z>0$ and if $f(n)=0$ for $n=1,2,\dots$, then $f(z)$ is identically zero.

(iii) If $f(z)$ and $g(z)$ are analytic and bounded for $\mathrm{Re}\,z>0$ and if $f(n)=g(n)$ for $n=1,2,\dots$, then $f(z)=g(z)$ holds for $\mathrm{Re}\,z>0$.

Proof.

The weaker version (i) can be found in [2, p.110]. As a matter of fact, the assumption $f(0)=0$ is superflous: the proof given there does not use $f(0)=0$. Anyway $n=0$ is removed in (ii), which is proved in the following way. Let $f(z)$ be analytic and bounded for $\mathrm{Re}\,z>0$ and assume $f(n)=0$ for $n=1,2,\dots$ If (i) is already known, we can apply it to $g(z)=f(z+1)$ and show that $f(z)$ is identically zero for $\mathrm{Re}\,z\geq 1$. It follows that $g(z)$ is identically zero for $\mathrm{Re}\,z\geq 0$ by analytic continuation. (This argument shows that it is enough to assume $f(z)=0$ for sufficiently large integers.) Now (iii) is a trivial consequence of (ii). ∎

5 Fourier-Ehrenpreis integral formula

Proposition 3 was used in [8]. Here we sketch its main result, which has some similarities with (1).

Students of math, physics and engineering are taught how to solve ordinary differential equations of the form $y^{\prime\prime}+ay^{\prime}+by=0\,(a,b:\text{const.})$. Any solution is a superposition (a linear combination) of exponential solutions (if $a^{2}-4b\neq 0$). Such a statement holds true for a very general class of (systems of) linear partial differential equations with constant coefficients. It is known as the Fundamental Principle of Ehrenpreis ([3]). In the PDE case, roughly speaking, any solution is obtained from a superposition of exponential solutions, where a superposition means an integral with respect to a measure. The original finding was that such a measure just exists. Later developments concentrated on expressing such a measure in terms of differential forms in an more or less concrete way (e.g. [6]). In the case of harmonic functions, a very explicit formula was obtained in [8].

Let $z=(z_{1},\dots,z_{n})$ be the coordinate of $\mathbb{C}^{n}\,(n\geq 3)$. Set $y=(y_{1},\dots,y_{n})$, where $y_{j}$ is the imaginary part of $z_{j}$. We consider V=\bigl{\{}z\in\mathbb{C}^{n};z^{2}=\sum_{j=1}^{n}z_{j}^{2}=0\bigr{\}} . The exponential function $f_{z}(t)=\exp(-iz\cdot t)$ is harmonic in $t$. The main result of [8] states that any harmonic function is a superposition of such harmonic exponentials. Let $B_{n}$ be the open unit ball of $\mathbb{R}^{n}\,(n\geq 3)$. Assume that $u(t)$ is harmonic in $B_{n}$ and is continuous up to the boundary $S^{n-1}$. Let $f=u|_{\partial B_{n}}$ be its Dirichlet boundary value. Then in $B_{n}$, we have

[TABLE]

In particular, $u(t)$ is a superposition of the exponentials $\exp(-iz\cdot t)$ with $z^{2}=\sum_{j=1}^{n}z_{j}^{2}=0$. and $y/|y|\in\mathrm{supp\,}f$.