Revisiting the orthogonality of Bessel functions of the first kind on an infinite interval

The Bessel differential equation appears in a wide variety of physical problems, including static potentials in electromagnetism, propagation of waves in cylindrical waveguides, solutions of the radial Schrödinger equation, modes of oscillation of thin plates and membranes, and heat conduction [1].

In cylindrical coordinates $\left( \rho ,\;\phi ,\;z \right)$ the assumption that the solution of the Laplace equation can be found as the superposition of functions of the form $\Phi (\rho ,\;\phi ,\;z)=F(\rho )Q(\phi )Z(z)$ leads to three ordinary differential equations, one for each function. The equation for the radial function $F(\rho )$ is given by

$$\begin{eqnarray}&&{{\rho }^{2}}\;\frac{{{{\rm d}}^{2}}F}{{\rm d}{{\rho }^{2}}}+\rho \;\frac{{\rm d}F}{{\rm d}\rho }+\left( {{k}^{2}}{{\rho }^{2}}-{{\nu }^{2}} \right)F=0,\end{eqnarray} \tag{ 1 }$$

where k and ν are separation constants, k has dimensions of (length)$^{-1},$ and ν is dimensionless. If the physical problem shows translational symmetry along the z-axis, then k = 0. Similarly, if it shows rotational symmetry about z, then $\nu =0$.

When k is real and positive, it is customary to introduce the dimensionless variable

$$\begin{eqnarray}&&x=k\rho ,\end{eqnarray} \tag{ 2 }$$

in terms of which (1) becomes

$$\begin{eqnarray}&&\frac{{{{\rm d}}^{2}}F}{{\rm d}{{x}^{2}}}+\frac{1}{x}\frac{{\rm d}F}{{\rm d}x}+\left( 1-\frac{{{\nu }^{2}}}{{{x}^{2}}} \right)F=0.\end{eqnarray} \tag{ 3 }$$

This is the well-known Bessel differential equation of order $\nu \geqslant 0$, whose general solution is customarily written as

$$\begin{eqnarray}&&F(x)={{c}_{1}}{{J}_{\nu }}(x)+{{c}_{2}}{{Y}_{\nu }}(x),\end{eqnarray} \tag{ 4 }$$

where ${{J}_{\nu }}(x)$ and ${{Y}_{\nu }}(x)$ are linearly independent solutions of (3) called Bessel functions of the first kind and Bessel functions of the second kind, respectively. These functions have been thoroughly studied in the literature [2–4] and, thus, one could hardly add something substantial to what is already known about them.

Therefore, our interest in this note is of a pedagogical nature. Here, we consider the problem of finding a simple analytical derivation of the integral relation

$$\begin{eqnarray}&&\int _{0}^{\infty }\rho \;{{J}_{\nu }}(k\rho ){{J}_{\nu }}(k^{\prime} \rho )\;{\rm d}\rho =\frac{\delta (k-k^{\prime} )}{k},\ \ \ \nu \geqslant -1,\ \ \ k,k^{\prime} \gt 0,\end{eqnarray} \tag{ 5 }$$

which is the statement that Bessel functions ${{J}_{\nu }}(k\rho )$, with different values of k, are orthogonal on the infinite range $0\leqslant \rho \lt \infty$ with respect to the weight $w(\rho )=\rho$. The proof of the orthogonality of these functions on a finite interval, $0\leqslant \rho \leqslant a$, is a standard procedure that is widely discussed in many books and textbooks. However, this is not so when the interval is infinite.

The above integral follows from Hankelʼs integral theorem¹ . In section 14.4 of [4], Watson provides an elaborate and thorough proof of Hankelʼs integral theorem for $\nu \geqslant -1/2$. The proof is quite long, almost eight pages in length, and is more difficult when the range of ν is extended to $\nu \gt -1$. The complexity of the proof is probably the reason why the discussion of the orthogonality of the Bessel functions ${{J}_{\nu }}(x)$ over an infinite interval is avoided.

In physics, generally we are not very much concerned with the details of the proofs of mathematical theorems. Usually, we are more interested in their application for analysing and solving physical problems. This happens, in particular, with the orthogonality relation (5). In the monograph Mathematical Methods for Physicists [5], the integral (5) with $\nu \gt -1/2$ is given, without proof, in section 11.2 (equation (11.59)). Later, in the same reference (problem $15.1.2$), the inverse Hankel transform is assumed in such a way that it allows one to obtain (5) in a line of calculation² . However, the truth of the matter is in the opposite direction: the inversion formula is a consequence of (5). Clearly, this is circular logic where the conclusion is part of the premises³ .

For several purposes it is useful when physics students, especially those of theoretical physics, have a clear notion of the mathematical foundations underlying the theory; it improves students' confidence and enhances their critical thinking. In Jacksonʼs Classical Electrodynamics [6], in problem $3.16(a)$, the verification of (5) is left as an exercise to the reader, with the hint of using Besselʼs differential equation and the appropriate limiting procedures. It should be noted that despite the large number of solutions books and websites (for example [7–11]) for many of the problems in Jackson, most of them either do not address the problem or do it in an incorrect way.

The paper is organized as follows. In section 2, we give a simple analytical derivation of the orthogonality integral (5) for the case where ν is an integer, zero, or half-integer not less than $-1/2$. In section 3, using the asymptotic behaviour of ${{J}_{\nu }}(k\rho )$, we obtain by means of direct calculation the orthogonality integral (5) for an arbitrary $\nu \geqslant -1$. In section 4, we give a summary and discussion.

To make the presentation self-consistent, we start from the Fourier–Bessel series over a finite interval $0\leqslant \rho \leqslant a$ and then consider the case where the interval becomes infinite $(a\to \infty )$. We will show that the orthogonality integral (5) is a consequence of the fact that $\int _{0}^{\infty }\;{{J}_{\nu }}(x){\rm d}x=1$.

Since ${{J}_{\nu }}(x)$ oscillates about 0, the equation ${{J}_{\nu }}(x)=0$ has an infinite number of roots: ${{x}_{\nu 1}}$, ${{x}_{\nu 2}}$, ${{x}_{\nu 3}}$.... When a is finite, these roots play an important role. First, the functions $\sqrt{\rho }{{J}_{\nu }}\left( {{x}_{\nu i}}\rho /a \right)$, with fixed $\nu \geqslant -1$ and different i, are mutually orthogonal in the interval $0\leqslant \rho \leqslant a$, i.e.

$$\begin{eqnarray}&&\int _{0}^{a}\rho {{J}_{\nu }}\left( {{x}_{\nu i}}\frac{\rho }{a} \right){{J}_{\nu }}\left( {{x}_{\nu j}}\frac{\rho }{a} \right){\rm d}\rho =\frac{{{a}^{2}}}{2}\;J_{\nu +1}^{2}\left( {{x}_{\nu i}} \right){{\delta }_{ij}},\ \ \ i,j=1,\;2,\;3,...\ \ \ \nu \geqslant -1.\end{eqnarray} \tag{ 6 }$$

Second, any piecewise smooth function $f(\rho )$ on the interval $0\leqslant \rho \leqslant a$ can be expanded in a Fourier–Bessel series⁴

$$\begin{eqnarray}&&f(\rho )=\mathop{\sum }\limits_{i=1}^{\infty }{{C}_{\nu i}}\;{{J}_{\nu }}\left( {{x}_{\nu i}}\frac{\rho }{a} \right),\ \ \ 0\leqslant \rho \leqslant a,\end{eqnarray} \tag{ 7 }$$

where the coefficients ${{C}_{\nu i}}$ are obtained with the help of the orthogonality condition (6) as

$$\begin{eqnarray}&&{{C}_{\nu i}}=\frac{2}{{{a}^{2}}J_{\nu +1}^{2}({{x}_{\nu i}})}\;\int _{0}^{a}\rho f(\rho ){{J}_{\nu }}\left( {{x}_{\nu i}}\frac{\rho }{a} \right){\rm d}\rho .\end{eqnarray} \tag{ 8 }$$

When $a\to \infty$, the discrete superposition in (7) is replaced by a continuous superposition. In this limit, the chain of formal transformations

$$\begin{eqnarray*}&&\begin{array}{l} \qquad \qquad \,{{k}_{i}}=\frac{{{x}_{\nu i}}}{a}\to k, \\ {{\delta }_{ij}}\to \delta (a{{k}_{i}}-a{{k}_{j}})\to \delta \left[ a\left( k-k^{\prime} \right) \right]=\frac{\delta (k-k^{\prime} )}{a}, \\ \qquad \qquad \mathop{\sum }\limits_{i}\to \int _{0}^{\infty }a\;{\rm d}k \\ \qquad \qquad \,a\;{{C}_{\nu i}}\to C(k) \\ \frac{aJ_{\nu +1}^{2}\left( {{k}_{i}}a \right)}{2}\to N(k,\nu ), \\ \end{array}\end{eqnarray*}$$

where the 'normalization factor' $N(k,\nu )$ is still unknown, changes the set of equations (6)–(8) into

$$\begin{eqnarray}&&\int _{0}^{\infty }\rho {{J}_{\nu }}\left( k\rho \right){{J}_{\nu }}\left( k^{\prime} \rho \right){\rm d}\rho =N(k,\nu )\delta (k-k^{\prime} ),\end{eqnarray} \tag{ 9 }$$

$$\begin{eqnarray}&&f(\rho )=\int _{0}^{\infty }\;{{C}_{\nu }}(k){{J}_{\nu }}(k\rho ){\rm d}k,\end{eqnarray} \tag{ 10 }$$

$$\begin{eqnarray}&&{{C}_{\nu }}(k)=\frac{1}{N(k,\nu )}\;\int _{0}^{\infty }\rho \;f(\rho ){{J}_{\nu }}(k\rho ){\rm d}\rho .\end{eqnarray} \tag{ 11 }$$

Since the norm of ${{J}_{\nu }}\left( {{k}_{i}}\rho \right)$ on a finite interval depends on ν, we assume that N is also a function of ν. To check the compatibility of these equations, we multiply both sides of (10) by $\rho {{J}_{\nu }}(k^{\prime} \rho )$ and integrate over ρ from zero to infinity. Then, using (11), we obtain

$$\begin{eqnarray*}&&C(k^{\prime} )N(k^{\prime} ,\nu )=\int _{0}^{\infty }C(k)\left[ \int _{0}^{\infty }\rho {{J}_{\nu }}(k\rho ){{J}_{\nu }}(k^{\prime} \rho ){\rm d}\rho \right]\;{\rm d}k,\end{eqnarray*}$$

which is an identity by virtue of (9).

Finally, the closure relation

$$\begin{eqnarray}&&\int _{0}^{\infty }\frac{{{J}_{\nu }}(k\rho ){{J}_{\nu }}(k\rho ^{\prime} )}{N(k,\nu )}\;{\rm d}k=\frac{\delta (\rho -\rho ^{\prime} )}{\rho }\end{eqnarray} \tag{ 12 }$$

is obtained when we substitute (11) into (10).

To find the unknown normalization factor $N(k,\nu )$, we integrate both sides of (9) over $k^{\prime}$ from zero to infinity and obtain

$$\begin{eqnarray}&&N(k,\nu )=\int _{0}^{\infty }\rho {{J}_{\nu }}\left( k\rho \right)\left[ \int _{0}^{\infty }{{J}_{\nu }}\left( k^{\prime} \rho \right){\rm d}k^{\prime} \right]\;{\rm d}\rho .\end{eqnarray} \tag{ 13 }$$

To give an analytical derivation of (5), as promised in the introduction, we need to calculate these integrals manually, without the aid of computers or tables of integrals. Below, we do this for the case where ν is an integer, zero, or half-integer.

2.1. Evaluating the Integrals

From the recurrence relation

$$\begin{eqnarray*}&&{{J}_{\nu -1}}(x)-{{J}_{\nu +1}}(x)=2\;\frac{{\rm d}{{J}_{\nu }}(x)}{{\rm d}x},\end{eqnarray*}$$

it follows that

$$\begin{eqnarray}&&\int _{0}^{\infty }{{J}_{\nu }}(x)\ {\rm d}x-\int _{0}^{\infty }{{J}_{\nu +2}}(x){\rm d}x=-2\;{{J}_{\nu +1}}(x){{|}_{x=0}},\end{eqnarray} \tag{ 14 }$$

where we have used that ${{J}_{\nu }}\left( \infty \right)=0$, for an arbitrary ν. Since

$$\begin{eqnarray*}&&{{J}_{\nu +1}}(x)\to \frac{1}{\Gamma (\nu +2)}{{\left( \frac{x}{2} \right)}^{\nu +1}},\ \ \ {\rm as}\;{\rm x}\to 0\end{eqnarray*}$$

the rhs of (14) is zero for $\nu \gt -1$. Consequently

$$\begin{eqnarray}&&\int _{0}^{\infty }{{J}_{\nu +2}}(x){\rm d}x=\int _{0}^{\infty }{{J}_{\nu }}(x)\ {\rm d}x,\ \ \ \nu \gt -1.\end{eqnarray} \tag{ 15 }$$

Odd integer order: when $\nu =-1$, the rhs of (14) is equal to −2, because ${{J}_{0}}(0)=1$. Thus, using that ${{J}_{-1}}(x)=-{{J}_{1}}(x)$, we obtain

$$\begin{eqnarray*}&&\int _{0}^{\infty }{{J}_{1}}(x){\rm d}x=1.\end{eqnarray*}$$

From this expression and (15), we get

$$\begin{eqnarray}&&\int _{0}^{\infty }{{J}_{2n+1}}(x){\rm d}x=1,\ \ \ n=0,\;1,\;2...\;.\end{eqnarray} \tag{ 16 }$$

Even integer order: when ν is an even number, to obtain a similar relation, we need to calculate $\int _{0}^{\infty }{{J}_{0}}(x){\rm d}x$. Since

$$\begin{eqnarray}&&{{J}_{0}}(x)=\mathop{\sum }\limits_{n=0}^{\infty }\frac{{{\left( -1 \right)}^{n}}}{{{\left( n! \right)}^{2}}}{{\left( \frac{x}{2} \right)}^{2n}}=1-\frac{{{x}^{2}}}{4}+\frac{{{x}^{4}}}{64}+\cdots ,\end{eqnarray} \tag{ 17 }$$

we cannot integrate term by term. Instead, we use the Parseval integral

$$\begin{eqnarray}&&{{J}_{0}}(x)=\frac{1}{\pi }\;\int _{0}^{\pi }{\rm cos} \left( x{\rm sin} \theta \right){\rm d}\theta ,\end{eqnarray} \tag{ 18 }$$

which is easy to verify; expanding in series around x = 0 and integrating over θ, we obtain (17).

Thus

$$\begin{eqnarray}&&\int _{0}^{\infty }{{J}_{0}}(x){\rm d}x=\mathop{{\rm lim} }\limits_{\lambda \to \infty }\;\frac{1}{\pi }\;\int _{0}^{\lambda }\;{\rm d}x\;\int _{0}^{\pi }{\rm cos} \left( x{\rm sin} \theta \right){\rm d}\theta .\end{eqnarray} \tag{ 19 }$$

• Straightforward evaluation of $\int _{0}^{\infty }\;{{J}_{0}}(x){\rm d}x$: assuming that we can change the order of integration, we obtain

$$\begin{eqnarray}&&\int _{0}^{\infty }{{J}_{0}}(x){\rm d}x=\mathop{{\rm lim} }\limits_{\lambda \to \infty }\;\frac{1}{\pi }\;\int _{0}^{\pi }\frac{{\rm sin} \left( \lambda {\rm sin} \theta \right)}{{\rm sin} \theta }\;{\rm d}\theta .\end{eqnarray} \tag{ 20 }$$

The function under the integral, in the rhs of (20), possesses mirror symmetry with respect to $\theta =\pi /2$. It tends to λ when $\theta \to 0$, as well as when $\theta \to \pi$, and in between it becomes zero $2\left \lfloor \mathcal{N}\right \rfloor$ times, where $\left \lfloor \mathcal{N}\right \rfloor ={\rm floor}(\mathcal{N})$ represents the largest integer not greater than

$$\begin{eqnarray*}&&\mathcal{N}=\frac{\lambda }{\pi }.\end{eqnarray*}$$

Therefore, in the limit $\lambda \to \infty$ this function vanishes everywhere except for the two infinitely high and infinitely narrow peaks at $\theta =0$ and $\theta =\pi$. This suggests that, in this limit the integrand in the rhs of (20) can be expressed as a combination of Dirac delta functions.

A useful representation of the delta function comes from the theory of the Fourier integral. Namely⁵

$$\begin{eqnarray}\begin{array}{rcl} \delta (x) & = & \frac{1}{2\pi }\;\int _{-\infty }^{\infty }{{{\rm e}}^{{\rm i}kx}}\;{\rm d}k=\frac{1}{\pi }\;\int _{0}^{\infty }{\rm cos} kx\;{\rm d}k=\mathop{{\rm lim} }\limits_{\lambda \to \infty }\;\frac{1}{\pi }\;\int _{0}^{\lambda }{\rm cos} kx\;{\rm d}k \\ {} & = & \mathop{{\rm lim} }\limits_{\lambda \to \infty }\;\frac{{\rm sin} \lambda x}{\pi x}. \\ \end{array}\end{eqnarray} \tag{ 21 }$$

From this expression, it follows that the rhs of (20) is $\int _{0}^{\pi }\delta \left( {\rm sin} \theta \right){\rm d}\theta$. At this point, we recall that when the Dirac delta function has as argument a function f(s), with simple zeros located at $s={{s}_{i}}$, it reduces to

$$\begin{eqnarray*}&&\delta \left[ f(s) \right]=\mathop{\sum }\limits_{i}\frac{\delta \left( s-{{s}_{i}} \right)}{|{{\left( {\rm d}f/{\rm d}s \right)}_{{{s}_{i}}}}|}.\end{eqnarray*}$$

Consequently

$$\begin{eqnarray*}\begin{array}{rcl} \mathop{{\rm lim} }\limits_{\lambda \to \infty }\;\frac{{\rm sin} \left( \lambda {\rm sin} \theta \right)}{{\rm sin} \theta } & = & \delta ({\rm sin} \theta ) \\ {} & = & \delta (\theta )+\delta (\theta -\pi ), \\ \end{array}\end{eqnarray*}$$

in agreement with the above discussion. Thus, we obtain

$$\begin{eqnarray}&&\int _{0}^{\infty }{{J}_{0}}(x){\rm d}x=\int _{0}^{\pi }\delta \left( {\rm sin} \theta \right){\rm d}\theta =\int _{0}^{\pi }\left[ \delta (\theta )+\delta (\theta -\pi ) \right]\;{\rm d}\theta =1.\end{eqnarray} \tag{ 22 }$$

• Alternative evaluation of $\int _{0}^{\infty }\;{{J}_{0}}(x){\rm d}x$: for pedagogical reasons, it is worth mentioning that (19) can be evaluated, without the explicit use of Diracʼs delta function, by solving the integral

$$\begin{eqnarray}&&I(\alpha )=\frac{1}{\pi }\;\int _{0}^{\pi }{\rm d}\theta \;\int _{0}^{\infty }{\rm cos} \left( x{\rm sin} \theta \right){{{\rm e}}^{-\alpha x}}\;{\rm d}x,\ \ \ \alpha \gt 0,\end{eqnarray} \tag{ 23 }$$

which in the limit $\alpha \to 0$ yields $\int _{0}^{\infty }\;{{J}_{0}}(x){\rm d}x$.

The integral over x in (23) is easily done after using Eulerʼs formula ${{{\rm e}}^{{\rm i}\phi }}={\rm cos} \phi +{\rm i}\;{\rm sin} \phi$ to express the cosine in terms of exponents. The result is

$$\begin{eqnarray}&&\frac{1}{\pi }\int _{0}^{\infty }{\rm cos} \left( x{\rm sin} \theta \right){{{\rm e}}^{-\alpha x}}\;{\rm d}x=\frac{\alpha }{\pi \left( {{\alpha }^{2}}+{{{\rm sin} }^{2}}\theta \right)}.\end{eqnarray} \tag{ 24 }$$

Thus

$$\begin{eqnarray*}&&I(\alpha )=\frac{1}{\pi }\;\int _{0}^{\pi }\frac{\alpha }{{{\alpha }^{2}}+{{{\rm sin} }^{2}}\;\theta }\;{\rm d}\theta =\frac{1}{\pi }\;\int _{-\pi /2}^{\pi /2}\;\frac{\alpha }{{{\alpha }^{2}}+{{{\rm cos} }^{2}}\;\theta }\;{\rm d}\theta =\frac{2}{\pi }\;\int _{0}^{\pi /2}\;\frac{\alpha }{{{\alpha }^{2}}+{{{\rm cos} }^{2}}\;\theta }\;{\rm d}\theta .\end{eqnarray*}$$

To evaluate the last integral, we use the variable $u=\frac{\alpha }{\sqrt{{{\alpha }^{2}}+1}}\;{\rm tan} \theta$, in terms of which

$$\begin{eqnarray*}&&I(\alpha )=\frac{2}{\pi \sqrt{{{\alpha }^{2}}+1}}\;\int _{0}^{\infty }\frac{{\rm d}u}{1+{{u}^{2}}}=\frac{2}{\pi \sqrt{{{\alpha }^{2}}+1}}\;{{{\rm tan} }^{-1}}u|_{u=0}^{\infty }=\frac{1}{\sqrt{{{\alpha }^{2}}+1}}.\end{eqnarray*}$$

In the limit $\alpha \to 0$, we get $\int _{0}^{\infty }\;{{J}_{0}}(x){\rm d}x=1$, in agreement with (22).

It may seem somewhat surprising that we obtain the expected result without using the Dirac delta function. However, the delta function appears in the limit of $\alpha \to 0$ since

$$\begin{eqnarray*}&&\delta (x)=\frac{1}{\pi }\;\mathop{{\rm lim} }\limits_{\alpha \to 0}\;\frac{\alpha }{{{\alpha }^{2}}+{{x}^{2}}},\end{eqnarray*}$$

and the rhs of (24) becomes $\delta ({\rm sin} \theta )$. Thus, in this limit (23) reduces to (22).

In conclusion, we have shown that $\int _{0}^{\infty }\;{{J}_{0}}(x){\rm d}x=1$. Then, from (15), we get

$$\begin{eqnarray}&&\int _{0}^{\infty }{{J}_{2n}}(x){\rm d}x=1,\ \ \ n=0,\;1,\;2...\;.\end{eqnarray} \tag{ 25 }$$

Half-integer order: Bessel functions of half-integer order are elementary functions. For $\nu =1/2$, we have

$$\begin{eqnarray}&&\int _{0}^{\infty }\;{{J}_{1/2}}(x){\rm d}x=\sqrt{\frac{2}{\pi }}\;\int _{0}^{\infty }\frac{{\rm sin} x}{\sqrt{x}}\;{\rm d}x.\end{eqnarray} \tag{ 26 }$$

To evaluate this integral, we first introduce the variable

$$\begin{eqnarray*}&&x={{u}^{2}},\end{eqnarray*}$$

in terms of which (26) becomes

$$\begin{eqnarray}&&\sqrt{\frac{2}{\pi }}\;\int _{0}^{\infty }\frac{{\rm sin} x}{\sqrt{x}}\;{\rm d}x=2\;\sqrt{\frac{2}{\pi }}\;\int _{0}^{\infty }\;{\rm sin} {{u}^{2}}\;{\rm d}u=2\sqrt{\frac{2}{\pi }}\ {\rm Im}\left( \int _{0}^{\infty }\;{{{\rm e}}^{{\rm i}{{u}^{2}}}}\ {\rm d}u \right),\qquad \quad \end{eqnarray} \tag{ 27 }$$

where we have used Eulerʼs formula and ${\rm Im}(\alpha )$ means the complex part of the complex number α.

Next, we make a second change of variables, i.e. ${\rm i}{{u}^{2}}=-{{v}^{2}}$ so that $u={{{\rm e}}^{{\rm i}\pi /4}}\;v$. In terms of this new variable, (27) becomes

$$\begin{eqnarray}&&\sqrt{\frac{2}{\pi }}\;\int _{0}^{\infty }\frac{{\rm sin} x}{\sqrt{x}}\;{\rm d}x=2\sqrt{\frac{2}{\pi }}\ {\rm Im}\left( {{{\rm e}}^{{\rm i}\pi /4}}\int _{0}^{\infty }\;{{{\rm e}}^{-{{v}^{2}}}}\;{\rm d}v \right).\end{eqnarray} \tag{ 28 }$$

Since

$$\begin{eqnarray*}&&\int _{0}^{\infty }\;{{{\rm e}}^{-{{v}^{2}}}}\;{\rm d}v=\frac{\sqrt{\pi }}{2},\end{eqnarray*}$$

we obtain

$$\begin{eqnarray}&&\int _{0}^{\infty }{{J}_{1/2}}(x){\rm d}x=\sqrt{\frac{2}{\pi }}\;\int _{0}^{\infty }\frac{{\rm sin} x}{\sqrt{x}}\;{\rm d}x=1.\end{eqnarray} \tag{ 29 }$$

In a similar way, we get

$$\begin{eqnarray}&&\int _{0}^{\infty }\;{{J}_{-1/2}}(x){\rm d}x=\sqrt{\frac{2}{\pi }}\;\int _{0}^{\infty }\frac{{\rm cos} x}{x}\;{\rm d}x=2\sqrt{\frac{2}{\pi }}\ \operatorname{Re}\left( {{{\rm e}}^{{\rm i}\pi /4}}\int _{0}^{\infty }\;{{{\rm e}}^{-{{v}^{2}}}}\;{\rm d}v \right)=1,\end{eqnarray} \tag{ 30 }$$

where Re $(\alpha )$ means the real part of the complex number α.

Consequently, using (15), we get

$$\begin{eqnarray}&&\int _{0}^{\infty }\;{{J}_{n-1/2}}(x)\;{\rm d}x=1,\ \ \ n=0,\;1,\;2,...\;.\end{eqnarray} \tag{ 31 }$$

• Collecting the results given by (16), (25), and (31), we have shown that

$$\begin{eqnarray}&&\int _{0}^{\infty }\;{{J}_{\nu }}(x)\;{\rm d}x=1,\end{eqnarray} \tag{ 32 }$$

for any integer, zero, or half-integer $\nu \geqslant -1/2$.

To extend this result to an arbitrary ν, we would have to use an integral representation for ${{J}_{\nu }}(x)$ valid for all values of ν. However, this is beyond the scope of the present note.

2.2. Finding the factor $N(k,\nu )$

As mentioned earlier (2), in physical problems $x=k\rho$. Therefore, when x varies from 0 to infinity, either k or ρ should be kept fixed. If k remains fixed and $0\leqslant \rho \lt \infty$, then (32) yields

$$\begin{eqnarray}&&\int _{0}^{\infty }{{J}_{\nu }}\left( k\rho \right)\;{\rm d}\rho =\frac{1}{k}.\end{eqnarray} \tag{ 33 }$$

Similarly, if ρ remains fixed and $0\leqslant k\lt \infty$, then

$$\begin{eqnarray}&&\int _{0}^{\infty }{{J}_{\nu }}\left( k\rho \right)\;{\rm d}k=\frac{1}{\rho }.\end{eqnarray} \tag{ 34 }$$

With these two integrals at hand, the task of finding of $N(k,\nu )$ is straightforward. Namely, from (13) we get

$$\begin{eqnarray}&&N(k,\nu )=\frac{1}{k}.\end{eqnarray} \tag{ 35 }$$

Thus, N is independent of ν.

As a consequence of (35), the orthogonality integral (9) and the closure integral (12) become completely symmetric under the change $\rho \leftrightarrow k$, i.e.

$$\begin{eqnarray}\begin{array}{rcl} \int _{0}^{\infty }\rho {{J}_{\nu }}\left( k\rho \right){{J}_{\nu }}\left( k^{\prime} \rho \right){\rm d}\rho & = & \frac{\delta (k-k^{\prime} )}{k}, \\ \int _{0}^{\infty }k{{J}_{\nu }}\left( k\rho \right){{J}_{\nu }}\left( k\rho ^{\prime} \right){\rm d}k & = & \frac{\delta (\rho -\rho ^{\prime} )}{\rho }. \\ \end{array}\end{eqnarray} \tag{ 36 }$$

Similarly, the Fourier–Bessel transform is also symmetric if we change ${{C}_{\nu }}(k)=k{{A}_{\nu }}(k)$, namely,

$$\begin{eqnarray}&&f(\rho )=\int _{0}^{\infty }k\;{{A}_{\nu }}(k){{J}_{\nu }}(k\rho ){\rm d}k,\end{eqnarray} \tag{ 37 }$$

$$\begin{eqnarray}&&{{A}_{\nu }}(k)=\int _{0}^{\infty }\rho f(\rho ){{J}_{\nu }}(k\rho ){\rm d}\rho .\end{eqnarray} \tag{ 38 }$$

In this notation, ${{A}_{\nu }}(k)$ is the νth-order Hankel transform of $f(\rho )$. Reciprocally, $f(\rho )$ is the νth-order Hankel transform of ${{A}_{\nu }}(k)$; in short, it is called the inverse Hankel transform.

To finish this section, we would like to remark that we are perfectly aware that the integrals considered here are tabulated and can be found, for example, in [12]. If we use that table of integrals, or some computational software program, we can readily verify the validity of (5), for any ν. However, although fast and simple, such a verification would have no, or little, educational value because it teaches nothing about the Bessel functions nor develops technical or analytical skills⁶ .

In this section, we use the asymptotic behaviour of Bessel functions ${{J}_{\nu }}(x)$ to demonstrate that $N(k)=1/k$, for all $\nu \geqslant -1$. In addition, we explain why this factor is independent of ν.

To this end, we consider the limit

$$\begin{eqnarray}&&\mathop{{\rm lim} }\limits_{a\to \infty }\int _{0}^{a}\rho {{J}_{\nu }}(k\rho ){{J}_{\nu }}(k^{\prime} \rho ){\rm d}\rho =\int _{0}^{\infty }\rho {{J}_{\nu }}(k\rho ){{J}_{\nu }}(k^{\prime} \rho ){\rm d}\rho ,\end{eqnarray} \tag{ 39 }$$

where the integral in the lhs is given by⁷

$$\begin{eqnarray}&&\int _{0}^{a}\rho {{J}_{\nu }}(k\rho ){{J}_{\nu }}(k^{\prime} \rho ){\rm d}\rho =\left. \frac{\rho \left[ k\;{{J}_{\nu }}\left( k^{\prime} \rho \right){{J}_{\nu +1}}\left( k\rho \right)-k^{\prime} \;{{J}_{\nu }}\left( k\rho \right){{J}_{\nu +1}}\left( k^{\prime} \rho \right) \right]}{{{k}^{2}}-k{{^{\prime} }^{2}}} \right|_{\rho =0}^{a}.\end{eqnarray} \tag{ 40 }$$

• To evaluate this expression at $\rho =0$, we use the asymptotic expression of ${{J}_{\nu }}(k\rho )$ for small values of ρ,

$$\begin{eqnarray*}&&{{J}_{\nu }}(k\rho )\to \frac{1}{\Gamma (\nu +1)}{{\left( \frac{k\rho }{2} \right)}^{\nu }},\ \ \ {{J}_{-1}}(k\rho )\to -\frac{k\rho }{2}.\end{eqnarray*}$$

Substituting into (40), we get

$$\begin{eqnarray}\rho \to 0:{\rm rhs}\;{\rm of}\;{\rm (40)}\to \left\{ \begin{array}{ccccccccccccccc} \frac{1}{2\left( \nu +1 \right){{\Gamma }^{2}}(\nu +1)}{{\left( \frac{kk^{\prime} }{4} \right)}^{\nu }}\;{{\rho }^{2(\nu +1)}} & {\rm for}\;\nu \ne -1, \\ \frac{kk^{\prime} }{16}\;{{\rho }^{4}} & {\rm for}\;\nu =-1. \\ \end{array} \right.\end{eqnarray} \tag{ 41 }$$

Consequently, for any $\nu \geqslant -1$ the lower limit of (40) is zero.

• To compute the limit $a\to \infty$, we first evaluate (40) for $a\gg 1$. To this end, we use the large ρ behaviour of ${{J}_{\nu }}(k\rho )$, i.e.

$$\begin{eqnarray}&&{{J}_{\nu }}\left( k\rho \right)\to \sqrt{\frac{2}{\pi k\rho }}\;{\rm cos} \left[ k\rho -\frac{\left( 2\nu +1 \right)\pi }{4} \right].\end{eqnarray} \tag{ 42 }$$

Substituting this into the rhs of (40), we find

$$\begin{eqnarray}\begin{array}{rll} {\rm rhs}\;{\rm of}\;{\rm (40)} & {} & \to \frac{1}{\pi \sqrt{kk^{\prime} }}\left\{ \frac{{\rm sin} \left[ a\left( k-k^{\prime} \right) \right]}{k-k^{\prime} } \right. \\ {} & {} & -\left. \frac{{\rm sin} \pi \nu {\rm sin} \left[ a\left( k+k^{\prime} \right) \right]}{k+k^{\prime} }-\frac{{\rm cos} \pi \nu {\rm cos} \left[ a\left( k+k^{\prime} \right) \right]}{k+k^{\prime} } \right\}. \\ \end{array}\end{eqnarray} \tag{ 43 }$$

We note that the denominator in the last two expressions never vanishes, because k and $k^{\prime}$ are positive.

Next, in the limit $a\to \infty$ we have

$$\begin{eqnarray}&&\mathop{{\rm lim} }\limits_{a\to \infty }\frac{{\rm sin} a\left( k\pm k^{\prime} \right)}{\pi \left( k\pm k^{\prime} \right)}=\delta (k\pm k^{\prime} ),\end{eqnarray} \tag{ 44 }$$

$$\begin{eqnarray}&&\mathop{{\rm lim} }\limits_{a\to \infty }\frac{{\rm cos} a\left( k+k^{\prime} \right)}{\pi \left( k+k^{\prime} \right)}\propto \delta (k+k^{\prime} ).\end{eqnarray} \tag{ 45 }$$

The first expression follows from (21). To verify (45) we take the limit in two steps. First, we introduce the small parameter ξ as follows:

$$\begin{eqnarray*}&&{\rm cos} a\left( k+k^{\prime} \right)={\rm sin} \left[ a\left( k+k^{\prime} \right)+\pi /2 \right]={\rm sin} \left[ a\left( k+k^{\prime} +\pi \xi /2 \right) \right],\ \ \ \xi =\frac{1}{a};\qquad \quad \end{eqnarray*}$$

then, we expand in series around $\xi =0$ and rearrange terms to obtain

$$\begin{eqnarray*}&&\left( 1-\frac{\pi }{2}+\frac{{{\pi }^{3}}}{48}-\cdots \right)\frac{{\rm cos} a\left( k+k^{\prime} \right)}{\pi \left( k+k^{\prime} \right)}=\left( 1-\frac{{{\pi }^{2}}}{8}+\frac{{{\pi }^{4}}}{384}-\cdots \right)\frac{{\rm sin} a\left( k+k^{\prime} \right)}{\pi \left( k+k^{\prime} \right)},\ \ \ a\gg 1.\end{eqnarray*}$$

Finally, in the limit $a\to \infty$ the rhs yields $\delta (k+k^{\prime} )$, which justifies (45).

Thus, in the limit $a\to \infty$ the second and third terms in (43) are proportional to $\delta (k+k^{\prime} )$, which is nonzero only when $(k+k^{\prime} )=0$. Given that k and $k^{\prime}$ are positive, in this limit their contribution to (39) is zero. The only term that survives is the first one, which is proportional to $\delta (k-k^{\prime} )$. Consequently, from (9) and (39) we get

$$\begin{eqnarray}&&N(k)=\frac{1}{k},\ \ \ \nu \geqslant -1.\end{eqnarray} \tag{ 46 }$$

3.1. Role of phase shift

To elucidate why N(k) is independent of ν, we write (42) as

$$\begin{eqnarray*}&&{{J}_{\nu }}\left( k\rho \right)\to \sqrt{\frac{2}{\pi k\rho }}\;{\rm cos} \left( k\rho -{{\phi }_{\nu }} \right),\end{eqnarray*}$$

where the explicit form of the phase ${{\phi }_{\nu }}=\phi (\nu )$ is irrelevant. Using this, we obtain

$$\begin{eqnarray*}\begin{array}{rll} {\rm rhs}\;{\rm of}\;{\rm (40)} & {} & \to \frac{1}{\pi \sqrt{kk^{\prime} }}\left\{ {\rm sin} \delta \;\frac{{\rm sin} \left[ a\left( k-k^{\prime} \right) \right]}{k-k^{\prime} } \right. \\ {} & {} & +\left. \frac{{\rm sin} \sigma {\rm sin} \left[ a\left( k+k^{\prime} \right) \right]}{k+k^{\prime} }+\frac{{\rm cos} \sigma {\rm cos} \left[ a\left( k+k^{\prime} \right) \right]}{k+k^{\prime} }+\frac{{\rm cos} \delta {\rm cos} \left[ a\left( k-k^{\prime} \right) \right]}{k+k^{\prime} } \right\}, \\ \end{array}\end{eqnarray*}$$

where

$$\begin{eqnarray*}\begin{array}{rcl} \delta & = & {{\phi }_{\nu +1}}-{{\phi }_{\nu }}, \\ \sigma & = & {{\phi }_{\nu +1}}+{{\phi }_{\nu }}. \\ \end{array}\end{eqnarray*}$$

In the limit $a\to \infty$, the second and third terms are proportional to $\delta (k+k^{\prime} )$, and therefore they do not contribute to (39) regardless of σ. Meanwhile, the first and fourth terms are proportional to $\delta (k-k^{\prime} )$. The fourth term vanishes when $\delta =\pi /2$. Consequently, the normalizing factor (46) turns out to be independent of ν because, for large arguments, ${{J}_{\nu }}(x)$ and ${{J}_{\nu +1}}(x)$ are shifted by a phase $\delta ={{\phi }_{\nu +1}}-{{\phi }_{\nu }}=\pi /2$, i.e. ${{J}_{\nu }}(k\rho )\to \sqrt{2/\pi k\rho }\;{\rm cos} \left( k\rho -{{\phi }_{\nu }} \right)$ and ${{J}_{\nu +1}}(k\rho )\to {{Y}_{\nu }}(k\rho )=\sqrt{2/\pi k\rho }\;{\rm sin} \left( k\rho -{{\phi }_{\nu }} \right)$.

The rigorous proof of the orthogonality integral (5) is laborious and requires the use of mathematical techniques that, probably, are unfamiliar to most physics students, even at the graduate level.

In physics, to make calculations more feasible, we often reduce the problems by making simplifying hypotheses, which have reasonable physical meaning but are not always well supported mathematically. This is epitomized by problem $15.1.2$ of [5]. In that problem, the orthogonality integral (5) is obtained, in one line of calculation, from the Hankel transform and its inverse (see footnote ²).

However, this is an exercise of circular reasoning (see footnote 3) that does not prove the orthogonality integral. In practice, the technique of orthogonal functions and expansions works in the opposite way: first we identify an appropriate set of orthonormal functions and then use it as a basis to expand other functions.

It should be noted that the orthogonality relationship (5) discussed here is between Bessel functions of the same order ν, which have the same functional form but have distinct constants in the arguments. This is different from other common orthogonal functions used in physics (e.g., Legendre/Hermite/Laguerre polynomials), where the orthogonality is between functions of different order—i.e., corresponding to different eigenvalues—which have different functional forms but the same argument.

In section 2, we used elementary properties of Bessel functions ${{J}_{\nu }}(x)$ to compute $\int _{0}^{\infty }{{J}_{\nu }}(x){\rm d}x$. For ν integer, zero, or half-integer $\geqslant -1/2$, we obtained $\int _{0}^{\infty }{{J}_{\nu }}(x){\rm d}x=1$. From this, we found the normalization factor $N(k)=1/k$. As a consequence, the orthogonality integral and the closure integral (36) become symmetrical under the change $\rho \leftrightarrow k$. Also, the integrals (10) and (11) turn into Hankel transforms. The merit of this approach is its simplicity. The shortcoming is that our calculation is limited to the case where ν is an integer, zero, or a half-integer no less than $-1/2$.

In section 3, using the asymptotic behaviour of ${{J}_{\nu }}(k\rho )$, we obtained the factor N(k) by means of direct calculation. The advantage of this approach, which is probably the one suggested by Jackson in his problem 3.16, is that it works for any $\nu \geqslant -1$. Additionally it allows us to understand why this factor is independent of ν. The disadvantage, however, could be the level of mathematical rigor used to evaluate the limit $a\to \infty$ in (43). Apart of this, one is left with the false sense that one has succeeded in proving Hankelʼs integral theorem, when the truth is that many of the details of the demonstration are skipped when using the asymptotic form (42).

Although the techniques used in sections 2 and 3 appear to be mutually different, they are complementary; the bridge between them is attained at a higher mathematical level when using contour integral representations of ${{J}_{\nu }}(x)$, which is beyond the scope of this work. From a pedagogical point of view, they serve not only as a source of validation of Hankelʼs integral theorem but also as a way to develop valuable technical and analytical skills that can be applied to broader issues.