Bessel function integral: $\int_{0}^{\infty} e^{-ax}x^{m}\left(J_{0}(bx) \right)^{2} \, \mathrm dx$

Question

I am working on a problem. Solving the PDE for my problem, these Bessel integrals arise:

$$\int^\infty_0 e^{-ax}x^m(J_0(bx))^2dx,\quad \int^\infty_0 e^{-ax}x^m(J_1(bx))^2dx\qquad \text{and} \qquad\int^\infty_0 e^{-ax}x^mJ_0(bx)J_1(bx)dx$$

where $~J_0~$ and $~J_1~$ are the Bessel functions of first kind.

I haven't found the solution in any table or book, and due to my limited background in applied mathematics I don't know how to integrate it by myself.

Does anybody know the solution?

Thanks a lot in advance

Answer 1

Sorry this is not a clear answer. But I myself is in the process of following the derivations of some of the well known integrals involving Bessel functions, and you may find your answers in Luke's, "Integrals of Bessel Functions", p.314 and after. The pity is that you can't find in detail how they are derived. Maybe you should consult the original papers if you are interested in the derivations. It seems to me that the multiplications of the Bessel functions sre transformed to the single one by the additional theorem, which then is followed by transforming into elliptic integrals.

Answer 2

I have no access to Luke's book "Integrals of Bessel Functions" that 萬雄彦 recommended in his/her answer.

Using a CAS, I got some results. Naming $$I_{i,j}=\int_{0}^\infty e^{-ax}\, x^m\, J_i(bx)\, J_j(bx)\,dx$$ $$\color{blue}{I_{0,0}=a^{-(m+1)} \Gamma (m+1) \, _3F_2\left(\frac{1}{2},\frac{m+1}{2},\frac{m+2}{2};1,1;-\frac{4b^2}{a^2}\right)}$$ provided $b\geq 0\land \Re(m)+1>0\land ((\Re(a)=0\land \Re(m)<1)\lor \Re(a)>0)$ $$\color{blue}{I_{1,1}=\frac{b^2}{4} a^{-(m+3)} \Gamma (m+3) \, _3F_2\left(\frac{3}{2},\frac{m+3}{2},\frac{m+4}{2};2,3;-\frac{4b^2}{a^2}\right)}$$ provided $b\geq 0\land \Re(m)+1>0\land ((\Re(a)=0\land \Re(m)<1)\lor \Re(a)>0)$ $$\color{blue}{I_{0,1}=\frac{b}{2} a^{-(m+2)} \Gamma (m+2) \, _3F_2\left(\frac{3}{2},\frac{m+2}{2},\frac{m+3}{2};2,2;-\frac{4b^2}{a^2}\right)}$$ provided $b\geq 0\land \Re(m)+2>0\land ((\Re(a)=0\land \Re(m)<1)\lor \Re(a)>0)$

A few numerical checks have been done.

Edit

Making the problem more general, it seems that $$J_{i,j}= \frac{ 2^{(i+j)}\, \Gamma (i+1)\, \Gamma (j+1) \, a^{(m+i+j+1)}}{\Gamma (m+i+j+1)\, b^{(i+j)}} \, I_{i,j}$$ can write $$\color{blue}{J_{i,j}=\, _4F_3\left(\frac{i+j+1}{2},\frac{i+j+2}{2},\frac{m+i+j+1 }{2},\frac{m+i+j+2}{2};i +1,j+1,i+j+1;-\frac{4 b^2}{a^2}\right)}$$ provided $\Re(m+i+j)>-1\land b\geq 0\land ((\Re(a)=0\land \Re(m)<1)\lor \Re(a)>0)$

Answer 3

Similar to my answer here, we can use a generalization of Ramanujan's master theorem called the method of brackets to derive some of the results in Claude Leibovici's answer.

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Assume that $b \ge0 $, $a >2b$, and $m \in \mathbb{Z} >-1$.

The exponential function has the series representation $$e^{- ax} = \sum_{j=0}^{\infty} \frac{(-1)^{j}}{j!}(ax)^{j} = \sum_{j=0}^{\infty} \phi{_j}(ax)^{j}. $$

And the square of the Bessel function of the first kind of order zero has the series representation $$J_{0}(bx)^{2} = {_2F_1} \left(\frac{1}{2};1,1;-(bz)^{2} \right) = \sum_{k=0}^{\infty} \phi_{k} \frac{\Gamma \left(k+ \frac{1}{2} \right)}{\Gamma \left(\frac{1}{2} \right)} \frac{1}{\Gamma(k+1)^{2}} \, (bx)^{2k}.$$

Therefore, the integral $\int_{0}^{\infty}e^{-ax} x^{m} J_{0}(bx)^{2} \, \mathrm dx$ has the corresponding bracket series $$\sum_{j=0}^{\infty} \sum_{k=0}^{\infty} \phi_{j,k} \, a^{j} \, b^{2k} \, \frac{\Gamma \left(k+ \frac{1}{2} \right)}{\Gamma \left(\frac{1}{2} \right)} \frac{1}{\Gamma(k+1)^{2}} \, \langle j+2k+m+1 \rangle.$$

Since there are $2$ indices but only $1$ bracket, there are two cases to examine.

If we let $k$ be a free parameter, then the bracket vanishes if $j=-2k-m-1$, and we have the contribution $$\begin{align} S_{1} &= \sum_{k=0}^{\infty} \frac{(-1)^{k}}{k!} a^{-2k-m-1} \, b^{2k} \frac{\Gamma \left(k+ \frac{1}{2} \right)}{\Gamma \left(\frac{1}{2} \right)} \frac{1}{\Gamma(k+1)^{2}} \, \Gamma \left( 2k+m+1 \right) \\ &= \sum_{k=0}^{\infty} \frac{1}{k!} a^{-(m+1)} \, \left( -\frac{b^{2}}{a^{2}} \right)^{k} \, \frac{\Gamma \left(k+ \frac{1}{2} \right)}{\Gamma \left(\frac{1}{2} \right)} \frac{1}{\Gamma(k+1)^{2}} \, \frac{2^{2k+m}}{\Gamma \left(\frac{1}{2} \right)} \, \Gamma \left(k+ \frac{m}{2} + \frac{1}{2} \right) \Gamma \left(k+ \frac{m}{2}+1 \right) \\ &= a^{-(m+1)} \, \frac{2^{m}}{\Gamma \left(\frac{1}{2} \right)} \, \Gamma \left(\frac{m}{2}+ \frac{1}{2} \right) \Gamma \left( \frac{m}{2}+1 \right) \, {_3F_2} \left(\frac{1}{2}, \frac{m+1}{2}, \frac{m+2}{2}; 1,1; -\frac{4b^{2}}{a^{2}} \right) \\ &= a^{-(m+1)} \, 2^{m} \, \Gamma \left(2 \left(\frac{m}{2}+ \frac{1}{2} \right) \right) 2^{-2(m/2+1/2)+1} \, {_3F_2} \left(\frac{1}{2}, \frac{m+1}{2}, \frac{m+2}{2}; 1,1; -\frac{4b^{2}}{a^{2}} \right) \\ &= a^{-(m+1)} \, \Gamma(m+1) \, {_3F_2} \left(\frac{1}{2}, \frac{m+1}{2}, \frac{m+2}{2}; 1,1; -\frac{4b^{2}}{a^{2}} \right). \end{align}$$

On the second and fourth lines I used Legendre's duplication formula.

If we let $j$ be a free parameter, then the bracket vanishes if $k=- \frac{j}{2} - \frac{m}{2} - \frac{1}{2}$, and we have the contribution $$ S_{2} = \frac{1}{2} \sum_{j=0}^{\infty}\frac{(-1)^{j}}{j!} \, a^{j} \, b^{-j-m-1} \, \frac{\color{red}{\Gamma \left(- \frac{j+m}{2} \right)}}{\Gamma \left(\frac{1}{2} \right)} \frac{1}{\Gamma \left(\frac{-j-m+1}{2} \right)^{2}} \, \Gamma \left(\frac{j+m+1}{2} \right). $$

Since not all terms of this series are finite, we discard it.

Therefore, we have $$\int_{0}^{\infty}e^{-ax} x^{m} J_{0}(bx)^{2} \, \mathrm dx = S_{1} = \Gamma(m+1) \, a^{-(m+1)}\, {_3F_2} \left(\frac{1}{2}, \frac{m+1}{2}, \frac{m+2}{2}; 1,1; -\frac{4b^{2}}{a^{2}} \right)$$ for $b \ge 0$, $a>2b$, and $m \in \mathbb{Z} >-1$.

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Now assume that $b \ge 0$, $a>2b$, and $m \in \mathbb{Z} > -2$.

The product $J_{0}(bx) J_{1}(bx)$ has the series representation $$ \begin{align} J_{0}(bx)J_{1}(bx) &= \frac{bx}{2} \, {_2F_3} \left( \frac{3}{2}, 1; 1, 2, 2; (bx)^{2} \right) \\ &= \frac{bx}{2} \, {_1F_2} \left( \frac{3}{2}; 2, 2; (bx)^{2} \right) \\ &= \frac{bx}{2} \sum_{k=0}^{\infty} \phi_{k} \frac{\Gamma \left(k+ \frac{3}{2} \right)}{\Gamma \left(\frac{3}{2} \right)} \frac{1}{\Gamma \left(k+2 \right)^{2}} (bx)^{2k}. \end{align}$$

Therefore, the integral $\int_{0}^{\infty} e^{-ax} x^{m} J_{0}(bx)J_{1}(bx) \, \mathrm dx$ has the corresponding bracket series $$ \frac{b}{2} \sum_{j=0}^{\infty} \sum_{k=0}^{\infty} \phi_{j,k} \, a^{j} \, b^{2k} \, \frac{\Gamma \left(k+ \frac{3}{2} \right)}{\Gamma \left(\frac{3}{2} \right)} \frac{1}{\Gamma(k+2)^{2}} \, \langle j+2k+m+2 \rangle.$$

There are again two cases to examine.

If we let $j$ be a free parameter, then the bracket vanishes if $ j= -2k-m-2$, and we have the contribution $$ \begin{align} S_{3} &= \frac{b}{2} \sum_{k=0}^{\infty} \frac{(-1)^{k}}{k!} \, a^{-2k-m-2} \, b^{2k} \, \frac{\Gamma \left(k+ \frac{3}{2} \right)}{\Gamma \left(\frac{3}{2} \right)} \frac{1}{\Gamma(k+2)^{2}} \Gamma \left(2k+m+2 \right) \\ &= \small \frac{b}{2} \sum_{k=0}^{\infty} \frac{1}{k!} a^{-(m+2)} \left(-\frac{b^{2}}{a^{2}} \right)^{k} \, \frac{\Gamma \left(k+ \frac{3}{2} \right)}{\Gamma \left(\frac{3}{2} \right)} \frac{1}{\Gamma(k+2)^{2}} \frac{2^{2k+m+1}}{\Gamma \left(\frac{1}{2} \right)} \, \Gamma \left(k+ \frac{m}{2} + 1\right) \Gamma \left(k+ \frac{m}{2}+\frac{3}{2} \right) \\ &= \frac{b}{2} \frac{a^{-(m+2)} 2^{m+1}}{\Gamma\left(\frac{1}{2} \right)} \, \Gamma \left(\frac{m}{2}+1 \right) \Gamma \left(\frac{m}{2} + \frac{3}{2} \right) {_3F_2} \left(\frac{3}{2}, \frac{m+2}{2}; \frac{m+3}{2};2,2; - \frac{4b^{2}}{a^{2}} \right) \\ &= \frac{b}{2} \, \frac{a^{-(m+2)} 2^{m+1}}{\Gamma \left(\frac{1}{2} \right)} \, \Gamma \left(\frac{1}{2} \right) 2^{-m-1} \, \Gamma(m+2) \, {_3F_2} \left(\frac{3}{2}, \frac{m+2}{2}; \frac{m+3}{2};2,2; - \frac{4b^{2}}{a^{2}} \right) \\ &= \frac{b}{2} \, a^{-(m+2)} \, \Gamma(m+2) \, {_3F_2} \left(\frac{3}{2}, \frac{m+2}{2}; \frac{m+3}{2};2,2; - \frac{4b^{2}}{a^{2}} \right). \end{align}$$

If we let $k$ be a free parameter, we get a series like before that doesn't have all finite terms.

Therefore, we have $$\small \int_{0}^{\infty} e^{-ax} x^{m} J_{0}(bx) J_{1}(bx) \, \mathrm dx = S_{3} = \frac{b}{2} \, a^{-(m+2)} \, \Gamma(m+2) \, {_3F_2} \left(\frac{3}{2}, \frac{m+2}{2}, \frac{m+3}{2};2,2; - \frac{4b^{2}}{a^{2}} \right)$$ for $b \ge 0$, $a > 2b$, and $m \in \mathbb{Z} > -2$.

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Since the Laplace transform defines an analytic function in the half-plane that it converges absolutely, analytic continuation extends both results to $\Re(a) >0$.

Answer 4

The product of Bessel functions is expressible as a hypergeometric series:

$$J_{v}(x)J_{\mu}(x) = \frac{\left({\frac{x}{2}}\right)^{v+\mu}}{\Gamma(1+v)\Gamma(1+\mu)}\sum_{j=0}^{\infty} \frac{\left(\frac{1+v+\mu}{2}\right)_{j}\left(1+\frac{v+\mu}{2}\right)_{j}(-x^2)^j}{(1)_{j}(1+v)_{j}(1+\mu)_{j}(1+v+\mu)_{j}}$$

So if $v=\mu=0$

$$J_{0}^2(x) = \sum_{j=0}^{\infty} \frac{\left(\frac{1}{2}\right)_{j}(-x^2)^j}{(1)_{j}(1)_{j}(1)_{j}}$$

Then

\begin{align*} \int_{0}^{\infty} e^{-ax}J_{0}^2(bx)dx =& \int_{0}^{\infty}e^{-ax}\sum_{j=0}^{\infty} \frac{\left(\frac{1}{2}\right)_{j}(-b^2x^2)^j}{(1)_{j}(1)_{j}(1)_{j}}dx\\ =& \sum_{j=0}^{\infty} \frac{\left(\frac{1}{2}\right)_{j}(-1)^jb^{2j}}{(1)_{j}(1)_{j}(1)_{j}}\int_{0}^{\infty} e^{-xa}x^{2j} dx\\ =& \sum_{j=0}^{\infty} \frac{\left(\frac{1}{2}\right)_{j}(-1)^jb^{2j}}{(1)_{j}(1)_{j}(1)_{j}}\frac{1}{a^{2j+1}} \int_{0}^{\infty} e^{-w}w^{2j} dw\\ =& \frac{1}{a}\sum_{j=0}^{\infty} \frac{\left(\frac{1}{2}\right)_{j}\Gamma(2j+1)(-1)^j}{(1)_{j}(1)_{j}} \frac{\left(\frac{b}{a}\right)^{2j} }{j!}\\ =& \frac{1}{a} \sum_{j=0}^{\infty} \frac{\left(\frac{1}{2}\right)_{j}\left(\frac{1}{2}\right)_{j}}{(1)_{j}} \frac{\left(-\frac{4b^2}{a^2}\right)^{j} }{j!}\\ =& \frac{1}{a}{}_{2}F_{1}\left(\frac{1}{2},\frac{1}{2};1;-\frac{4b^2}{a^2}\right) \end{align*}

Recalll that

$$K(p) = \frac{\pi}{2} {}_{2}F_{1}\left(\frac{1}{2},\frac{1}{2};1;p^2\right)$$

where $K(p)$ is the complete elliptic integral of the first kind with modulus $p$.

So

$$\int_{0}^{\infty} e^{-ax}J_{0}^2(bx)dx = \frac{2}{a\pi}K\left(i\frac{2b}{a}\right)$$

since $a\in \mathbb{R}$ and if $b\in \mathbb{R}$ this modulus is purely imagnary and we can make the transformation

$$K(ip) = \frac{1}{\sqrt{1+p^2}}K\left(\frac{p}{\sqrt{1+p^2}}\right)$$ So

$$\int_{0}^{\infty} e^{-ax}J_{0}^2(bx)dx = \frac{2}{\pi\sqrt{a^2+4b^2}}K\left(\frac{2b}{\sqrt{a^2+4b^2}}\right)$$

Differentiating both sides with respect to $a$ $m$ times

$$\int_{0}^{\infty} (-1)^mx^me^{-ax}J_{0}^2(bx)dx = \frac{d^m}{da^m} \left[\frac{2}{\pi\sqrt{a^2+4b^2}}K\left(\frac{2b}{\sqrt{a^2+4b^2}}\right)\right]$$

Then

$$\boxed{\int_{0}^{\infty} x^me^{-ax}J_{0}^2(bx)dx = (-1)^m\frac{d^m}{da^m} \left[\frac{2}{\pi\sqrt{a^2+4b^2}}K\left(\frac{2b}{\sqrt{a^2+4b^2}}\right)\right]}$$

However, this is not an easy to calculate derivative and the complexity compounded fast.

We can do a similar process for the case $v=\mu=1$. See entry 6.612 in Gradshteyn and Ryzhik, Table of integrals, series, and products, 7th edition.