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Do you know how to do this integral? $$\int\limits_{0}^{2\pi}\mathrm{d}\phi\,\frac{J_2\left(\sqrt{a^2+b^2-2ab\cos(\phi)}\right)}{a^2+b^2-2ab\cos(\phi)}\,,$$ where $J_2$ is the Bessel function of the first kind of second order, and a and b are two positive constants.

I have tried various different tricks: using integral representation of the Bessel function, series expansion of the Bessel function, or converting the integral into complex integral over the unit circle, but I couldn't simplify the results I got afterward.

Thanks.

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For any $x>0$ we have $$ \frac{J_2(\sqrt{x})}{x}=\sum_{n\geq 0}\frac{(-1)^n x^n}{4^{n+1}n!(n+2)!} $$ and $$ \int_{0}^{2\pi}(a^2+b^2-2ab\cos\phi)^n\,d\phi = -i\oint_{|z|=1}(a-bz)^n(a-b/z)^n\frac{dz}{z}$$ by the residue theorem, equals $$ 2\pi \sum_{k=0}^{n}\binom{n}{k}^2 a^{2k} b^{2n-2k} $$ so the given integral can be represented as $$ 2\pi\sum_{n\geq 0}\frac{(-1)^n b^{2n}}{4^{n+1}n!(n+2)!}\sum_{k=0}^{n}\binom{n}{k}^2 \left(\frac{a}{b}\right)^{2k}$$ or as $$\frac{\pi}{2}\sum_{m,n\ge 0}\frac{(-1)^m a^{2m} (-1)^n b^{2n}}{4^m m!^2 4^n n!^2}\cdot\frac{1}{(m+n+2)(m+n+1)} $$ $$=\frac{\pi}{2}\sum_{m,n\ge 0}\frac{(-1)^m a^{2m} (-1)^n b^{2n}}{4^m m!^2 4^n n!^2}\int_{0}^{1}z^{m+n}(1-z)\,dz $$ or as $$\frac{\pi}{2}\int_{0}^{1}J_0(a\sqrt{z})J_0(b\sqrt{z})(1-z)\,dz = \pi\int_{0}^{1}J_0(az)J_0(bz)z(1-z^2)\,dz$$ I need a few time to check if this can be simplified further; in any case the numerical approximation of the involved integrals is simple by exploiting the Taylor series of $J_0$ or $J_2$ close to the origin and Tricomi's $J_0(z)\approx \frac{\sin(z)+\cos(z)}{\sqrt{\pi z}}$ far from the origin.

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From the Gegenbauer’s addition theorem: \begin{equation} \frac{J_{2}\left(w\right)}{w^{2}}=4 \sum_{k=0}^{\infty}(2+k)\frac{J_{2+k}\left(a\right)}{a^{2}}% \frac{J_{2+k}\left(b\right)}{b^{2}}C^{(2)}_{k}\left(\cos\varphi\right) \end{equation} where \begin{equation} w=\sqrt{a^{2}+b^{2}-2ab\cos\varphi} \end{equation} and where $C^{(2)}_{k}\left(\cos\varphi\right)$ is a Gegenbauer (or ultraspherical) polynomial. Then, \begin{align} I&=\int\limits_{0}^{2\pi}\mathrm{d}\varphi\,\frac{J_2\left(\sqrt{a^2+b^2-2ab\cos(\varphi)}\right)}{a^2+b^2-2ab\cos(\varphi)}\\ &=4 \sum_{k=0}^{\infty}(2+k)\frac{J_{2+k}\left(a\right)}{a^{2}}% \frac{J_{2+k}\left(b\right)}{b^{2}}\int\limits_{0}^{2\pi}\mathrm{d}\varphi C^{(2)}_{k}\left(\cos\varphi\right) \end{align} From the representationDLMF \begin{equation} C^{(2)}_{k}\left(\cos\varphi\right)=\sum_{\ell=0}^{k}\frac{{\left(2 \right)_{\ell}}{\left(2\right)_{k-\ell}}}{\ell!\;(k-\ell)!}\cos\left((k-% 2\ell)\varphi\right) \end{equation} the only non-vanishing integral terms in the summation correspond to $k=2\ell$, then $k$ is an even number: $k=2p$ \begin{equation} \int\limits_{0}^{2\pi}\mathrm{d}\varphi C^{(2)}_{2p}\left(\cos\varphi\right)=2\pi\left( p+1 \right)^2 \end{equation} and thus \begin{equation} I=\frac{16\pi}{a^2b^2} \sum_{p=1}^{\infty}p^3J_{2p}\left(a\right)J_{2p}\left(b\right) \end{equation}

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  • $\begingroup$ yeah, I got this sum too and don't know how to sum. I expect the result to be some difference between two Bessel functions of different arguments multiplied with some factors. It stems from a physics result which in the 3D case can be simplified thanks to the \sin(\phi) we get from spherical coordinates. In 2D there is no \sin(\phi) $\endgroup$ – Willi Tschau Jun 22 '18 at 23:53
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Here's one that's close:

From "Table of Integrals, Series, and Products" by I.S. Gradshteyn and I.M. Ryzhik, section 6.684, pg 726.

$$\int_0^{\pi} (\sin x)^{2v}\dfrac{J_v(\sqrt{a^2+b^2-2ab\cos x})}{(\sqrt{a^2+b^2-2ab\cos x})^{v}}dx =2^v\sqrt{\pi}\Gamma(v+\frac12) \dfrac{J_v(a)}{a^v} \dfrac{J_v(b)}{b^v} \quad[Re\ v > -\frac12] $$

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  • $\begingroup$ sadly, the integrand has (\sin(x))^{2\nu} there. $\endgroup$ – Willi Tschau Jun 22 '18 at 23:50
  • $\begingroup$ I know, but that's all I could find there. $\endgroup$ – marty cohen Jun 23 '18 at 1:21
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Too long to be a comment, but not an answer. P. Enta and the OP say they have reduced the integral to some factors multiplying $$ \sum_{m=1}^\infty m^3 J_{2m}(a)\,J_{2m}(b).$$ The desired form of 'a sum of a few Bessel functions' I think is impossible, and here's why: The Graf addition formula for Bessel functions can be written, for Bessel function index $\nu=0$ $$J_0\Big(\sqrt{a^2+b^2-2a\,b\,\cos(t)}\Big)= J_{0}(a)\,J_{0}(b) + 2 \sum_{m=1}^\infty J_{m}(a)\,J_{m}(b) \cos{(m\,t)}.$$ Differentiate this an even number of times and you'll get $\pm \,m^{2k} \cos{(m\,t)}$ within the sum on the right-hand side (RHS). You can get the index doubling by taking $t=0$ and $t=\pi$ and adding the results since $\cos(0) + \cos(\pi\,m)=2(1+(-1)^m).$ The LHS will obviously be a finite sum of Bessel functions. This trick cannot work for odd-order derivatives since on the RHS you will have $\pm \,m^{2k+1} \sin{(m\,t)}.$ You can't find a linear combination of $t$ that gives you $(1+(-1)^m)$ with the sine fucntion.

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