I'm trying to calculate the following integral:

$$ \int_{0}^{1}x\,\mathrm{J}_{2}\!\left(\,bx\,\right) \sin\left(\,a\,\sqrt{\,1 - x^{2}\,}\,\right)\,\mathrm{d}x $$where $a$ and $b$ are parameters ( independent of $x$ ).

Things I have tried so far, without success (but possibly not driven through far enough):

  1. look up in tables: no joy. The ''closest'' match I have found is eq. (6.738.1) in Gradshteyn and Ryzhik, but that result is for the first factor being $x^3$ instead of $x$;
  2. use WolframAlpha (Standard): does not give values for general $a$ and $b$; only for assigned selected $a$ and $b$. Since I need results for a wide range of values for both $a$ and $b$, this (or any numerical quadrature for arbitrary $a$ and $b$) is not practical;
  3. use an integral representation for Bessel functions: \begin{align} J_2(bx) = \frac{1}{\pi} \int^\pi_0 \cos(2\theta - bx \sin\theta) d\theta \end{align} then swap the order of integration. However, the outer integration (i.e., with respect to $\theta$) then becomes problematic;
  4. use a recurrence relation for Bessel functions: \begin{align} J_2(bx) = (2/(bx)) J_1(bx) - J_0(bx) \end{align} This does not seem to simplify matters, because the square root in sin() remains a difficulty.
  5. attempt partial integration: since the integrand contains three factors, the choice is not obvious. I tried grouping the first two factors and use the partial integral result \begin{align} \int x^m J_n(x) dx = -x^m J_{n-1}(x) + (m+n-1) \int x^{m-1} J_{n-1}(x) dx \end{align} for $m=1$ , $n=2$, i.e., \begin{align} \int x J_2(x) dx = -x J_{1}(x) - 2 J_{0}(x) + C\end{align} but the derivative of $\sin(a\sqrt{1-x^2})$ with respect to $x$ complicates the remaining integration;
  6. use a series representation of the Bessel function: this leads to a double summation of integrals (one sum is semi-infinite) of the form \begin{align} \int^1_0 u^{2m+1} \sin(a u) du \end{align} but this integral is a itself a difference of hypergeometric functions (or alternatively an additional series representation). Such a double or triple summation is again impractical for calculating for parametrized $a$ and $b$;
  7. converting the original integral to \begin{align} \int^1_0 y J_2 (b \sqrt{1-y^2}) \sin(a y) dx \end{align} where $y=\sqrt{1-x^2}$ and repeating the previous approaches;
  8. attempt a trigonometric substitution such as $x = \sin \alpha$ and repeat the previous approaches.

Can you find a solution or give further suggestions what could be attempted?

  • 1
    $\begingroup$ Have you tried the very-brute-force approach of writing $x J_2(bx)$ as its Taylor series, $\sin(a\sqrt{1-x^2})$ as a power series in $\sqrt{1-x^2}$, then computing the integral of the product by exploiting the fact that $\int_{0}^{1}x^m\sqrt{1-x^2}^n\,dx$ is given by Euler's Beta function? I am not sure this leads to something manageable, it just looks like a viable way. $\endgroup$ Jul 9, 2017 at 18:08
  • $\begingroup$ In the worst scenario the original integral is turned into a value of a hypergeometric function with not too many parameters. $\endgroup$ Jul 9, 2017 at 18:10
  • $\begingroup$ nicely written question (+1) $\endgroup$
    – tired
    Jul 9, 2017 at 22:05
  • $\begingroup$ What about $x=\cos\phi$ and then $\int_0^{\pi} \cos\phi \sin \phi J(b\cos\phi) \sin(a\sin\phi)d\phi$? Maybe it leads to a different expansion... $\endgroup$
    – orion
    Jul 10, 2017 at 8:07

2 Answers 2


This is not an answer either, but MSE wouldn't let me edit my comment.

Let $M(a,b)$ be your integral, and let $N(a,b)$ be the integral you found in the table with $x^3$. Then $M$ and $N$ are related by the ODE: $M(a,b)-N(a,b)=\frac{\partial^2 M}{\partial a^2}(a,b)$, which can be solved for $M(a,b)$ in terms of $N$ (which is in your table), $M(0,b)=0$, and $\frac{\partial M}{\partial a}(0,b)$ in the usual way. According to WolframAlpha, $$\frac{\partial M}{\partial a}(0,b)=\int_0^1xJ_2(bx)\sqrt{1-x^2}dx=\frac{b(2+\cos b)-3\sin b}{b^3}.$$ So if $N(a,b)$ is sufficiently nice, you might have a chance!


This is not an answer but it is too long for a comment.

Inspired by Jack D'Aurizio's first comment, let us write the series

$$J_2 (b x)=\sum_{m=0}^\infty \frac{(-1)^m}{m!(m+2)!} \left(\frac {b}2\right)^{2m+2}x^{2m+2}$$ $$\sin(a\sqrt{1-x^2})=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}a^{2n+1}\left(1-x^2\right)^{n+\frac 12}$$ which let us with a double summation of integrals $$I_{m,n}=\int_0^1 x^{2m+3}\left(1-x^2\right)^{n+\frac 12}\,dx=\frac{\Gamma (m+2) \Gamma \left(n+\frac{3}{2}\right)}{2\, \Gamma \left(m+n+\frac{7}{2}\right)}$$ which leads to

$$\int^1_0 x J_2 (b x) \sin(a\sqrt{1-x^2})\, dx=\sqrt{\pi }\sum _{m=0}^{\infty } \sum _{n=0}^{\infty } (-1)^{m+n}\frac{ a^{2 n+1}\, b^{2 m+2} }{2^{2 (m+n+2)}(m+2) \,m! \,n!\, \Gamma \left(m+n+\frac{7}{2}\right)}$$ which, looking at some numerical tests, seems to converge quite fast.

For illustration purposes, using $\sum _{m=0}^{p } \sum _{n=0}^{p }$, $a=3$ and $b=5$ the following decimal representations are obtained $$\left( \begin{array}{cc} p & \text{result} \\ 2 & 0.263943 \\ 3 & 0.083938 \\ 4 & 0.119953 \\ 5 & 0.114727 \\ 6 & 0.115299 \\ 7 & 0.115250 \\ 8 & 0.115253 \end{array} \right)$$ which is identical (at least for six significant figures) to the result of the numerical integration.


May be interesting is that, thanks to a CAS, summing over $n$, the integral can be expressed as $$\int^1_0 x J_2 (b x) \sin(a\sqrt{1-x^2})\, dx=\sqrt{\pi }\sum _{m=0}^{\infty }(-1)^m\frac{b^{2m+2} }{(2a)^{m+\frac 32}\, (m+2)\, m! }\,J_{m+\frac{5}{2}}(a)$$

$$\int^1_0 x J_2 (b x) \sin(a\sqrt{1-x^2})\, dx=2a\sum _{m=0}^{\infty }(-1)^m\frac{ (m+1) b^{2 m+2} }{\Gamma (2 m+6)}\, _0F_1\left(;m+\frac{7}{2};-\frac{a^2}{4}\right)$$

Using the same example as above $$\left( \begin{array}{cc} m & \sum_{k=0}^m \\ 2 & 0.249952 \\ 3 & 0.085208 \\ 4 & 0.119870 \\ 5 & 0.114731 \\ 6 & 0.115298 \\ 7 & 0.115250 \\ 8 & 0.115253 \end{array} \right)$$

  • $\begingroup$ Thanks Jack and Claude. A double sum result is indeed what I also had as a solution so far, although through a different route (item 6). The beta function approach is quite elegant. I gave up at the point of evaluating numerically, but it seems convergence is quite rapid. $\endgroup$
    – TeXCub
    Jul 10, 2017 at 13:52
  • $\begingroup$ @TeXCub. I have been running a bunch of these integrals using the simple summation over $J_{m+\frac{5}{2}}(a)$. Convergence is really fast. $\endgroup$ Jul 11, 2017 at 3:20
  • $\begingroup$ @Claude: can you provide some guidance as to how CAS can be used to reduce the double sum to a single sum, as you did? I have not much experience, but I tried inputting "simplify sum ((-1)^n * a^(2*n+1) / (2^(2*n+4)*n!) * sum ((-1)^(m) * b^(2m+2) / (2^(2*m) * (m+2) *m! * gamma(m+n+7/2)) from m=0 to infinity) from n=0 to infinity)" in WolframAlpha (or even without the operator "simplify", but it did not even incorporate the outer sum. Do you use this or another CAS package, and how did you do it? Any pointers will be gratefully received. Thanks. $\endgroup$
    – Lucozade
    Jul 22, 2017 at 11:54
  • $\begingroup$ [PS: my motivation: using pen and paper, I managed to re-derive the single-sum expression in your EDIT (involving Bessel function) from the double-sum solution you give previously. Now I want to use CAS to related integrals for which these steps cannot be trivially extended, but I'm still interested to find if there is an alternative single-sum representation.] $\endgroup$
    – Lucozade
    Jul 22, 2017 at 12:18
  • $\begingroup$ @Lucozade. I use another CAS. Could you try FullSimplify[...] and tell me if it improves (or not) ? $\endgroup$ Jul 22, 2017 at 13:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.