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I am trying to calculate this integral. I know it has an analytic expression when $a = 0$. But, is there any analytic expression for this case?

$$\int_{a}^{\infty}J_2(bx)J_1(cx)\,dx$$

Thanks in advance.

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  • $\begingroup$ The following may be helpful: For $a=0$ we have a known formula; writing $\int_0^\infty-\int_0^a$ we get a formula for your case; might be possible because $\int_0^a J_\mu(x)J_{\mu+1}(x)dx = \sum_{k \ge 0} J_{\mu+k+1}(a)^2$, though haven't given it more thought. $\endgroup$ – Suvrit Jan 27 '14 at 16:57
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    $\begingroup$ It is not clear what you mean by "analytic expression". The integral that you wrote is an analytic expression (in my vocabulary). $\endgroup$ – GH from MO Jan 27 '14 at 18:52
  • $\begingroup$ I mean analytic solution. Sorry for confusion. $\endgroup$ – bordart Jan 27 '14 at 19:03
  • $\begingroup$ It is not clear what you mean by "analytic solution". We are talking about an integral (not an equation), which is analytic (complex differentiable) in $a$. The word "solution" makes no sense in this context. $\endgroup$ – GH from MO Jan 27 '14 at 19:11
  • $\begingroup$ OK, let me explain it in this way. Can you solve the integral? Find any F(x) in an explicit way, that you may put it on the right side. $\endgroup$ – bordart Jan 27 '14 at 19:35
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For the case $b=c$ ... $$ \int \!{{\rm J}_2\left(bx\right)}{{\rm J}_1\left(bx\right)}{dx}= \frac{1}{2b}-{\frac { \left( {{\rm J}_0\left(bx\right)} \right) ^{2}}{2b}}-{\frac { \left( {{\rm J}_1\left(bx\right)} \right) ^{2}}{b }} $$ (I used Maple.)

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  • $\begingroup$ unfortunately in my case $b\neq c$ $\endgroup$ – bordart Jan 28 '14 at 8:45
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Using formula (18.17) at this link here, you can get a power-series for $J_2(bx)J_1(cx)$, which you can simplify and integrate term-by-term to obtain a "closed-form" expression for your integral. Maple or Mathematica might be able to simplify that even further.

Alternatively, you can look the book Integrals of Bessel Functions by Y. L. Luke, McGraw Hill. 1962.

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