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. $\endgroup$– SuvritCommented Jan 27, 2014 at 16:57
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1$\begingroup$ 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 MOCommented Jan 27, 2014 at 18:52
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$\begingroup$ I mean analytic solution. Sorry for confusion. $\endgroup$– bordartCommented Jan 27, 2014 at 19:03
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$\begingroup$ 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 MOCommented Jan 27, 2014 at 19:11
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$\begingroup$ 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. $\endgroup$– bordartCommented Jan 27, 2014 at 19:35
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2 Answers
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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.)
answered Jan 27, 2014 at 22:23
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$\begingroup$ unfortunately in my case $b\neq c$ $\endgroup$– bordartCommented Jan 28, 2014 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.
