Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
    
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. –  Suvrit Jan 27 at 16:57
1  
It is not clear what you mean by "analytic expression". The integral that you wrote is an analytic expression (in my vocabulary). –  GH from MO Jan 27 at 18:52
    
I mean analytic solution. Sorry for confusion. –  artalexan Jan 27 at 19:03
    
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. –  GH from MO Jan 27 at 19:11
    
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. –  artalexan Jan 27 at 19:35

2 Answers 2

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.)

share|improve this answer
    
unfortunately in my case $b\neq c$ –  artalexan Jan 28 at 8:45

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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