Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $n$ be half an odd integer, say $n=k+1/2, k \in \mathbb{N}$.

Let $q\geq 1$. I would like to calculate (or approximate) the following integral: $$ \int_0^{\infty}\left(\sqrt{\frac{\pi}{2}}\cdot 1\cdot 3\cdot 5\cdots (2k+1) \frac{J_{k+\frac 12}(t)}{t^{k+ \frac 12}}\right)^q t\ dt. $$

Any ideas or references will be very helpful.

Thank you.

share|improve this question
    
Maybe the explicit statement of the function $J$ could be helpful? –  awllower May 27 '12 at 7:54
    
@awllower: Which ine would you propose?I've tried few representations-did not work. Thank you. –  David May 27 '12 at 13:25
1  
@Jack: I don't know if bessel is a good tag name. I also don't think that adding new tags should be done without consulting the community via the Mathematics Meta site. –  Asaf Karagila Aug 30 at 22:30
    
@AsafKaragila: good point, I'll post a proposal on Meta to create a specific tag for Bessel-functions-related questions. –  Jack D'Aurizio Aug 30 at 22:32
1  

1 Answer 1

Due to Rayleigh's formulas we have:

$$\sqrt{\frac{\pi}{2}}\frac{J_{k+1/2}(t)}{t^{k+1/2}}=(-1)^k\left(\frac{1}{t}\frac{d}{dt}\right)^k \frac{\sin t}{t}\tag{1}$$ and since: $$\frac{\sin t}{t}=\sum_{m=0}^{+\infty}\frac{(-1)^m\,t^{2m}}{(2m+1)!}$$ we have: $$\left(\frac{1}{t}\frac{d}{dt}\right)\frac{\sin t}{t}=\sum_{m=1}^{+\infty}\frac{(-1)^m(2m)t^{2m-2}}{(2m+1)!}=(-1)\sum_{m=0}^{+\infty}\frac{(-1)^m (2m+2)t^{2m}}{(2m+3)!},$$ $$\left(\frac{1}{t}\frac{d}{dt}\right)^k\frac{\sin t}{t}=(-1)^k\sum_{m=0}^{+\infty}\frac{(-1)^m (2m+2k)\cdot\ldots\cdot(2m+2)t^{2m}}{(2m+2k+1)!}$$ so: $$\begin{eqnarray*}(2k+1)!!\cdot\sqrt{\frac{\pi}{2}}\frac{J_{k+1/2}(t)}{t^{k+1/2}}&=&\sum_{m=0}^{+\infty}\frac{(-1)^m (2m+2k)!!(2k+1)!!}{(2m+2k+1)!(2m)!!}\,t^{2m}\\&=&\sum_{m=0}^{+\infty}\frac{(-1)^m \binom{m+k}{m}}{\binom{2m+2k+1}{2m}}\cdot\frac{t^{2m}}{(2m)!}.\end{eqnarray*}\tag{2}$$ Now a really good approximation for the LHS of $(2)$ is simply given by: $$(2k+1)!!\cdot\sqrt{\frac{\pi}{2}}\frac{J_{k+1/2}(t)}{t^{k+1/2}}\approx \exp\left(-\frac{t^2}{4k+6}\right).\tag{3}$$ $\hspace2in$Approximation for $k=3$ $\hspace2in\qquad$ Approximation for $k=3$

Hence the starting integral can be approximated by:

$$\int_{0}^{+\infty}t\,\exp\left(-\frac{qt^2}{4k+6}\right)\,dt=\frac{2k+3}{q}.$$

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.