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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.

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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
@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 '14 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 '14 at 22:32

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:


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