asymptotic limit of $\int_0^{\infty}\left(1-\frac{t^2}{2(2k+3)}+\frac{t^4}{2\cdot 4\cdot(2k+3)\cdot(2k+5)}\right)^qdt$

Question

Help me please with the following integral. I've asked this question before https://math.stackexchange.com/q/140460/30554, but it turns out that it was incorrect question. I should get an asymptotic limit (with $k$ goes to $\infty$, and $q$ fixed) of it. For $q\ge 2, t\ge 0, k \in Z$ $$ \int_0^{\infty}\left(1-\frac{t^2}{2(2k+3)}+\frac{t^4}{2\cdot 4\cdot(2k+3)\cdot(2k+5)}\right)^qdt $$

Thank you.

Answer 1

comment

I'm guessing Alex wants hypergeometric $$\begin{align} {}_0F_1(;k+3/2;-t^2/4)&=1-\frac{t^2}{2(2k+3)}+\frac{t^4}{2\cdot 4\cdot(2k+3)\cdot(2k+5)}+\dots \\ &=\Gamma(k+3/2) t^{-k-1/2} 2^{k+1/2} J_{k+1/2}(t) \end{align}$$ inside the parentheses. Then, for example, with $k=10, q=3$ we get $$ \int_0^1 \frac{2147483648\;\Gamma(23/2)^3 \sqrt{2} J_{21/2}(t)^3\,dt}{t^{63/2}} = \frac{2043784548694122266058130317213\pi}{1869392021479944872009840721920} $$

added

With $f=\Gamma(k+3/2) t^{-k-1/2} 2^{k+1/2} J_{k+1/2}(t)$, Maple 16 has these asymptotic claims as $k \to \infty$: $$\begin{align} \int_0^\infty f\,dt &= \sqrt{\pi}\,k^{1/2} + \frac{3\sqrt{\pi}}{8}\,k^{-1/2} - \frac{7\sqrt{\pi}}{128}\,k^{-3/2}+O(k^{-5/2}) \\ \int_0^\infty f^2\,dt &=\frac{\sqrt{\pi}}{\sqrt{2}}\,k^{1/2}+ \frac{9\sqrt{\pi}}{16\sqrt{2}}\,k^{-1/2} -\frac{31\sqrt{\pi}}{512\sqrt{2}}\,k^{-3/2} + O(k^{-5/2}) \end{align}$$ and did not do $f^3$.