Timeline for Estimating an integral involving Bessel functions
Current License: CC BY-SA 3.0
7 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
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| Nov 5, 2016 at 17:23 | comment | added | Igor Khavkine | First of all, the integrand is bounded and even continuous, so the only divergence can come from $r\gg 1$. So, if convergence of the integral is the only information you care about, there is no reason to bother with integrating the asymptotic estimates near the origin. Second, the only information that I used except the asymptotic inequality you already provided is that $1/\|x-b\| \le C_b 1/\|x\|$ for some constant $C_b$ and sufficiently large $\|x\|$. Hence, the integrand is a product, where each factor can be estimated in the same way. I think the rest follows easily. | |
| Nov 5, 2016 at 13:48 | comment | added | user363087 | Forgive my ignorance, but I'm not sure I understand your answer. Firstly, could you explain how you can use the asymptotic bound to achieve that? Secondly, if you plug that bound into the integral with polar co-ordinates, doesn't $r$ range from $0$ to $\infty$ -- in which case, the integral diverges? | |
| Nov 5, 2016 at 0:04 | comment | added | Igor Khavkine | Using your asymptotic bound, you can get $J_1(\rho\|b-x\|)/\|b-x\| \le C_b r^{-3/2}$, where $r = \|x\|$ and the constant is now allowed to depend on $b$. Plugging this bound into the integral in polar coordinates shows that it is dominated by $C_b \int^\infty dr/r^2$, which seems absolutely convergent to me. Am I missing anything? | |
| Nov 4, 2016 at 23:12 | history | edited | user363087 | CC BY-SA 3.0 |
Changed J_{d/2} to J_1, since d = 2 in this case.
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| Nov 4, 2016 at 19:41 | history | asked | user363087 | CC BY-SA 3.0 |