asymptotic expansion of integral of Bessel function?
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The integral is
$$int_0^infty dxfrac{x J_1(sqrt{y} cdot x/pi)^2}{ysinh^2(x)}$$
I want to compute it in large $y$ expansion. I have checked numerically that the integral is convergent for all positive $y$. Numerically, the leading term seems to be $frac{1}{2y}$, i would like to compute at least the next few terms.
definite-integrals asymptotics
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up vote
1
down vote
favorite
The integral is
$$int_0^infty dxfrac{x J_1(sqrt{y} cdot x/pi)^2}{ysinh^2(x)}$$
I want to compute it in large $y$ expansion. I have checked numerically that the integral is convergent for all positive $y$. Numerically, the leading term seems to be $frac{1}{2y}$, i would like to compute at least the next few terms.
definite-integrals asymptotics
The same question has an answer here. (Just change $y$ into $sqrt{y}/pi$ and divide the result by $y$).
– Paul Enta
Nov 17 at 22:59
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The integral is
$$int_0^infty dxfrac{x J_1(sqrt{y} cdot x/pi)^2}{ysinh^2(x)}$$
I want to compute it in large $y$ expansion. I have checked numerically that the integral is convergent for all positive $y$. Numerically, the leading term seems to be $frac{1}{2y}$, i would like to compute at least the next few terms.
definite-integrals asymptotics
The integral is
$$int_0^infty dxfrac{x J_1(sqrt{y} cdot x/pi)^2}{ysinh^2(x)}$$
I want to compute it in large $y$ expansion. I have checked numerically that the integral is convergent for all positive $y$. Numerically, the leading term seems to be $frac{1}{2y}$, i would like to compute at least the next few terms.
definite-integrals asymptotics
definite-integrals asymptotics
edited Nov 17 at 21:51
Yuriy S
15.3k433115
15.3k433115
asked Sep 13 at 0:55
esches
212
212
The same question has an answer here. (Just change $y$ into $sqrt{y}/pi$ and divide the result by $y$).
– Paul Enta
Nov 17 at 22:59
add a comment |
The same question has an answer here. (Just change $y$ into $sqrt{y}/pi$ and divide the result by $y$).
– Paul Enta
Nov 17 at 22:59
The same question has an answer here. (Just change $y$ into $sqrt{y}/pi$ and divide the result by $y$).
– Paul Enta
Nov 17 at 22:59
The same question has an answer here. (Just change $y$ into $sqrt{y}/pi$ and divide the result by $y$).
– Paul Enta
Nov 17 at 22:59
add a comment |
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The same question has an answer here. (Just change $y$ into $sqrt{y}/pi$ and divide the result by $y$).
– Paul Enta
Nov 17 at 22:59