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.










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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

















up vote
1
down vote

favorite
1












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.










share|cite|improve this question
























  • 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















up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





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.










share|cite|improve this question















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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edited Nov 17 at 21:51









Yuriy S

15.4k433115




15.4k433115










asked Sep 13 at 0:55









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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




















  • 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

















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