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I am trying to find a closed form (in terms of known functions) for this integral:

$$ \int_{0}^\Lambda\!\!\! \text{d} k\, J_0( k x)\sin (k y)$$

where $x>0$, $\Lambda>0$, $y\in \mathbb{R}$

I was able to find a closed expression when $\Lambda\rightarrow\infty$:

$$ \int_{0}^\infty\!\!\! \text{d} k\, J_0( k x)\sin (k y)=\frac{-\text{sgn}( y)}{\sqrt{y^2-x^2}}\Theta\big[y^2-x^2\big]$$

But I wasn't able to find a closed expression for the case of finite $\Lambda$. Do you know if it admits a closed form and if so what it would be or how to get it?

Thank you so much in advance.

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  • $\begingroup$ Getting rid of every useless parameter, you are just asking if $J_0(x)\sin(nx)$ has a nice primitive for every $n\in\mathbb{R}$. Do you think that a hypergeometric $\phantom{}_2 F_3$ function is "nice" ? $\endgroup$ – Jack D'Aurizio Sep 2 '15 at 16:57
  • $\begingroup$ HI Jack, thanks for your answer. Yes, that's precisely what I'm asking. I would consider hypergeometric functions "nice" for the purposes of what I'm looking for, yes. I would appreciate any help you could provide on this regard. $\endgroup$ – Collector Sep 2 '15 at 20:08
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$\int_0^\Lambda J_0(kx)\sin ky~dk$

$=\int_0^\Lambda\sum\limits_{m=0}^\infty\dfrac{(-1)^m\left(\dfrac{kx}{2}\right)^{2m}}{(m!)^2}\sin ky~dk$

$=\int_0^\Lambda\left(\sin yk+\sum\limits_{m=1}^\infty\dfrac{(-1)^mx^{2m}k^{2m}\sin yk}{4^m(m!)^2}\right)dk$

$=\left[-\dfrac{\cos yk}{y}+\sum\limits_{m=1}^\infty\sum\limits_{n=0}^{m-1}\dfrac{(-1)^{m+n}(2m)!x^{2m}k^{2m-2n-1}\sin yk}{4^my^{2n+2}(m!)^2(2m-2n-1)!}+\sum\limits_{m=1}^\infty\sum\limits_{n=0}^m\dfrac{(-1)^{m+n+1}(2m)!x^{2m}k^{2m-2n}\cos yk}{4^my^{2n+1}(m!)^2(2m-2n)!}\right]_0^\Lambda$ (can be obtained from https://en.wikipedia.org/wiki/List_of_integrals_of_trigonometric_functions)

$=\left[\sum\limits_{m=1}^\infty\sum\limits_{n=1}^m\dfrac{(-1)^n(2m)!x^{2m}k^{2n-1}\sin yk}{4^my^{2m-2n+2}(m!)^2(2n-1)!}+\sum\limits_{m=0}^\infty\sum\limits_{n=0}^m\dfrac{(-1)^{m+n+1}(2m)!x^{2m}k^{2m-2n}\cos yk}{4^my^{2n+1}(m!)^2(2m-2n)!}\right]_0^\Lambda$

$=\left[\sum\limits_{m=1}^\infty\sum\limits_{n=1}^m\dfrac{(-1)^n(2m)!x^{2m}k^{2n-1}\sin yk}{4^my^{2m-2n+2}(m!)^2(2n-1)!}+\sum\limits_{m=0}^\infty\sum\limits_{n=0}^m\dfrac{(-1)^{n+1}(2m)!x^{2m}k^{2n}\cos yk}{4^my^{2m-2n+1}(m!)^2(2n)!}\right]_0^\Lambda$

$=\sum\limits_{m=1}^\infty\sum\limits_{n=1}^m\dfrac{(-1)^n(2m)!x^{2m}\Lambda^{2n-1}\sin y\Lambda}{4^my^{2m-2n+2}(m!)^2(2n-1)!}+\sum\limits_{m=0}^\infty\sum\limits_{n=0}^m\dfrac{(-1)^{n+1}(2m)!x^{2m}\Lambda^{2n}\cos y\Lambda}{4^my^{2m-2n+1}(m!)^2(2n)!}+\sum\limits_{m=0}^\infty\dfrac{(2m)!x^{2m}}{4^my^{2m+1}(m!)^2}$

$=\sum\limits_{m=1}^\infty\sum\limits_{n=1}^m\dfrac{(-1)^n(2m)!x^{2m}\Lambda^{2n-1}\sin y\Lambda}{4^my^{2m-2n+2}(m!)^2(2n-1)!}+\sum\limits_{m=0}^\infty\sum\limits_{n=0}^m\dfrac{(-1)^{n+1}(2m)!x^{2m}\Lambda^{2n}\cos y\Lambda}{4^my^{2m-2n+1}(m!)^2(2n)!}+\dfrac{1}{\sqrt{y^2-x^2}}$

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