Closed form of integral involving Bessel $J_0$ 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.
 A: $\int_0^\Lambda J_0(kx)\sin ky~dk$
$=\int_0^\Lambda\sum\limits_{m=0}^\infty\dfrac{(-1)^m\left(\dfrac{kx}{2}\right)^{2m}}{(m!)^2}\sin ky~dk$
$=\int_0^\Lambda\left(\sin yk+\sum\limits_{m=1}^\infty\dfrac{(-1)^mx^{2m}k^{2m}\sin yk}{4^m(m!)^2}\right)dk$
$=\left[-\dfrac{\cos yk}{y}+\sum\limits_{m=1}^\infty\sum\limits_{n=0}^{m-1}\dfrac{(-1)^{m+n}(2m)!x^{2m}k^{2m-2n-1}\sin yk}{4^my^{2n+2}(m!)^2(2m-2n-1)!}+\sum\limits_{m=1}^\infty\sum\limits_{n=0}^m\dfrac{(-1)^{m+n+1}(2m)!x^{2m}k^{2m-2n}\cos yk}{4^my^{2n+1}(m!)^2(2m-2n)!}\right]_0^\Lambda$ (can be obtained from https://en.wikipedia.org/wiki/List_of_integrals_of_trigonometric_functions)
$=\left[\sum\limits_{m=1}^\infty\sum\limits_{n=1}^m\dfrac{(-1)^n(2m)!x^{2m}k^{2n-1}\sin yk}{4^my^{2m-2n+2}(m!)^2(2n-1)!}+\sum\limits_{m=0}^\infty\sum\limits_{n=0}^m\dfrac{(-1)^{m+n+1}(2m)!x^{2m}k^{2m-2n}\cos yk}{4^my^{2n+1}(m!)^2(2m-2n)!}\right]_0^\Lambda$
$=\left[\sum\limits_{m=1}^\infty\sum\limits_{n=1}^m\dfrac{(-1)^n(2m)!x^{2m}k^{2n-1}\sin yk}{4^my^{2m-2n+2}(m!)^2(2n-1)!}+\sum\limits_{m=0}^\infty\sum\limits_{n=0}^m\dfrac{(-1)^{n+1}(2m)!x^{2m}k^{2n}\cos yk}{4^my^{2m-2n+1}(m!)^2(2n)!}\right]_0^\Lambda$
$=\sum\limits_{m=1}^\infty\sum\limits_{n=1}^m\dfrac{(-1)^n(2m)!x^{2m}\Lambda^{2n-1}\sin y\Lambda}{4^my^{2m-2n+2}(m!)^2(2n-1)!}+\sum\limits_{m=0}^\infty\sum\limits_{n=0}^m\dfrac{(-1)^{n+1}(2m)!x^{2m}\Lambda^{2n}\cos y\Lambda}{4^my^{2m-2n+1}(m!)^2(2n)!}+\sum\limits_{m=0}^\infty\dfrac{(2m)!x^{2m}}{4^my^{2m+1}(m!)^2}$
$=\sum\limits_{m=1}^\infty\sum\limits_{n=1}^m\dfrac{(-1)^n(2m)!x^{2m}\Lambda^{2n-1}\sin y\Lambda}{4^my^{2m-2n+2}(m!)^2(2n-1)!}+\sum\limits_{m=0}^\infty\sum\limits_{n=0}^m\dfrac{(-1)^{n+1}(2m)!x^{2m}\Lambda^{2n}\cos y\Lambda}{4^my^{2m-2n+1}(m!)^2(2n)!}+\dfrac{1}{\sqrt{y^2-x^2}}$
