A closed-form of an integral containing Bessel's function What is the closed-form of this integral? $$\int_0^\infty{J_0(\lambda a)\sin(\lambda a)d\lambda}$$where $J_0 $ is the Bessel function of the first kind of order zero, and $a$ is real. Based on the flow of the paper I am reading, this integral should have a closed form expression, but I couldn't find it everywhere I searched for.
Edit: Additional information about the origin of this integral, and why I think it should converge to some value:
I am working with the following function, written in cylindrical coordinate:$$\mathrm V(r,z)=-\frac{c}{2\pi a\sigma}\int_0^\infty\frac{\mathrm u(z, \lambda)}{\mathrm u'(0, \lambda)} \mathrm J_0(\lambda r)\sin(\lambda a)d\lambda$$, where $\mathrm u$ is of exponential form and monotonically decreasing as a function of $z$; and $a,\sigma$ are constants.
The paper says that this function has been constructed so that:$$\frac{\partial V}{\partial z}|_{z=0}=\begin{cases}
0,  & \text{if $r>a$} \\
\ne 0, & \text{if $r\leq a$}
\end{cases}$$
I couldn't understand why this is the case, given this construction. Here is my attempt: taking the derivative of $\mathrm V$ with respect to $z$:$$\frac{\partial V}{\partial z}=-\frac{c}{2\pi a\sigma}\int_0^\infty\frac{\mathrm u'(z, \lambda)}{\mathrm u'(0, \lambda)} \mathrm J_0(\lambda r)\sin(\lambda a)d\lambda$$
At $z=0$, this reduces to:$$\frac{\partial V}{\partial z}|_{z=0}=-\frac{c}{2\pi a\sigma}\int_0^\infty\mathrm J_0(\lambda r)\sin(\lambda a)d\lambda$$From here, I don't know what else to do to see why $\mathrm V$ satisfies the above conditions.
 A: Quoting DLMF 10.22.59,
When $\Re \mu > -1$,
$$\int_0^\infty e^{ibt}J_\mu(at) dt = \begin{cases}
\frac{\exp(i\mu \sin^{-1}(b/a))}{\sqrt{a^2-b^2}},& 0 \le b < a\\
\frac{ia^\mu \exp\left(\frac12\mu\pi i\right)}{\sqrt{b^2-a^2}\left(b + \sqrt{b^2-a^2}\right)^\mu}, & 0 < a < b\end{cases}$$
Setting $\mu$ to $0$ and taking imaginary part, one get
$$\int_0^\infty J_0(at)\sin(bt)dt = \begin{cases}
0,& 0 \le b < a\\
\frac{1}{\sqrt{b^2-a^2}},& b > a > 0
\end{cases}$$
Your integral corresponds to the case $a = b$.
As one can see, the integral tends to different value ( $0$ and $\infty$ ) as $b$ approaches $a$ from the left and from the right. There is a big chance
your integral diverges.
A: For any $\lambda>0$
$$ \int_{0}^{+\infty}J_0(x)\,\sin(x)\,e^{-\lambda x}\,dx = \frac{2}{\pi}\int_{0}^{+\infty}\int_{0}^{\pi/2}\cos(x\cos u)\sin(x)\,e^{-\lambda x}\,du\,dx$$
can be written (through Fubini's theorem) as
$$ \frac{2}{\pi}\int_{0}^{\pi/2}\frac{1+\lambda^2-\cos^2 u}{(1+\lambda^2+\cos^2 u)^2-4\cos^2 u}\,du \\= \frac{2}{\pi}\int_{0}^{1}\frac{t^2+\lambda^2 \left(1+t^2\right)}{t^4+\lambda^4 \left(1+t^2\right)^2+2 \lambda^2 \left(2+3 t^2+t^4\right)}\,dt $$
and as $\lambda\to 0^+$, the last integrand function tends to $\frac{1}{t^2}\not\in L^1(0,1)$.
It follows that the original integral is simply divergent.
