Evaluation of the integral $I=\int_0^{\infty}[1-J_0(kr)]k^{1-\alpha}dk$ How to compute
\begin{equation}
I(r) = \int_0^{\infty}[1-J_0(kr)]k^{1-\alpha}dk,\quad 2<\alpha<4
\end{equation}
where $J_0(x)$ is the zeroth order Bessel function of the first kind.
A paper gives the following result
\begin{equation}
I(r)=cr^{\alpha-2}
\end{equation}
and 
\begin{equation}
c = \frac{4\Gamma(3-\alpha)}{\alpha-2}\sin\left(\frac{\pi}{2}(3-\alpha)\right)B(1/2,(\alpha-1)/2)
\end{equation}
with $B(x,y)$ being Beta function.
It seems the above result does not give a defined value for $\alpha=3$.
Can someone show how to evaluate $I(r)$?  
Thanks.
 A: The first identity just follows from setting $k=\frac{1}{r}u$, so the point is just to compute:
$$ I=\int_{0}^{+\infty}(1-J_0(u))\,u^{1-\alpha}\,du $$
that depends on the Beta function since the Fourier transform of $J_0(u)$ is simply given by $\frac{\mathbb{1}_{(-1,1)}}{\sqrt{1-t^2}}$:
$$I = \frac{1}{\alpha-2}\int_{0}^{+\infty} J_1(u)\,u^{2-\alpha}\,du =\frac{2}{\pi(\alpha-2)}\int_{0}^{1}\int_{0}^{+\infty}\frac{t \sin(st)}{\sqrt{1-t^2}}s^{2-\alpha}\,ds\,dt$$
and since for any $t>0$:
$$ \int_{0}^{+\infty}\sin(st)\,s^{2-\alpha}\,ds = \Gamma(3-\alpha)\cos\frac{\pi\alpha}{2} t^{\alpha-3}$$
by the residue theorem, the Beta function arises from:
$$ \int_{0}^{1}\frac{t^{\alpha-2}}{\sqrt{1-t^2}}\,dt = \frac{1}{2}\int_{0}^{1}z^{\frac{\alpha-3}{2}}(1-z)^{-\frac{1}{2}}\,dz.$$
For $\alpha\geq 4$, the function $(1-J_0(u))\,u^{1-\alpha}$ has a non-integrable singularity in a right neighbourhood of zero, since over there:
$$ 1-J_0(u) = \frac{u^2}{4}-\frac{u^4}{64}+o(u^5).$$
When $\alpha=3$, notice that the zero of the sine function and the pole of the Gamma function cancel out, giving:
$$ \lim_{\alpha\to 3}\Gamma(3-\alpha)\sin\left(\frac{\pi}{2}(3-\alpha)\right) = \frac{\pi}{2}.$$
