How to evaluate the integral involving two Bessel functions and following elementary function? Let's have the integral 
$$
\int \limits_{0}^{\infty} J_{0}(kr) J_{0}(kr')\frac{kdk}{k^2 + a^2}.
$$
How to evaluate it? I failed when was trying using integral representations for the Bessel functions.
 A: Let me first state the answer: for $r<r'$
$$\mathcal{I}(r,r',a)=\int_0^{\infty}J_0(kr)J_0(kr')\frac{kdk}{k^2+a^2}=I_0(ar)K_0(ar').\tag{1}$$
More general formulas can be found in Prudnikov-Brychkov-Marychev, Vol. 2.
To show (1) (for generalizations the procedure is similar), note that
\begin{align}
&\left(\frac{d^2}{dr^2}+\frac{1}{r}\frac{d}{dr}-a^2\right)\mathcal{I}(r,r',a)=\\&=
\int_0^{\infty}\left(\frac{d^2}{dr^2}+\frac{1}{r}\frac{d}{dr}-a^2\right)J_0(kr)J_0(kr')\frac{kdk}{k^2+a^2}=\\
&=-\int_0^{\infty}(k^2+a^2)J_0(kr)J_0(kr')\frac{kdk}{k^2+a^2}=\\
&=-\int_0^{\infty}kJ_0(kr)J_0(kr')dk=\\
&=0.\tag{2}
\end{align}
At the last step we have used the known orthogonality relations for Bessel functions.
Thus we see that $\mathcal{I}(r,r',a)$ solves the modified Bessel equation w.r.t. $ar$ and should therefore be a linear combination of its two independent solutions. Moreover since it is regular at $r=0$, it can only be proportional to $I_0(ar)$.
Similarly, $\mathcal{I}(r,r',a)$ solves the modified Bessel equation w.r.t. $ar'$ and vanishes at infinity, hence it can only be proportional to $K_0(ar')$. Therefore,
$$\mathcal{I}(r,r',a)=\operatorname{const}\cdot I_0(ar) K_0(ar').$$
The remaining constant can be fixed in several ways - e.g. using the known asymptotic behavior of Bessel functions.
