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.

  • $\begingroup$ You want the answer or the technique? $\endgroup$ – Igor Rivin Dec 5 '13 at 17:01
  • $\begingroup$ @IgorRivin : of course, the second, according to the title. $\endgroup$ – John Taylor Dec 5 '13 at 17:06

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.