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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.

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  • $\begingroup$ 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
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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.

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