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Any idea how to calculate the following integral? $$ \int_0^{2\pi} K_1\left(\sqrt{r^2+r'^2-2rr'\cos\theta}\right)d\theta $$ This integral appears when using perturbation theory for the electron-electron interaction in cylindrical coordinates, and should have been addressed by someone in the literature, but somehow I cannot manage to find anything related. Any ideas? Thanks!

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Leafing through Gradshteyn & Rizhik, I found 6.684: $$ \int_0^\pi (\sin x)^{2\nu} \frac{J_\nu\big(\sqrt{\alpha^2+\beta^2-2\alpha\beta\cos x}\;\big)} {\big(\sqrt{\alpha^2+\beta^2-2\alpha\beta\cos x}\;\big)^\nu} \;dx = 2^\nu\sqrt{\pi}\;\Gamma\left(\nu+\frac{1}{2}\right) \frac{J_\nu(\alpha)}{\alpha^\nu}\;\frac{J_\nu(\beta)}{\beta^\nu} $$ and something similar with $Y_\nu$.

But that doesn't seem to help with yours.

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