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$\begingroup$ Thank you for your time. This approach was also proposed by Jack in Mse. I'm not familiar with contour integration, but Mathematica can do the integral $J_k$. It gives: $$\frac{1}{2} \pi \sec (\pi k) \left(\frac{\left(\frac{c^2}{b^2}\right)^{k+\frac{1}{2}} \left(1-\frac{c^2}{b^2}\right)^{-k}}{c^2}-\frac{\sqrt{\pi } \, _2\tilde{F}_1\left(\frac{1}{2},1;\frac{3}{2}-k;\frac{c^2}{b^2}\right)}{b^2 \Gamma (k)}\right)$$ It seems not useful to me :( $\endgroup$

Dinesh Shankar
– Dinesh Shankar

2018-08-22 09:14:39 +00:00
Commented Aug 22, 2018 at 9:14

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