How to calculate this integral? Note that, $K_{c-m}$ is the modified Bessel function of the second kind.

  • $\begingroup$ May we assume $b$ is positive? $\endgroup$ – David H Jan 16 '15 at 20:53
  • $\begingroup$ Yes, you can assume that. $\endgroup$ – Kira Jan 16 '15 at 22:04

Let $t = 2b\sqrt{y}$, then $\frac{t}{2b^2} dt = dy$, so $$ \int_0^\infty \left(\frac{t}{2b}\right)^{2+m+c} K_{c-m}(t) \frac{t}{2b^2} dt $$ $$ \frac{(2b)^{-3-m-c}}{b} \int_0^\infty t^{3+m+c} K_{c-m}(t) dt $$ Now applying the integral identity here, $$ \frac{(2b)^{-3-m-c}}{b} 2^{2+m+c}\Gamma(m+2)\Gamma(c+2) $$ $$ \frac{b^{-4-m-c}}{2} \Gamma(m+2)\Gamma(c+2) $$

  • $\begingroup$ The link you provided seems to be using the first kind, no? $\endgroup$ – Kira Jan 17 '15 at 0:51
  • $\begingroup$ @Kira The identity you are supposed to apply is equation 10.43.19 in section (iv). You may have to scroll down to find it, but it is definitely using modified Bessel functions of the second kind. $\endgroup$ – David H Jan 17 '15 at 6:12

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