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I'd like to calculate $$ \int_0^R r^2 \left[ I_0(r)K_1(r) + I_1(r)K_0(r) \right] \mathrm{d}r $$

Earlier, I have used the property $$\int x I_0(x) \,\mathrm{d}x = x I_1(x)$$ but I can't find a similar standard integral in books and tables for my problem. From another source I am hoping the answer to be like $\frac{R^2}{2}$, but Mathematica gives a complicated answer involving Meijer G-functions. Integrating by parts doesn't seem to help.

bacchicaa
  • 31

1 Answers1

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The integrand can be rewritten. Firstly, using these relations for the derivatives: $$ \frac{\mathrm{d}I_0}{dx}=I_1 $$ $$ \frac{\mathrm{d}K_0}{dx}=-K_1 $$ I can write $$ I_0 K_1 + K_0 I_1 = -I_0 K_0' + K_0 I_0' = - W(I_0, K_0) $$ and then using the identity from this question, $W(I_0, K_0) = - \frac{1}{x}$, the integral above becomes

$$ \int_0^R\! r\, \mathrm{d} r = \frac{R^2}{2} $$ as expected.

bacchicaa
  • 31