How to integrate product of BesselJ and BesselY with different orders?

Question

I need help with analytic solution for the following integral

$Assumptions = 
 n ∈ Integers && a ∈ Reals && b ∈ Reals && 
  k ∈ Reals && r ∈ Reals && b > a && a > 0 && 
  b > 0 && k > 0 && r > 0 && n >= 0    
Integrate[BesselJ[n - 1, k*r]*BesselY[n, k*r], {r, a, b}]

As suggested by @Akku14 in How to find analytical solution for integral from product of Bessel functions of second kind with different order? I use Mathematica 8.0.1 to find solutions for different orders of Bessel functions. A comparison of analytic integration result with numeric one reveals the problem:

tmpVal = {a -> 0.2, b -> 0.4, k -> 1.8}; 
{N[Integrate[BesselJ[2, k*r]*BesselY[3, k*r], {r, a, b}] /.tmpVal],
 NIntegrate[BesselJ[2, k*r]*BesselY[3, k*r] /. tmpVal, {r, a /. tmpVal,b /. tmpVal}]}
(*{-0.125636, -0.248211}*)

Answer 1

First define assumptions:

$Assumptions = 
 n ∈ Integers && a ∈ Reals && b ∈ Reals && 
  k ∈ Reals && r ∈ Reals && b > a && a > 0 && b > 0 && 
  k > 0 && r > 0 && n >= 0 && s ∈ Integers;

Transform integrand using recursion formula (http : // functions.wolfram.com/03.01 .17 .0001 .01) in order to lift order of BesselJ to positive numbers:

ex = Expand[
  BesselJ[n - 1, k*r]*BesselY[n, k*r] /. 
   HoldPattern[
     BesselJ[\[Nu]_, z_]] :> ((2 (\[Nu] + 1)) BesselJ[\[Nu] + 1, z])/z - 
     BesselJ[\[Nu] + 2, z]]

As a result we receive 2 summands:

(2 n BesselJ[n, k r] BesselY[n, k r])/(k r) - 
 BesselJ[1 + n, k r] BesselY[n, k r]

Integrating each summand using approach shown by Akku14, and defining workaround rules for this integrals we receive:

iJnYnr := Integrate[
   z_.*BesselJ[n_, k_*r_]*BesselY[n_, k_*r_]/r_, {r_, a_, b_}] :>
  RuleCondition[
   z*(1/(2 Sqrt[\[Pi]]))(MeijerG[{{1/2}, {-(1/2), 1}}, {{0, 0, 
         n}, {-n, -(1/2)}}, a k, 1/2] - 
      MeijerG[{{1/2}, {-(1/2), 1}}, {{0, 0, n}, {-n, -(1/2)}}, b k, 1/2]), 
   Refine[FreeQ[z, r] && k ∈ Reals && r ∈ Reals && k > 0 && 
     r > 0 && n >= 0 && n ∈ Integers && a ∈ Reals && 
     b ∈ Reals && a > 0 && b > 0 && b > a]];

iJn1Yn := Integrate[
   z_.*BesselJ[1 + n_, k_*r_]*BesselY[n_, k_*r_], {r_, a_, b_}] :>
  RuleCondition[
   z*(1/(2 Sqrt[\[Pi]]))(-a MeijerG[{{1/2, 1/2}, {0}}, {{1/2, 
          1/2 + n}, {-(1/2), -(1/2), -(1/2) - n}}, a k, 1/2] + 
      b MeijerG[{{1/2, 1/2}, {0}}, {{1/2, 
          1/2 + n}, {-(1/2), -(1/2), -(1/2) - n}}, b k, 1/2]), 
   Refine[FreeQ[z, r] && k ∈ Reals && r ∈ Reals && k > 0 && 
     r > 0 && n >= 0 && n ∈ Integers && a ∈ Reals && 
     b ∈ Reals && a > 0 && b > 0 && b > a]];

Using defined rules on expanded integrand

solut = ReleaseHold[Hold@Integrate[#, {r, a, b}] /. {iJnYnr, iJn1Yn}] & /@ ex

we obtain solution:

(1/(k Sqrt[\[Pi]]))
 n (MeijerG[{{1/2}, {-(1/2), 1}}, {{0, 0, n}, {-(1/2), -n}}, a k, 1/2] - 
    MeijerG[{{1/2}, {-(1/2), 1}}, {{0, 0, n}, {-(1/2), -n}}, b k, 1/2]) - (1/(
 2 Sqrt[\[Pi]]))(-a MeijerG[{{1/2, 1/2}, {0}}, {{1/2, 
       1/2 + n}, {-(1/2), -(1/2), -(1/2) - n}}, a k, 1/2] + 
   b MeijerG[{{1/2, 1/2}, {0}}, {{1/2, 
       1/2 + n}, {-(1/2), -(1/2), -(1/2) - n}}, b k, 1/2])

Comparison with numeric integration now shows correct result for arbitrary parameters:

tmpVal = {a -> 0.2, b -> 0.4, k -> 1.8}; nRule = 
 n -> 1; {N[solut /. nRule /. tmpVal], 
 NIntegrate[
  BesselJ[n - 1, k*r]*BesselY[n, k*r] /. nRule /. tmpVal, {r, a /. tmpVal, 
   b /. tmpVal}]}

(*{-0.264519, -0.264519}*)

Be especially carefull about obtained fragile solution. Application of FunctionExpand and therefore FullSimplify to this solution may lead to non-numeric results. Define the the workaround rule for the integral in question:

iJn1mYn := 
 Integrate[z_. BesselJ[n - 1, k_*r_]*BesselY[n, k_*r_], {r_, a_, b_}] :>
  RuleCondition[
   z*1/(k Sqrt[\[Pi]])
      n (MeijerG[{{1/2}, {-(1/2), 1}}, {{0, 0, n}, {-(1/2), -n}}, a k, 1/2] - 
       MeijerG[{{1/2}, {-(1/2), 1}}, {{0, 0, n}, {-(1/2), -n}}, b k, 1/2]) - 
    1/(2 Sqrt[\[Pi]]) (-a MeijerG[{{1/2, 1/2}, {0}}, {{1/2, 
           1/2 + n}, {-(1/2), -(1/2), -(1/2) - n}}, a k, 1/2] + 
       b MeijerG[{{1/2, 1/2}, {0}}, {{1/2, 
           1/2 + n}, {-(1/2), -(1/2), -(1/2) - n}}, b k, 1/2]), 
   Refine[FreeQ[z, r] && k ∈ Reals && r ∈ Reals && k > 0 && 
     r > 0 && n >= 0 && n ∈ Integers && a ∈ Reals && 
     b ∈ Reals && a > 0 && b > 0 && b > a]]

P.S. I'm newbie in pattern matching, so this rule probably will not fit to every expression resembling integral in question. Any comments about workaround rules will be appreciated.