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While trying to compute the following integrals, Mathematica is not able to do that. Any suggestion will be helpful.

Assuming[{R ∈ Reals, α ∈ Reals}, 
 Integrate[q^4 BesselJ[1, q ]/(1 - α^2 q^2)^2, {q, 0, ∞}]
 ]

Assuming[{R ∈ Reals, α ∈ Reals}, 
 Integrate[q^4 BesselJ[1, q ]/(1 - α^2 q^2)^3, {q, 0, ∞}]
 ]
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  • 1
    $\begingroup$ What makes you expect that the integrals can be written down in closed-form? $\endgroup$ – Henrik Schumacher Sep 12 '18 at 7:30
  • $\begingroup$ In the "Table of Integrals, Series, and Products" by Gradshteyn and Ryzhik, (seventh edition), there is a similar integral given in page no 679, which reads $\endgroup$ – ark Sep 12 '18 at 8:30
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    $\begingroup$ @ark I do not have the seventh edition, but in the fourth edition, which I have, this integral does not exist. Can you publish a copy of the page or write out the integral itself here? $\endgroup$ – Alex Trounev Sep 12 '18 at 15:06
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Don't know about the non-convergence messages for NIntegrate, but this seems to work.

$Assumptions = \[Alpha] \[Element] Reals

int = Integrate[
  q^4 BesselJ[1, q]/(1 - \[Alpha]^2 q^2)^n, {q, 0, \[Infinity]}];

ConditionalExpression[((1/(\[Alpha]^4*Gamma[n - 1]*Gamma[n]))*Pi*Csc[Pi*n]*
    Abs[\[Alpha]]^(-n - 3)*(Abs[\[Alpha]]^3*Gamma[n - 1]*
      ((1 - 4*\[Alpha]^2*(n^2 - 3*n + 2))*BesselI[n - 2, -(I/Abs[\[Alpha]])] - 
       BesselI[4 - n, -(I/Abs[\[Alpha]])]) - 
     2*I*\[Alpha]^4*(2*Gamma[n - 1]*BesselI[3 - n, -(I/Abs[\[Alpha]])] - 
       Gamma[n]*BesselI[n - 1, -(I/Abs[\[Alpha]])])))/
   (2^n*E^((1/2)*I*Pi*n)), Re[n] > 7/4]

Limit[int, n -> 2] // FunctionExpand

(*Piecewise[{{(I*Pi*BesselJ[0, 1/\[Alpha]])/(4*\[Alpha]^6) - (Pi*BesselY[0, 1/\[Alpha]])/
      (4*\[Alpha]^6) + (I*Pi*BesselJ[1, 1/\[Alpha]])/(2*\[Alpha]^5) - 
     (Pi*BesselY[1, 1/\[Alpha]])/(2*\[Alpha]^5), \[Alpha] >= 0}}, 
  -((Pi*BesselY[0, 1/\[Alpha]])/(4*\[Alpha]^6)) + 
   ((Log[1/\[Alpha]] - Log[I/\[Alpha]])*BesselJ[0, 1/\[Alpha]])/(2*\[Alpha]^6) - 
   (Pi*BesselY[1, 1/\[Alpha]])/(2*\[Alpha]^5) + 
   ((Log[1/\[Alpha]] - Log[I/\[Alpha]])*BesselJ[1, 1/\[Alpha]])/\[Alpha]^5]*)

Limit[int, n -> 3] // FunctionExpand

(*Piecewise[{{(-4*I*Pi*\[Alpha]*BesselJ[0, 1/\[Alpha]] + I*Pi*BesselJ[1, 1/\[Alpha]] + 
      4*Pi*\[Alpha]*BesselY[0, 1/\[Alpha]] - Pi*BesselY[1, 1/\[Alpha]])/(16*\[Alpha]^7), 
    \[Alpha] >= 0}}, (-12*I*Pi*\[Alpha]*BesselJ[0, 1/\[Alpha]] + 
    3*I*Pi*BesselJ[1, 1/\[Alpha]] + 4*Pi*\[Alpha]*BesselY[0, 1/\[Alpha]] - 
    Pi*BesselY[1, 1/\[Alpha]])/(16*\[Alpha]^7)]*)

Without taking limits, the results are ComplexInfinity, which may explain why NIntegrate has trouble.

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  • $\begingroup$ Thank You. @Bill $\endgroup$ – ark Sep 13 '18 at 9:12
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          Quiet[Integrate[
         q^4 BesselJ[1, q]/(1 - \[Alpha]^2 q^2)^n, {q, 0, \[Infinity]}]]

     (*ConditionalExpression[(1/(\[Alpha]^4 Gamma[n]))
     4^-n (-\[Alpha]^2)^(-(1/2) - 
      n) (2^n \[Pi] (-\[Alpha]^2)^(
      n/2) (4 \[Alpha]^2 BesselI[3 - n, 1/Sqrt[-\[Alpha]^2]] - 
        Sqrt[-\[Alpha]^2] BesselI[4 - n, 1/Sqrt[-\[Alpha]^2]]) Csc[
        n \[Pi]] + 
      16 (-1 + n) \[Alpha]^4 Sqrt[-\[Alpha]^2]
      Gamma[3 - n] HypergeometricPFQ[{n}, {-2 + n, -1 + n}, -(1/(
      4 \[Alpha]^2))]), 
       4 Re[n] > 
       7 && (Re[\[Alpha]^2] <= 0 || \[Alpha]^2 \[NotElement] Reals)]*)
      for real \[Alpha]^2 intégral is divergent.
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  • 1
    $\begingroup$ I was looking for a compact analytical form of the mentioned integral not the Numerical integration, if possible. Thanks $\endgroup$ – ark Sep 12 '18 at 9:05
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    $\begingroup$ It is dangerous to switch of the MMA error messages, especially in the case of not converging integrals! $\endgroup$ – Ulrich Neumann Sep 12 '18 at 9:52
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If you try to solve the integrals numerically

int[\[Alpha]_?NumericQ, n_?NumericQ] :=NIntegrate[q^4 BesselJ[1, q]/(1 - \[Alpha]^2 q^2)^n, {q, 0, \[Infinity]} ]

the evaluation of for example

int[  1, 2]

gives error messages NIntegrate::ncvb: NIntegrate failed to converge to prescribed accuracy after 9 recursive bisections in q near {q} = {1.00277}. NIntegrate obtained 6648.614253396791 and 6097.681026759085for the integral and error estimates.

Elaborate the comment Henrik Schumacher gave...

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