x = kr
ψ = x*Sqrt[π/(2*x)]*BesselJ[n + 1/2, x];
Q = 1/(I + 2*Z); 
G = I*Q*Exp[-I*Q*(r^2*(Sin[θ])^2)/w0^2]*
  Exp[-I*x*Cos[θ]]*(1 - (2*Q)/(k*w0^2)*r*Cos[θ])*ψ*
  LegendreP[1, n, Cos[θ]]*(Sin[θ])^2;
G1 = Sum[(2*n + 1)/(π*n*(n + 1))*(1/((-1)^n*I^(n - 1)))*G, {n, 1, 

I want to integrate G1 first w.r.t. x from 0 to Infinity and then w.r.t. theta from 0 to pi.

Cp = Sum[((2*n + 1)/(n (n + 1)))*G1*s1 + ((n (n + 2))/(n + 1))*
      Re[G2*s2], {n, 1, Infinity}];

here, G2 is same as G1 but value of n change from n to n+1.s1 and s2 are constant number.

Can you please give me a suitable solution for that ASAP?? thank you

  • $\begingroup$ Please use the editing buttons available to format code. It's not clear that you've actually run the code because of some spelling mistakes. Are you getting no output or are there errors? Giving such details will get you much quicker and more accurate help. For example, you might want to check on your use of n. You have n as a variable in the definition of $\psi$ and also as the index variable in the sum. $\endgroup$ – JimB Sep 12 '16 at 4:28
  • $\begingroup$ All terms in the sum for G1 are zero for $n>1$. Is that what you want? You also use the term "integrate" but you show Mathematica code using Sum. This doesn't make it easy to assist with this question. $\endgroup$ – JimB Sep 12 '16 at 4:54
  • $\begingroup$ i want to integrate G1 that is the bigest problm $\endgroup$ – Anita Maheshwari Sep 12 '16 at 5:15
  • $\begingroup$ there is no x in G1 (since you eliminated x on the first line.) $\endgroup$ – george2079 Sep 12 '16 at 18:32
  • $\begingroup$ i defined above that my x=k*r so i want to differentiate w.r.t. kr (d(kr)) $\endgroup$ – Anita Maheshwari Sep 13 '16 at 4:14

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