Hi there,
this you can possibly solve: First of all, one sets
q = 0, k2 Cos[?]= x, k2 Sin[?] = y
because the q is a strong obstacle and the ?-terms are unimportant for the integral. The simplified integral
(* q = 0, k2 Cos[\[Theta]] \[Rule] x, k2 Sin[\[Theta]]\[Rule] y *)
Integrate[
Cos[((k2 - x) \[Rho] Cos[\[Phi]] - y \[Rho] Sin[\[Phi]])] \[Rho], {\[Phi], 0, 2 \[Pi]},
Assumptions -> {k2 > 0 && \[Theta] \[Element] Reals && \[Rho] > 0}]
ConditionalExpression[
2 \[Pi] \[Rho] BesselJ[0, Sqrt[(k2 - x)^2 + y^2] \[Rho]],
Sqrt[(k2 - x)^2 + y^2] \[Element] Reals]
The conditional means
In[8]:= Reduce[(k2 - k2 Cos[\[Theta]])^2 + (k2 Sin[\[Theta]])^2 >= 0, \[Theta], Reals]
Out[8]= (C[1] \[Element]
Integers && \[Theta] == \[Pi] +
2 \[Pi] C[1]) || (-\[Pi] + \[Theta])/(2 \[Pi]) \[NotElement] Integers
and one finds
In[7]:= S
Integrate[Sin[((k2 - x) \[Rho] Cos[\[Phi]] - y \[Rho] Sin[\[Phi]])] \[Rho], {\[Phi], 0, 2 \[Pi]},
Assumptions -> {k2 > 0 && \[Theta] \[Element] Reals && \[Rho] > 0}]
Out[7]= 0
Why this is interesting? If you develop the orginal integral into a seris in q and integrate
Integrate[Series[
Cos[q + ((k2 - k2 Cos[\[Theta]]) \[Rho] Cos[\[Phi]] -
k2 Sin[\[Theta]] \[Rho] Sin[\[Phi]])] \[Rho], {q, 0,
12}], {\[Phi], 0, 2 \[Pi]}]
you are left with only the both integrals shown above, one is zero, so the result is
(Cosh[q]- 1) ConditionalExpression[
2 \[Pi] \[Rho] BesselJ[0, Sqrt[(k2 - x)^2 + y^2] \[Rho]],
Sqrt[(k2 - x)^2 + y^2] \[Element] Reals]
x and y do depend on \ because of
In[13]:= Sum[q^(2 o)/(2 o)!, {o, 1, Infinity}]
Out[13]= -1 + Cosh[q]
Best regards
Udo.
