How to evaluate the integral of exponentials

Question

I want to evaluate the following integral wrt $\phi$ from 0 to $2 \pi$:

Integrate[
 Exp[(Γ^2 (Cos[ϕ]^2 Sin[θ]^2 σx^2 + Sin[ϕ]^2 Sin[θ]^2 σy^2) 
   - 2 c Γ (x0 Cos[ϕ] Sin[θ] + y0 Sin[ϕ] Sin[θ]))/(2 c^2)], 
 {ϕ, 0, 2 π}
]

Currently Mathematica returns the input without evaluating anything. I have tried to Simplify the expression before evaluating the integral as well.

This can most likely be done by instead using NIntegrate and by using specific values for each of the parameters. However, I am really interested in obtaining the analytical result of this evaluation. Is anyone aware of a method of forcing Mathematica to generate a symbolic output for this integral.

I will also post this question on Mathematics Stack Exchange to try to source a substitution or equivalent method to break this problem down.

Answer 1

You can get an approximate answer by doing a series expansion of the integrand. As always, the more terms used in the expansion, the better the approximation.

expr = Exp[(Γ^2 (Cos[ϕ]^2 Sin[θ]^2 σx^2 + 
         Sin[ϕ]^2 Sin[θ]^2 σy^2) - 
      2 c Γ (x0 Cos[ϕ] Sin[θ] + 
         y0 Sin[ϕ] Sin[θ]))/(2 c^2)];

f[n_Integer, ϕ0_] := 
 Integrate[Series[expr, {ϕ, ϕ0, n}] // Normal, {ϕ, 0, 2 Pi}]

Expanding about the middle of the integration interval,

f[6, Pi] // Simplify

(*  (1/2520)E^((Γ Sin[θ] (2 c x0 + Γ \
σx^2 Sin[θ]))/(
 2 c^2)) π (5040 + (
   840 π^2 Γ Sin[θ] (-c x0 + Γ (y0^2 - \
σx^2 + σy^2) Sin[θ]))/c^2 - (1/(
   c^6))π^6 Γ Sin[θ] (c^5 x0 + 
      c^4 Γ (15 x0^2 - 
         16 (y0^2 - σx^2 + σy^2)) Sin[θ] + 
      15 c^3 x0 Γ^2 (x0^2 - 
         5 (y0^2 - σx^2 + σy^2)) Sin[θ]^2 - 
      5 c^2 Γ^3 (9 x0^2 (y0^2 - σx^2 + σy^2) - 
         4 (y0^4 - 6 y0^2 (σx^2 - σy^2) + 
            3 (σx^2 - σy^2)^2)) Sin[θ]^3 + 
      15 c x0 Γ^4 (y0^4 - 6 y0^2 (σx^2 - σy^2) + 
         3 (σx^2 - σy^2)^2) Sin[θ]^4 - Γ^5 \
(y0^6 - 15 y0^4 (σx^2 - σy^2) + 
         45 y0^2 (σx^2 - σy^2)^2 - 
         15 (σx^2 - σy^2)^3) Sin[θ]^5) + (1/(c^4))
   42 π^4 Γ Sin[θ] (c^3 x0 + Γ Sin[\
θ] (c^2 (3 x0^2 - 
            4 (y0^2 - σx^2 + σy^2)) + Γ Sin[\
θ] (-6 c x0 (y0^2 - σx^2 + σy^2) + Γ (y0^4 \
+ 3 (σx^2 - σy^2)^2 + 
               6 y0^2 (-σx^2 + σy^2)) Sin[θ]))))  *)