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I have following integration. $$I=\int_{0}^{\frac{\pi}{2}} \int_{0}^{\infty} \gamma e^{- \lambda \left(\gamma^2+2d\gamma\cos\theta -d^2 + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 2d\cos\Theta(d\cos\Theta - \sqrt{d^2\cos^2\Theta + 2d\gamma\cos\theta+\gamma^2}) d\Theta \right) } \,d\gamma d\theta$$ Since this may not have a closed-form solution, I tried to evaluate it numerically as below:

PoNum[λ_, d_] :=NIntegrate[
NIntegrate[
 x Exp[-λ (x^2 + 2 d x Cos[θ] - d^2 + 
      NIntegrate[
       2 d Cos[Θ] (d Cos[Θ] - 
          Sqrt[(d Cos[Θ])^2 + 2 d x  Cos[θ] +
            x^2]), {Θ, -(π/2), π/2}])], {x, 
  0, ∞}], {θ, 0, π/2}];

Here $\lambda$ and $d$ are positive constant. E.g.

PoNum[2.3, 1.1]

However, my code may not give correct answer as I have seen series of warnings.

Can someone please help me?

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You should split your integral.

First the inner integral (exponent) is

int[d_?NumericQ, θ_?NumericQ, x_?NumericQ] :=NIntegrate[2 d Cos[Θ] (d Cos[Θ] -Sqrt[(d Cos[Θ])^2 + 2 d x Cos[θ] + x^2]), {Θ, -(π/2), π/2}]

With this function the complete integral you are asking for is

PoNum[λ_?NumericQ, d_?NumericQ] := NIntegrate[x Exp[-λ (x^2 + 2 d x Cos[θ] - d^2 + int[λ, θ, x])], {x, 0, ∞}, {θ,0, π/2}];

PoNum[2.3, 1.1]
(*5.46735*10^16*)

addendum:

The inner integral int[] can be converted to

2 d^2Integrate[ Cos[Θ] ( Cos[Θ] -Sqrt[( Cos[Θ])^2 + (2 d x Cos[θ] + x^2)/d^2]),{Θ, -(π/2), π/2}]

and solved analytically

Simplify[Integrate[Cos[\[CurlyTheta]] (Cos[\[CurlyTheta]] - Sqrt[Cos[\[CurlyTheta]] + c^2]), {\[CurlyTheta], -Pi/2, Pi/2}],Re[c^2] > 0]
(* 1/2 (\[Pi] - (1/(3 Sqrt[1 + 1/c^2] Sqrt[c^2]))2 (2 (c^2 + c^4) EllipticE[2/(1 + c^2)] -2 (-1 + c^4) EllipticK[2/(1 + c^2)] +6 Sqrt[1 + 1/c^2]c^2 HypergeometricPFQ[{-(1/4), 1/4, 1}, {1/2, 3/2}, 1/c^4])) *)

Unfortunately this transformation doesn't lead to performance increase...

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  • $\begingroup$ thanks @Ulrich, it works smoothly. $\endgroup$ – Frey Sep 10 '18 at 10:37
  • $\begingroup$ @ Frey Thanks. Just one hint: The integral (exponent) could be solved analytically, but the evaluation of the complete integral doesn't finish... $\endgroup$ – Ulrich Neumann Sep 10 '18 at 10:47
  • $\begingroup$ Can you please share the solution of that exponent integral? Thanks! $\endgroup$ – Frey Sep 10 '18 at 12:13
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    $\begingroup$ My answer is edited. $\endgroup$ – Ulrich Neumann Sep 10 '18 at 12:54
  • $\begingroup$ thanks @Ulrich, that simplification helps me. $\endgroup$ – Frey Sep 11 '18 at 3:33

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