1
$\begingroup$

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?

$\endgroup$

1 Answer 1

3
$\begingroup$

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...

$\endgroup$
5
  • $\begingroup$ thanks @Ulrich, it works smoothly. $\endgroup$
    – Frey
    Sep 10, 2018 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... $\endgroup$ Sep 10, 2018 at 10:47
  • $\begingroup$ Can you please share the solution of that exponent integral? Thanks! $\endgroup$
    – Frey
    Sep 10, 2018 at 12:13
  • 1
    $\begingroup$ My answer is edited. $\endgroup$ Sep 10, 2018 at 12:54
  • $\begingroup$ thanks @Ulrich, that simplification helps me. $\endgroup$
    – Frey
    Sep 11, 2018 at 3:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.