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I am studying the result of a Feynman Diagram, trying to apply the Renormalization Group to a Field theory for Active Matter. I have arrived at a point where I want to integrate with Mathematica a complicated expression, and I would need an analytical result, rather than a numerical one.

Using Integrate, the code has been stuck on "running" for a day and it does not give a result nor an error of any kind. Does it mean that there is not an exact solution or does it mean something else? Is there something I can try to get a result?

The code with the function I need to integrate is the following:

Integrand = (64 a w Cos[θ]^2 (f^3 κ^2 (w + κ + 
     2 w κ) λx0^5 + 
  4 w^3 (1 + w) Λ^6 λx0^5 + 
  2 w^2 (1 + w) Λ^6 λx0^4 (λx0 - 
     h λx0) + 
  f^3 κ^2 λx0^5 (-ht (1 + w) - κ - 
     w κ + h (1 + κ) + w μy + 
     w κ μy) + 
  f^3 κ^2 λx0^5 (-1 + μy) (-ht (2 + w) - 
     3 κ - 2 w κ + h (2 + 3 κ) + w μy + 
     2 w κ μy) Cos[θ]^2 + 
  f^2 κ Λ^2 λx0^5 (2 (-1 + 
        ht) w^3 - 2 κ (-2 + h + μy) + 
     w^2 (-15 + 9 ht - 18 κ + h (4 + 6 κ) + 
        2 μy + 12 κ μy) + 
     w (-4 + 4 ht + 
        5 κ (-2 + h + μy))) Cos[θ]^2 + 
  f^2 κ Λ^2 λx0^4 (2 κ ( λx0 - h λx0) (-2 + h + μy) - 
     w^2 λx0 (h + 3 μy - 4 h μy - 
        2 κ (-1 + μy) (-3 + 2 h + μy) + 
        ht (-4 + 3 h + μy)) + 
     w λx0 (6 κ + h^2 (2 + κ) - 
        2 ht (-3 + μy) - 2 μy - 
        6 κ μy + κ μy^2 + 
        h (-4 - 4 ht - 6 κ + 4 μy + 
           4 κ μy))) Cos[θ]^2 - 
  f^2 (-1 +  h) κ Λ^2 λx0^5 (6 κ + 
     h^2 κ - 6 w κ - 3 w^2 κ + 
     h w (2 + ht - 3 κ (-1 + μy) - 3 μy) + 
     w μy - 6 κ μy + 9 w κ μy + 
     6 w^2 κ μy + κ μy^2 - 
     3 w κ μy^2 - 3 w^2 κ μy^2 + 
     ht w (-3 + 2 μy) + 
     2 h κ (-3 + 2 μy)) Cos[θ]^4 + 
  f (-1 + h) w Λ^4 λx0^5 (2 (-1 + 
        ht) w^3 - 8 κ (-1 + μy) + 
     w (-12 + 10 ht - 27 κ + h (2 + 20 κ) + 
        7 κ μy) + 
     w^2 (-18 + 15 ht - 20 κ + h (3 + 8 κ) + 
        12 κ μy)) Cos[θ]^4 + 
  f (-1 + h)^2 w Λ^4 λx0^5 (-2 ht w - 
     2 h (6 κ + 
      w (-1 + 2 κ)) + κ (-6 (-3 + μy) + 
       2 w^2 (-1 + μy) + w (3 + μy))) Cos[θ]^4 - 
 2 (-1 + h)^4 w^2 (4 + 
    w) Λ^6 λx0^5 Cos[θ]^6 + 
 2 (-1 + h)^3 w^2 (6 + 11 w + 
    2 w^2) Λ^6 λx0^5 Cos[θ]^6 - 
 f (-1 + h)^3 w κ Λ^4 λx0^5 (-12 + 4 h + w (-1 + μy) + 8 μy) Cos[θ]^6 + 2 w^2 Λ^6 (λx0 - h λx0)^5 Cos[θ]^8 - f^2 κ^2 λx0^4 (λx0 - h λx0) (-1 + μy)^2 Cos[θ]^6 (f κ (-1 + μy) - (-1 + h)^2 Λ^2 Cos[θ]^2) - w Λ^4 λx0^5 (-f (1 + w) (-ht w (2 + w) + h (2 w + w^2 - 4 κ) + κ (3 + 2 w (-1 + μy) + μy)) + 2 (-1 + h)^2 w (4 + 3 w) Λ^2 Cos[θ]^2) +  2 w Λ^4 λx0^5 (f (w + w^3 - κ + 2 w κ + w^2 (3 + 4 κ)) + (-1 + h) w (1 + 9 w + 6 w^2) Λ^2 Cos[θ]^2) - f κ λx0^5 (1 - μy) Cos[θ]^6 (f^2 κ (1 - μy) (h + w - ht (1 + w) + κ + w κ - κ μy - w κ μy) +f (-1 + h)^2 Λ^2 (κ + w (-1 + ht + κ (-1 + μy)) - κ μy) Cos[θ]^2 + (-1 + h)^4 w Λ^4 Cos[θ]^4) + f κ λx0^5 (1 - μy) Cos[θ]^4 (f^2  κ (h - ht + 3 h κ + κ (-3 + w (-1 + μy))) (1 -μy) + 2 f (-1 + h)^2 κ Λ^2 (-2 + h + w + μy - w μy) Cos[θ]^2 + 3 (-1 + h)^4 w Λ^4 Cos[θ]^4) +  Λ^2 λx0^5 (f^2 κ (-ht w (3 + 4 w + w^2) + κ - 2 w κ - 3 w^2 κ + h (-κ + w^2 (1 + κ) + w (2 + κ)) + w μy + 3 w^2 μy + w^3 μy + w κ μy + 2 w^2 κ μy) + f (-1 + h) w Λ^2 (-ht w (4 + 3 w) + h (3 w^2 + w (4 - 8 κ) - 12 κ) + κ (12 + 4 w^2 (-1 + μy) + w (3 + 5 μy))) Cos[θ]^2 - 6 (-1 + h)^3 w^2 (2 +w) Λ^4 Cos[θ]^4) + Λ^2 λx0^5 (f^2 κ (w + w^3 - κ + 3 w κ + w^2 (5 + 7 κ)) + f w Λ^2 ((-4 + h + 3 ht) w^3 + κ (7 - 4 h - 3 μy) + w (-8 + 4 ht - 17 κ + 4 h (1 + 4 κ) + κ μy) + w^2 (-18 + 9 ht - 22 κ + h (9 + 16 κ) + 6 κ μy)) Cos[θ]^2 + 2 (-1 + h)^2 w^2 (4 + 15 w + 6 w^2) Λ^4 Cos[θ]^4) - (1/λtx0)λx0^5 Cos[θ]^2 (-f^3 κ^2 λtx0 (-h + ht - 3 w + 2 ht w - 3 κ - 5 w κ + w μy + 3 κ μy + 5 w κ μy) + f^2 κ Λ^2 (w^3 (λtx0 - ht λtx0) (-1 + μy) -  w^2 λtx0 (15 + h (-9 + μy) - 7 μy + 5 κ (-1 + μy) (-3 + 2 h + μy) + 2 ht (-7 + 4 h + 3 μy)) + κ λtx0 (6 + h^2 - 6 μy + μy^2 + h (-6 + 4 μy)) - w λtx0 (6 + 12 κ + h^2 (-1 + 2 κ) + 3 ht (-3 + μy) - μy - 12 κ μy + 2 κ μy^2 + 2 h (-1 + 3 ht - 6 κ - μy + 4 κ μy))) Cos[θ]^2 - f (-1 + h)^2 w Λ^4 λtx0 (6 w^2 (-1 + ht + κ (-1 + μy)) + 2 κ (1 + 2 h - 3 μy) + w (-8 + 8 ht + κ (-19 + 8 h + 11 μy))) Cos[θ]^4 - 2 (-1 + h)^4 w^2 (4 + 3 w) Λ^6 λtx0 Cos[θ]^6) +  λx0^4 Cos[θ]^4 (f^3 κ^2 λx0 (-1 +  μy) (-2 h - 3 w + ht (2 + 3 w) - 3 κ - 4 w κ + 3 κ μy + 4 w κ μy) + f^2 κ Λ^2 (λx0 - h λx0) (w^2 (5 - 5 ht - 4 κ (-1 + μy)) (-1 + μy) +  2 κ (-1 + μy) (-2 + h + μy) + w (-3 κ (-1 + μy) (-2 + h + μy) -  2 ht (-3 + h + 2 μy) + 2 (-2 + μy + h μy))) Cos[θ]^2 + f (-1 + h)^3 w Λ^4 λx0 (2 (-1 + 
           h) κ + 
        w (-2 + 2 ht + 
           5 κ (-1 + μy))) Cos[θ]^4 + 
     2 (-1 + h)^5 w^2 Λ^6 λx0 Cos[θ]^6)) Sin[θ]^2)/(h π λx0^5 (1 + h + 2 w + (-1 + h) Cos[2 θ])^3 ((1 + h) w Λ^2 + f κ (1 + μy) + ((-1 + h) w Λ^2 + f κ (-1 + μy)) Cos[2 θ])^3);

Integrate[Integrand, {θ, 0, π}, 
 Assumptions -> {w > 0, h > 0, h != 1, μy != 1, μy > 0, ht > 0, Λ > 0, κ > 0}]
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    $\begingroup$ Integrand doesn't evaluate, please check your code! $\endgroup$ May 23, 2023 at 13:17
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    $\begingroup$ It doesn't mean anything special. Integration is a hard problem. There is no algorithm that is guaranteed to finish. You have a very large number of symbolic parameters, and a very large expression, so it is not surprising that Integrate would take a long time. My advice here is to think about why you need this integral in symbolic form. Symbolic forms are primarily useful for humans (as opposed to computers) to understand. The integrand you have here is already too large for that. Suppose you had a result: what would you do with it, how does it get you closer to your actual goal? $\endgroup$
    – Szabolcs
    May 23, 2023 at 13:22
  • $\begingroup$ But how do you use it? Do you just plot it? Is it part of some further equations? $\endgroup$
    – Domen
    May 23, 2023 at 13:42
  • $\begingroup$ I need this result as part of the Beta function of the problem I am studying. This means that I have to sum this result to 12 similar ones and to find the values of the parameters I am using such that certain combination of these results is zero. This is rather difficult to explain, but basically I can't specify the values of the parameters because in the end I need to look for certain values of these parameters $\endgroup$ May 23, 2023 at 13:56
  • $\begingroup$ Since I am rather new to mathematica I wanted to know if there was a way to obtain a symbolic result. What I understand is that it may not be possible, so I will look for a way to simplify the integrand $\endgroup$ May 23, 2023 at 14:00

1 Answer 1

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the expression probably integrates, but it will take a long time. I expanded it and got about 440 individual terms where each one (I picked random ones to test) will take about 10 mins to give a symbolic result on a quite fast computer. So it may take over 70 hours in total and give a VERY large expression...

Edit:

I found a way to get at least to the Antiderivate in reasonable time:

After expansion of the integrand collect and simplify all terms of equal theta dependence. After that 6 different integrals remain which can be done in about 2000 secs. The result and the calculation of the definite integral should be checked numerically by giving numerical values to the numerous parameters. The antiderivative contains terms with

ArcTan[(Sqrt[f*\[Kappa] + w*\[CapitalLambda]^2]*Tan[\[Theta]])/
Sqrt[h*w*\[CapitalLambda]^2 + f*\[Kappa]*\[Mu]y]]

which are zero for both theta equal 0 or pi. And it is not clear whether the antiderivative is continuous within the integration range.

If the limits are plugged in without check the following expression remains:

(1/(2*(-1 + h)*
 h*((-1 + h)*w*\[CapitalLambda]^2 + 
    f*\[Kappa]*(-1 + \[Mu]y))^4))*(a*
w*\[CapitalLambda]^2*(2*(-1 + h)^3*(1 + 3*h)*
  w^3*\[CapitalLambda]^6 - 
 2*f^3*\[Kappa]^2*(w*(-1 + ht + \[Kappa]*(-1 + \[Mu]y)) - 
    h*\[Kappa]*(-1 + \[Mu]y))*(-1 + \[Mu]y)^2 - 
     
 f^2*(-1 + h)*
  w*\[Kappa]*\[CapitalLambda]^2*(-1 + \[Mu]y)*(6*
     w*(-1 + ht + \[Kappa]*(-1 + \[Mu]y)) + \[Kappa]*(1 + 
       7*h + \[Mu]y - 9*h*\[Mu]y)) + 
 f*(-1 + h)^2*
  w^2*\[CapitalLambda]^4*(-4*
     w*(-1 + ht + \[Kappa]*(-1 + \[Mu]y)) + \[Kappa]*(-3 - 5*h - 
       5*\[Mu]y + 13*h*\[Mu]y))))
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    $\begingroup$ @guido I have now an expression which seems to agree with the numerical integral, but its size blasts the code size limit of 30000 characters, so I cannot post it here. If you let me know your email, I can send it to you $\endgroup$
    – Andreas
    May 24, 2023 at 15:21
  • $\begingroup$ pastebin.com and link in answer is another way to share it. $\endgroup$
    – Michael E2
    May 26, 2023 at 16:37

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