Approximating integral with small parameter












2












$begingroup$


I want to approximately compute integral $$I =int_0^1 dx frac{x(2-x)(1-x)}{(1-x)^2+mu x}$$ assuming that $mu$ is small. I tried



Integrate [(2 - x) (1 - x) x/((1 - x)^2 + A x), {x, 0, 1}]


My Mathematica for some reason fails to do this integral explicitly (which is strange, since it is an integral of a rational function; I guess Mathematica obtains divergent answer after decomposing the fraction, but can see that the integral is convergent in fact), so that I could just approximate the exact result.



On the other hand, the point of $mu x$ term is to "regulate" divergence of this integral (i.e. integral without it, $int_0^1 dx frac{x(2-x)(1-x)}{(1-x)^2}$, is logarithmically divergent on the upper limit). Therefore, the whole effect of this term can be accounted by shifting the upper limit:



$$I approx int_0^{1-mu} dx frac{x(2-x)(1-x)}{(1-x)^2}$$



Now Mathematica can compute this integral easily



Integrate [(2 - x) (1 - x) x/(1 - x)^2, {x, 0, 1 - A}, Assumptions -> A > 0]


giving $I = -frac{1}{2}(1+ln(mu^2)-mu^2)$, which can be approximated as $I approx -frac{1}{2}(1+ln(mu^2))$



I wonder if there is any function, smth like ApproximateIntegral[f[x,s],{x,a,b},{s,0}], which could do this whole manipulation for me.






calculus-and-analysis approximation







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  • $begingroup$
    Please, always remember to give copyable code.
    $endgroup$
    – Henrik Schumacher
    2 hours ago






  • 1




    $begingroup$
    @HenrikSchumacher did that. Thanks for reminding!
    $endgroup$
    – Kolya Terziev
    2 hours ago
















2












$begingroup$


I want to approximately compute integral $$I =int_0^1 dx frac{x(2-x)(1-x)}{(1-x)^2+mu x}$$ assuming that $mu$ is small. I tried



Integrate [(2 - x) (1 - x) x/((1 - x)^2 + A x), {x, 0, 1}]


My Mathematica for some reason fails to do this integral explicitly (which is strange, since it is an integral of a rational function; I guess Mathematica obtains divergent answer after decomposing the fraction, but can see that the integral is convergent in fact), so that I could just approximate the exact result.



On the other hand, the point of $mu x$ term is to "regulate" divergence of this integral (i.e. integral without it, $int_0^1 dx frac{x(2-x)(1-x)}{(1-x)^2}$, is logarithmically divergent on the upper limit). Therefore, the whole effect of this term can be accounted by shifting the upper limit:



$$I approx int_0^{1-mu} dx frac{x(2-x)(1-x)}{(1-x)^2}$$



Now Mathematica can compute this integral easily



Integrate [(2 - x) (1 - x) x/(1 - x)^2, {x, 0, 1 - A}, Assumptions -> A > 0]


giving $I = -frac{1}{2}(1+ln(mu^2)-mu^2)$, which can be approximated as $I approx -frac{1}{2}(1+ln(mu^2))$



I wonder if there is any function, smth like ApproximateIntegral[f[x,s],{x,a,b},{s,0}], which could do this whole manipulation for me.






calculus-and-analysis approximation







share|improve this question











$endgroup$












  • $begingroup$
    Please, always remember to give copyable code.
    $endgroup$
    – Henrik Schumacher
    2 hours ago






  • 1




    $begingroup$
    @HenrikSchumacher did that. Thanks for reminding!
    $endgroup$
    – Kolya Terziev
    2 hours ago














2












2








2





$begingroup$


I want to approximately compute integral $$I =int_0^1 dx frac{x(2-x)(1-x)}{(1-x)^2+mu x}$$ assuming that $mu$ is small. I tried



Integrate [(2 - x) (1 - x) x/((1 - x)^2 + A x), {x, 0, 1}]


My Mathematica for some reason fails to do this integral explicitly (which is strange, since it is an integral of a rational function; I guess Mathematica obtains divergent answer after decomposing the fraction, but can see that the integral is convergent in fact), so that I could just approximate the exact result.



On the other hand, the point of $mu x$ term is to "regulate" divergence of this integral (i.e. integral without it, $int_0^1 dx frac{x(2-x)(1-x)}{(1-x)^2}$, is logarithmically divergent on the upper limit). Therefore, the whole effect of this term can be accounted by shifting the upper limit:



$$I approx int_0^{1-mu} dx frac{x(2-x)(1-x)}{(1-x)^2}$$



Now Mathematica can compute this integral easily



Integrate [(2 - x) (1 - x) x/(1 - x)^2, {x, 0, 1 - A}, Assumptions -> A > 0]


giving $I = -frac{1}{2}(1+ln(mu^2)-mu^2)$, which can be approximated as $I approx -frac{1}{2}(1+ln(mu^2))$



I wonder if there is any function, smth like ApproximateIntegral[f[x,s],{x,a,b},{s,0}], which could do this whole manipulation for me.






calculus-and-analysis approximation







share|improve this question











$endgroup$




I want to approximately compute integral $$I =int_0^1 dx frac{x(2-x)(1-x)}{(1-x)^2+mu x}$$ assuming that $mu$ is small. I tried



Integrate [(2 - x) (1 - x) x/((1 - x)^2 + A x), {x, 0, 1}]


My Mathematica for some reason fails to do this integral explicitly (which is strange, since it is an integral of a rational function; I guess Mathematica obtains divergent answer after decomposing the fraction, but can see that the integral is convergent in fact), so that I could just approximate the exact result.



On the other hand, the point of $mu x$ term is to "regulate" divergence of this integral (i.e. integral without it, $int_0^1 dx frac{x(2-x)(1-x)}{(1-x)^2}$, is logarithmically divergent on the upper limit). Therefore, the whole effect of this term can be accounted by shifting the upper limit:



$$I approx int_0^{1-mu} dx frac{x(2-x)(1-x)}{(1-x)^2}$$



Now Mathematica can compute this integral easily



Integrate [(2 - x) (1 - x) x/(1 - x)^2, {x, 0, 1 - A}, Assumptions -> A > 0]


giving $I = -frac{1}{2}(1+ln(mu^2)-mu^2)$, which can be approximated as $I approx -frac{1}{2}(1+ln(mu^2))$



I wonder if there is any function, smth like ApproximateIntegral[f[x,s],{x,a,b},{s,0}], which could do this whole manipulation for me.






calculus-and-analysis approximation




calculus-and-analysis approximation






share|improve this question















share|improve this question













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edited 2 hours ago







Kolya Terziev

















asked 3 hours ago









Kolya TerzievKolya Terziev

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  • $begingroup$
    Please, always remember to give copyable code.
    $endgroup$
    – Henrik Schumacher
    2 hours ago






  • 1




    $begingroup$
    @HenrikSchumacher did that. Thanks for reminding!
    $endgroup$
    – Kolya Terziev
    2 hours ago


















  • $begingroup$
    Please, always remember to give copyable code.
    $endgroup$
    – Henrik Schumacher
    2 hours ago






  • 1




    $begingroup$
    @HenrikSchumacher did that. Thanks for reminding!
    $endgroup$
    – Kolya Terziev
    2 hours ago
















$begingroup$
Please, always remember to give copyable code.
$endgroup$
– Henrik Schumacher
2 hours ago




$begingroup$
Please, always remember to give copyable code.
$endgroup$
– Henrik Schumacher
2 hours ago




1




1




$begingroup$
@HenrikSchumacher did that. Thanks for reminding!
$endgroup$
– Kolya Terziev
2 hours ago




$begingroup$
@HenrikSchumacher did that. Thanks for reminding!
$endgroup$
– Kolya Terziev
2 hours ago










2 Answers
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$begingroup$

You could use AsymptoticIntegrate, although I change the $mu x$ term to just $mu$, as the latter version is easier for AsymptoticIntegrate to handle:



AsymptoticIntegrate[(x(2-x)(1-x))/((1-x)^2+μ), {x,0,1}, {μ,0,2}]



-1/2 + μ^2/4 + μ (1/2 - Log[μ]/2) - Log[μ]/2







share|improve this answer









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  • $begingroup$
    Sorry, but this is a surrogate, not the true answer.
    $endgroup$
    – user64494
    2 hours ago



















3












$begingroup$

The true answer in version 12 is as follows.



Integrate[(x*(2 - x)*(1 - x))/((1 - x)^2 +[Mu]*x),{x, 0, 1},Assumptions-> [Mu] > 0 && [Mu] < 1]



(1/(2 (-4 + [Mu])^(
3/2)))Sqrt[[Mu]] (4 I Sqrt[-1 + 4/[Mu]] +
7 I Sqrt[(4 - [Mu]) [Mu]] - 2 I Sqrt[(4 - [Mu]) [Mu]^3] +
I (4 Sqrt[-1 + 4/[Mu]] + 3 Sqrt[(4 - [Mu]) [Mu]] -
5 Sqrt[(4 - [Mu]) [Mu]^3] +
Sqrt[-(-4 + [Mu]) [Mu]^5]) Log[[Mu]] + [Mu] (12 -
7 [Mu] + [Mu]^2) Log[1 - I Sqrt[-([Mu]/(-4 + [Mu]))]] -
12 [Mu] Log[1 + I Sqrt[-([Mu]/(-4 + [Mu]))]] +
7 [Mu]^2 Log[1 + I Sqrt[-([Mu]/(-4 + [Mu]))]] - [Mu]^3 Log[
1 + I Sqrt[-([Mu]/(-4 + [Mu]))]] +
4 Log[1 - (I [Mu])/Sqrt[-(-4 + [Mu]) [Mu]]] - [Mu] Log[
1 - (I [Mu])/Sqrt[-(-4 + [Mu]) [Mu]]] -
4 Log[1 + (I [Mu])/Sqrt[-(-4 + [Mu]) [Mu]]] + [Mu] Log[
1 + (I [Mu])/Sqrt[-(-4 + [Mu]) [Mu]]] -
4 Log[(2 I - I [Mu] + Sqrt[-(-4 + [Mu]) [Mu]])/
Sqrt[-(-4 + [Mu]) [Mu]]] -
11 [Mu] Log[(2 I - I [Mu] + Sqrt[-(-4 + [Mu]) [Mu]])/
Sqrt[-(-4 + [Mu]) [Mu]]] +
7 [Mu]^2 Log[(2 I - I [Mu] + Sqrt[-(-4 + [Mu]) [Mu]])/
Sqrt[-(-4 + [Mu]) [Mu]]] - [Mu]^3 Log[(
2 I - I [Mu] + Sqrt[-(-4 + [Mu]) [Mu]])/
Sqrt[-(-4 + [Mu]) [Mu]]] +
4 Log[(-2 I + I [Mu] + Sqrt[-(-4 + [Mu]) [Mu]])/
Sqrt[-(-4 + [Mu]) [Mu]]] +
11 [Mu] Log[(-2 I + I [Mu] + Sqrt[-(-4 + [Mu]) [Mu]])/
Sqrt[-(-4 + [Mu]) [Mu]]] -
7 [Mu]^2 Log[(-2 I + I [Mu] + Sqrt[-(-4 + [Mu]) [Mu]])/
Sqrt[-(-4 + [Mu]) [Mu]]] + [Mu]^3 Log[(-2 I + I [Mu] +
Sqrt[-(-4 + [Mu]) [Mu]])/Sqrt[-(-4 + [Mu]) [Mu]]])




Series[%, {[Mu], 0, 2}, Assumptions -> [Mu] > 0 && [Mu] < 1]



$$ frac{1}{2} (-log (mu )-1)+frac{pi sqrt{mu }}{4}+frac{1}{4} mu (-2 log (mu )-5)+frac{25}{32} pi mu ^{3/2}+frac{1}{24} mu ^2 (12 log (mu )-19)+Oleft(mu ^{5/2}right)$$




Addition. In order to complete the answer, let us consider the asymptotics for negative values of $mu$:



Integrate[(x*(2 - x)*(1 - x))/((1 - x)^2 + [Mu]*x), {x, 0, 1}, 

PrincipalValue -> True, Assumptions -> [Mu] < 0 && [Mu] > -1]



1/2 (-1 - 2 [Mu] + (-1 - [Mu] + [Mu]^2) Log[-[Mu]] +
Sqrt[[Mu]/(-4 + [Mu])] (-1 - 3 [Mu] + [Mu]^2) Log[
1 + Sqrt[[Mu]/(-4 + [Mu])]] +
Sqrt[[Mu]/(-4 + [Mu])] Log[1 + [Mu]/Sqrt[(-4 + [Mu]) [Mu]]] +
3 [Mu] Sqrt[[Mu]/(-4 + [Mu])]
Log[1 + [Mu]/
Sqrt[(-4 + [Mu]) [Mu]]] - [Mu]^2 Sqrt[[Mu]/(-4 + [Mu])]
Log[1 + [Mu]/Sqrt[(-4 + [Mu]) [Mu]]] -
Sqrt[[Mu]/(-4 + [Mu])]
Log[(2 - [Mu] - Sqrt[(-4 + [Mu]) [Mu]])/
Sqrt[(-4 + [Mu]) [Mu]]] -
3 [Mu] Sqrt[[Mu]/(-4 + [Mu])]
Log[(2 - [Mu] - Sqrt[(-4 + [Mu]) [Mu]])/
Sqrt[(-4 + [Mu]) [Mu]]] + [Mu]^2 Sqrt[[Mu]/(-4 + [Mu])]
Log[(2 - [Mu] - Sqrt[(-4 + [Mu]) [Mu]])/
Sqrt[(-4 + [Mu]) [Mu]]] +
Sqrt[[Mu]/(-4 + [Mu])]
Log[(2 - [Mu] + Sqrt[(-4 + [Mu]) [Mu]])/
Sqrt[(-4 + [Mu]) [Mu]]] +
3 [Mu] Sqrt[[Mu]/(-4 + [Mu])]
Log[(2 - [Mu] + Sqrt[(-4 + [Mu]) [Mu]])/
Sqrt[(-4 + [Mu]) [Mu]]] - [Mu]^2 Sqrt[[Mu]/(-4 + [Mu])]
Log[(2 - [Mu] + Sqrt[(-4 + [Mu]) [Mu]])/
Sqrt[(-4 + [Mu]) [Mu]]])




Series[%, {[Mu], 0, 2}, Assumptions -> [Mu] < 0 && [Mu] > -1]



$$frac{1}{2} (-log (-mu )-1)+frac{1}{4} mu (-2 log (-mu )-5)+frac{1}{24} mu ^2 (12 log (-mu )-19)+Oleft(mu ^{5/2}right) $$







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    2 Answers
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    oldest

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    2 Answers
    2






    active

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    $begingroup$

    You could use AsymptoticIntegrate, although I change the $mu x$ term to just $mu$, as the latter version is easier for AsymptoticIntegrate to handle:



    AsymptoticIntegrate[(x(2-x)(1-x))/((1-x)^2+μ), {x,0,1}, {μ,0,2}]



    -1/2 + μ^2/4 + μ (1/2 - Log[μ]/2) - Log[μ]/2







    share|improve this answer









    $endgroup$













    • $begingroup$
      Sorry, but this is a surrogate, not the true answer.
      $endgroup$
      – user64494
      2 hours ago
















    2












    $begingroup$

    You could use AsymptoticIntegrate, although I change the $mu x$ term to just $mu$, as the latter version is easier for AsymptoticIntegrate to handle:



    AsymptoticIntegrate[(x(2-x)(1-x))/((1-x)^2+μ), {x,0,1}, {μ,0,2}]



    -1/2 + μ^2/4 + μ (1/2 - Log[μ]/2) - Log[μ]/2







    share|improve this answer









    $endgroup$













    • $begingroup$
      Sorry, but this is a surrogate, not the true answer.
      $endgroup$
      – user64494
      2 hours ago














    2












    2








    2





    $begingroup$

    You could use AsymptoticIntegrate, although I change the $mu x$ term to just $mu$, as the latter version is easier for AsymptoticIntegrate to handle:



    AsymptoticIntegrate[(x(2-x)(1-x))/((1-x)^2+μ), {x,0,1}, {μ,0,2}]



    -1/2 + μ^2/4 + μ (1/2 - Log[μ]/2) - Log[μ]/2







    share|improve this answer









    $endgroup$



    You could use AsymptoticIntegrate, although I change the $mu x$ term to just $mu$, as the latter version is easier for AsymptoticIntegrate to handle:



    AsymptoticIntegrate[(x(2-x)(1-x))/((1-x)^2+μ), {x,0,1}, {μ,0,2}]



    -1/2 + μ^2/4 + μ (1/2 - Log[μ]/2) - Log[μ]/2








    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered 3 hours ago









    Carl WollCarl Woll

    75.7k3100197




    75.7k3100197












    • $begingroup$
      Sorry, but this is a surrogate, not the true answer.
      $endgroup$
      – user64494
      2 hours ago


















    • $begingroup$
      Sorry, but this is a surrogate, not the true answer.
      $endgroup$
      – user64494
      2 hours ago
















    $begingroup$
    Sorry, but this is a surrogate, not the true answer.
    $endgroup$
    – user64494
    2 hours ago




    $begingroup$
    Sorry, but this is a surrogate, not the true answer.
    $endgroup$
    – user64494
    2 hours ago











    3












    $begingroup$

    The true answer in version 12 is as follows.



    Integrate[(x*(2 - x)*(1 - x))/((1 - x)^2 +[Mu]*x),{x, 0, 1},Assumptions-> [Mu] > 0 && [Mu] < 1]



    (1/(2 (-4 + [Mu])^(
    3/2)))Sqrt[[Mu]] (4 I Sqrt[-1 + 4/[Mu]] +
    7 I Sqrt[(4 - [Mu]) [Mu]] - 2 I Sqrt[(4 - [Mu]) [Mu]^3] +
    I (4 Sqrt[-1 + 4/[Mu]] + 3 Sqrt[(4 - [Mu]) [Mu]] -
    5 Sqrt[(4 - [Mu]) [Mu]^3] +
    Sqrt[-(-4 + [Mu]) [Mu]^5]) Log[[Mu]] + [Mu] (12 -
    7 [Mu] + [Mu]^2) Log[1 - I Sqrt[-([Mu]/(-4 + [Mu]))]] -
    12 [Mu] Log[1 + I Sqrt[-([Mu]/(-4 + [Mu]))]] +
    7 [Mu]^2 Log[1 + I Sqrt[-([Mu]/(-4 + [Mu]))]] - [Mu]^3 Log[
    1 + I Sqrt[-([Mu]/(-4 + [Mu]))]] +
    4 Log[1 - (I [Mu])/Sqrt[-(-4 + [Mu]) [Mu]]] - [Mu] Log[
    1 - (I [Mu])/Sqrt[-(-4 + [Mu]) [Mu]]] -
    4 Log[1 + (I [Mu])/Sqrt[-(-4 + [Mu]) [Mu]]] + [Mu] Log[
    1 + (I [Mu])/Sqrt[-(-4 + [Mu]) [Mu]]] -
    4 Log[(2 I - I [Mu] + Sqrt[-(-4 + [Mu]) [Mu]])/
    Sqrt[-(-4 + [Mu]) [Mu]]] -
    11 [Mu] Log[(2 I - I [Mu] + Sqrt[-(-4 + [Mu]) [Mu]])/
    Sqrt[-(-4 + [Mu]) [Mu]]] +
    7 [Mu]^2 Log[(2 I - I [Mu] + Sqrt[-(-4 + [Mu]) [Mu]])/
    Sqrt[-(-4 + [Mu]) [Mu]]] - [Mu]^3 Log[(
    2 I - I [Mu] + Sqrt[-(-4 + [Mu]) [Mu]])/
    Sqrt[-(-4 + [Mu]) [Mu]]] +
    4 Log[(-2 I + I [Mu] + Sqrt[-(-4 + [Mu]) [Mu]])/
    Sqrt[-(-4 + [Mu]) [Mu]]] +
    11 [Mu] Log[(-2 I + I [Mu] + Sqrt[-(-4 + [Mu]) [Mu]])/
    Sqrt[-(-4 + [Mu]) [Mu]]] -
    7 [Mu]^2 Log[(-2 I + I [Mu] + Sqrt[-(-4 + [Mu]) [Mu]])/
    Sqrt[-(-4 + [Mu]) [Mu]]] + [Mu]^3 Log[(-2 I + I [Mu] +
    Sqrt[-(-4 + [Mu]) [Mu]])/Sqrt[-(-4 + [Mu]) [Mu]]])




    Series[%, {[Mu], 0, 2}, Assumptions -> [Mu] > 0 && [Mu] < 1]



    $$ frac{1}{2} (-log (mu )-1)+frac{pi sqrt{mu }}{4}+frac{1}{4} mu (-2 log (mu )-5)+frac{25}{32} pi mu ^{3/2}+frac{1}{24} mu ^2 (12 log (mu )-19)+Oleft(mu ^{5/2}right)$$




    Addition. In order to complete the answer, let us consider the asymptotics for negative values of $mu$:



    Integrate[(x*(2 - x)*(1 - x))/((1 - x)^2 + [Mu]*x), {x, 0, 1}, 
    
PrincipalValue -> True, Assumptions -> [Mu] < 0 && [Mu] > -1]



    1/2 (-1 - 2 [Mu] + (-1 - [Mu] + [Mu]^2) Log[-[Mu]] +
    Sqrt[[Mu]/(-4 + [Mu])] (-1 - 3 [Mu] + [Mu]^2) Log[
    1 + Sqrt[[Mu]/(-4 + [Mu])]] +
    Sqrt[[Mu]/(-4 + [Mu])] Log[1 + [Mu]/Sqrt[(-4 + [Mu]) [Mu]]] +
    3 [Mu] Sqrt[[Mu]/(-4 + [Mu])]
    Log[1 + [Mu]/
    Sqrt[(-4 + [Mu]) [Mu]]] - [Mu]^2 Sqrt[[Mu]/(-4 + [Mu])]
    Log[1 + [Mu]/Sqrt[(-4 + [Mu]) [Mu]]] -
    Sqrt[[Mu]/(-4 + [Mu])]
    Log[(2 - [Mu] - Sqrt[(-4 + [Mu]) [Mu]])/
    Sqrt[(-4 + [Mu]) [Mu]]] -
    3 [Mu] Sqrt[[Mu]/(-4 + [Mu])]
    Log[(2 - [Mu] - Sqrt[(-4 + [Mu]) [Mu]])/
    Sqrt[(-4 + [Mu]) [Mu]]] + [Mu]^2 Sqrt[[Mu]/(-4 + [Mu])]
    Log[(2 - [Mu] - Sqrt[(-4 + [Mu]) [Mu]])/
    Sqrt[(-4 + [Mu]) [Mu]]] +
    Sqrt[[Mu]/(-4 + [Mu])]
    Log[(2 - [Mu] + Sqrt[(-4 + [Mu]) [Mu]])/
    Sqrt[(-4 + [Mu]) [Mu]]] +
    3 [Mu] Sqrt[[Mu]/(-4 + [Mu])]
    Log[(2 - [Mu] + Sqrt[(-4 + [Mu]) [Mu]])/
    Sqrt[(-4 + [Mu]) [Mu]]] - [Mu]^2 Sqrt[[Mu]/(-4 + [Mu])]
    Log[(2 - [Mu] + Sqrt[(-4 + [Mu]) [Mu]])/
    Sqrt[(-4 + [Mu]) [Mu]]])




    Series[%, {[Mu], 0, 2}, Assumptions -> [Mu] < 0 && [Mu] > -1]



    $$frac{1}{2} (-log (-mu )-1)+frac{1}{4} mu (-2 log (-mu )-5)+frac{1}{24} mu ^2 (12 log (-mu )-19)+Oleft(mu ^{5/2}right) $$







    share|improve this answer











    $endgroup$


















      3












      $begingroup$

      The true answer in version 12 is as follows.



      Integrate[(x*(2 - x)*(1 - x))/((1 - x)^2 +[Mu]*x),{x, 0, 1},Assumptions-> [Mu] > 0 && [Mu] < 1]



      (1/(2 (-4 + [Mu])^(
      3/2)))Sqrt[[Mu]] (4 I Sqrt[-1 + 4/[Mu]] +
      7 I Sqrt[(4 - [Mu]) [Mu]] - 2 I Sqrt[(4 - [Mu]) [Mu]^3] +
      I (4 Sqrt[-1 + 4/[Mu]] + 3 Sqrt[(4 - [Mu]) [Mu]] -
      5 Sqrt[(4 - [Mu]) [Mu]^3] +
      Sqrt[-(-4 + [Mu]) [Mu]^5]) Log[[Mu]] + [Mu] (12 -
      7 [Mu] + [Mu]^2) Log[1 - I Sqrt[-([Mu]/(-4 + [Mu]))]] -
      12 [Mu] Log[1 + I Sqrt[-([Mu]/(-4 + [Mu]))]] +
      7 [Mu]^2 Log[1 + I Sqrt[-([Mu]/(-4 + [Mu]))]] - [Mu]^3 Log[
      1 + I Sqrt[-([Mu]/(-4 + [Mu]))]] +
      4 Log[1 - (I [Mu])/Sqrt[-(-4 + [Mu]) [Mu]]] - [Mu] Log[
      1 - (I [Mu])/Sqrt[-(-4 + [Mu]) [Mu]]] -
      4 Log[1 + (I [Mu])/Sqrt[-(-4 + [Mu]) [Mu]]] + [Mu] Log[
      1 + (I [Mu])/Sqrt[-(-4 + [Mu]) [Mu]]] -
      4 Log[(2 I - I [Mu] + Sqrt[-(-4 + [Mu]) [Mu]])/
      Sqrt[-(-4 + [Mu]) [Mu]]] -
      11 [Mu] Log[(2 I - I [Mu] + Sqrt[-(-4 + [Mu]) [Mu]])/
      Sqrt[-(-4 + [Mu]) [Mu]]] +
      7 [Mu]^2 Log[(2 I - I [Mu] + Sqrt[-(-4 + [Mu]) [Mu]])/
      Sqrt[-(-4 + [Mu]) [Mu]]] - [Mu]^3 Log[(
      2 I - I [Mu] + Sqrt[-(-4 + [Mu]) [Mu]])/
      Sqrt[-(-4 + [Mu]) [Mu]]] +
      4 Log[(-2 I + I [Mu] + Sqrt[-(-4 + [Mu]) [Mu]])/
      Sqrt[-(-4 + [Mu]) [Mu]]] +
      11 [Mu] Log[(-2 I + I [Mu] + Sqrt[-(-4 + [Mu]) [Mu]])/
      Sqrt[-(-4 + [Mu]) [Mu]]] -
      7 [Mu]^2 Log[(-2 I + I [Mu] + Sqrt[-(-4 + [Mu]) [Mu]])/
      Sqrt[-(-4 + [Mu]) [Mu]]] + [Mu]^3 Log[(-2 I + I [Mu] +
      Sqrt[-(-4 + [Mu]) [Mu]])/Sqrt[-(-4 + [Mu]) [Mu]]])




      Series[%, {[Mu], 0, 2}, Assumptions -> [Mu] > 0 && [Mu] < 1]



      $$ frac{1}{2} (-log (mu )-1)+frac{pi sqrt{mu }}{4}+frac{1}{4} mu (-2 log (mu )-5)+frac{25}{32} pi mu ^{3/2}+frac{1}{24} mu ^2 (12 log (mu )-19)+Oleft(mu ^{5/2}right)$$




      Addition. In order to complete the answer, let us consider the asymptotics for negative values of $mu$:



      Integrate[(x*(2 - x)*(1 - x))/((1 - x)^2 + [Mu]*x), {x, 0, 1}, 
      
PrincipalValue -> True, Assumptions -> [Mu] < 0 && [Mu] > -1]



      1/2 (-1 - 2 [Mu] + (-1 - [Mu] + [Mu]^2) Log[-[Mu]] +
      Sqrt[[Mu]/(-4 + [Mu])] (-1 - 3 [Mu] + [Mu]^2) Log[
      1 + Sqrt[[Mu]/(-4 + [Mu])]] +
      Sqrt[[Mu]/(-4 + [Mu])] Log[1 + [Mu]/Sqrt[(-4 + [Mu]) [Mu]]] +
      3 [Mu] Sqrt[[Mu]/(-4 + [Mu])]
      Log[1 + [Mu]/
      Sqrt[(-4 + [Mu]) [Mu]]] - [Mu]^2 Sqrt[[Mu]/(-4 + [Mu])]
      Log[1 + [Mu]/Sqrt[(-4 + [Mu]) [Mu]]] -
      Sqrt[[Mu]/(-4 + [Mu])]
      Log[(2 - [Mu] - Sqrt[(-4 + [Mu]) [Mu]])/
      Sqrt[(-4 + [Mu]) [Mu]]] -
      3 [Mu] Sqrt[[Mu]/(-4 + [Mu])]
      Log[(2 - [Mu] - Sqrt[(-4 + [Mu]) [Mu]])/
      Sqrt[(-4 + [Mu]) [Mu]]] + [Mu]^2 Sqrt[[Mu]/(-4 + [Mu])]
      Log[(2 - [Mu] - Sqrt[(-4 + [Mu]) [Mu]])/
      Sqrt[(-4 + [Mu]) [Mu]]] +
      Sqrt[[Mu]/(-4 + [Mu])]
      Log[(2 - [Mu] + Sqrt[(-4 + [Mu]) [Mu]])/
      Sqrt[(-4 + [Mu]) [Mu]]] +
      3 [Mu] Sqrt[[Mu]/(-4 + [Mu])]
      Log[(2 - [Mu] + Sqrt[(-4 + [Mu]) [Mu]])/
      Sqrt[(-4 + [Mu]) [Mu]]] - [Mu]^2 Sqrt[[Mu]/(-4 + [Mu])]
      Log[(2 - [Mu] + Sqrt[(-4 + [Mu]) [Mu]])/
      Sqrt[(-4 + [Mu]) [Mu]]])




      Series[%, {[Mu], 0, 2}, Assumptions -> [Mu] < 0 && [Mu] > -1]



      $$frac{1}{2} (-log (-mu )-1)+frac{1}{4} mu (-2 log (-mu )-5)+frac{1}{24} mu ^2 (12 log (-mu )-19)+Oleft(mu ^{5/2}right) $$







      share|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        The true answer in version 12 is as follows.



        Integrate[(x*(2 - x)*(1 - x))/((1 - x)^2 +[Mu]*x),{x, 0, 1},Assumptions-> [Mu] > 0 && [Mu] < 1]



        (1/(2 (-4 + [Mu])^(
        3/2)))Sqrt[[Mu]] (4 I Sqrt[-1 + 4/[Mu]] +
        7 I Sqrt[(4 - [Mu]) [Mu]] - 2 I Sqrt[(4 - [Mu]) [Mu]^3] +
        I (4 Sqrt[-1 + 4/[Mu]] + 3 Sqrt[(4 - [Mu]) [Mu]] -
        5 Sqrt[(4 - [Mu]) [Mu]^3] +
        Sqrt[-(-4 + [Mu]) [Mu]^5]) Log[[Mu]] + [Mu] (12 -
        7 [Mu] + [Mu]^2) Log[1 - I Sqrt[-([Mu]/(-4 + [Mu]))]] -
        12 [Mu] Log[1 + I Sqrt[-([Mu]/(-4 + [Mu]))]] +
        7 [Mu]^2 Log[1 + I Sqrt[-([Mu]/(-4 + [Mu]))]] - [Mu]^3 Log[
        1 + I Sqrt[-([Mu]/(-4 + [Mu]))]] +
        4 Log[1 - (I [Mu])/Sqrt[-(-4 + [Mu]) [Mu]]] - [Mu] Log[
        1 - (I [Mu])/Sqrt[-(-4 + [Mu]) [Mu]]] -
        4 Log[1 + (I [Mu])/Sqrt[-(-4 + [Mu]) [Mu]]] + [Mu] Log[
        1 + (I [Mu])/Sqrt[-(-4 + [Mu]) [Mu]]] -
        4 Log[(2 I - I [Mu] + Sqrt[-(-4 + [Mu]) [Mu]])/
        Sqrt[-(-4 + [Mu]) [Mu]]] -
        11 [Mu] Log[(2 I - I [Mu] + Sqrt[-(-4 + [Mu]) [Mu]])/
        Sqrt[-(-4 + [Mu]) [Mu]]] +
        7 [Mu]^2 Log[(2 I - I [Mu] + Sqrt[-(-4 + [Mu]) [Mu]])/
        Sqrt[-(-4 + [Mu]) [Mu]]] - [Mu]^3 Log[(
        2 I - I [Mu] + Sqrt[-(-4 + [Mu]) [Mu]])/
        Sqrt[-(-4 + [Mu]) [Mu]]] +
        4 Log[(-2 I + I [Mu] + Sqrt[-(-4 + [Mu]) [Mu]])/
        Sqrt[-(-4 + [Mu]) [Mu]]] +
        11 [Mu] Log[(-2 I + I [Mu] + Sqrt[-(-4 + [Mu]) [Mu]])/
        Sqrt[-(-4 + [Mu]) [Mu]]] -
        7 [Mu]^2 Log[(-2 I + I [Mu] + Sqrt[-(-4 + [Mu]) [Mu]])/
        Sqrt[-(-4 + [Mu]) [Mu]]] + [Mu]^3 Log[(-2 I + I [Mu] +
        Sqrt[-(-4 + [Mu]) [Mu]])/Sqrt[-(-4 + [Mu]) [Mu]]])




        Series[%, {[Mu], 0, 2}, Assumptions -> [Mu] > 0 && [Mu] < 1]



        $$ frac{1}{2} (-log (mu )-1)+frac{pi sqrt{mu }}{4}+frac{1}{4} mu (-2 log (mu )-5)+frac{25}{32} pi mu ^{3/2}+frac{1}{24} mu ^2 (12 log (mu )-19)+Oleft(mu ^{5/2}right)$$




        Addition. In order to complete the answer, let us consider the asymptotics for negative values of $mu$:



        Integrate[(x*(2 - x)*(1 - x))/((1 - x)^2 + [Mu]*x), {x, 0, 1}, 
        
PrincipalValue -> True, Assumptions -> [Mu] < 0 && [Mu] > -1]



        1/2 (-1 - 2 [Mu] + (-1 - [Mu] + [Mu]^2) Log[-[Mu]] +
        Sqrt[[Mu]/(-4 + [Mu])] (-1 - 3 [Mu] + [Mu]^2) Log[
        1 + Sqrt[[Mu]/(-4 + [Mu])]] +
        Sqrt[[Mu]/(-4 + [Mu])] Log[1 + [Mu]/Sqrt[(-4 + [Mu]) [Mu]]] +
        3 [Mu] Sqrt[[Mu]/(-4 + [Mu])]
        Log[1 + [Mu]/
        Sqrt[(-4 + [Mu]) [Mu]]] - [Mu]^2 Sqrt[[Mu]/(-4 + [Mu])]
        Log[1 + [Mu]/Sqrt[(-4 + [Mu]) [Mu]]] -
        Sqrt[[Mu]/(-4 + [Mu])]
        Log[(2 - [Mu] - Sqrt[(-4 + [Mu]) [Mu]])/
        Sqrt[(-4 + [Mu]) [Mu]]] -
        3 [Mu] Sqrt[[Mu]/(-4 + [Mu])]
        Log[(2 - [Mu] - Sqrt[(-4 + [Mu]) [Mu]])/
        Sqrt[(-4 + [Mu]) [Mu]]] + [Mu]^2 Sqrt[[Mu]/(-4 + [Mu])]
        Log[(2 - [Mu] - Sqrt[(-4 + [Mu]) [Mu]])/
        Sqrt[(-4 + [Mu]) [Mu]]] +
        Sqrt[[Mu]/(-4 + [Mu])]
        Log[(2 - [Mu] + Sqrt[(-4 + [Mu]) [Mu]])/
        Sqrt[(-4 + [Mu]) [Mu]]] +
        3 [Mu] Sqrt[[Mu]/(-4 + [Mu])]
        Log[(2 - [Mu] + Sqrt[(-4 + [Mu]) [Mu]])/
        Sqrt[(-4 + [Mu]) [Mu]]] - [Mu]^2 Sqrt[[Mu]/(-4 + [Mu])]
        Log[(2 - [Mu] + Sqrt[(-4 + [Mu]) [Mu]])/
        Sqrt[(-4 + [Mu]) [Mu]]])




        Series[%, {[Mu], 0, 2}, Assumptions -> [Mu] < 0 && [Mu] > -1]



        $$frac{1}{2} (-log (-mu )-1)+frac{1}{4} mu (-2 log (-mu )-5)+frac{1}{24} mu ^2 (12 log (-mu )-19)+Oleft(mu ^{5/2}right) $$







        share|improve this answer











        $endgroup$



        The true answer in version 12 is as follows.



        Integrate[(x*(2 - x)*(1 - x))/((1 - x)^2 +[Mu]*x),{x, 0, 1},Assumptions-> [Mu] > 0 && [Mu] < 1]



        (1/(2 (-4 + [Mu])^(
        3/2)))Sqrt[[Mu]] (4 I Sqrt[-1 + 4/[Mu]] +
        7 I Sqrt[(4 - [Mu]) [Mu]] - 2 I Sqrt[(4 - [Mu]) [Mu]^3] +
        I (4 Sqrt[-1 + 4/[Mu]] + 3 Sqrt[(4 - [Mu]) [Mu]] -
        5 Sqrt[(4 - [Mu]) [Mu]^3] +
        Sqrt[-(-4 + [Mu]) [Mu]^5]) Log[[Mu]] + [Mu] (12 -
        7 [Mu] + [Mu]^2) Log[1 - I Sqrt[-([Mu]/(-4 + [Mu]))]] -
        12 [Mu] Log[1 + I Sqrt[-([Mu]/(-4 + [Mu]))]] +
        7 [Mu]^2 Log[1 + I Sqrt[-([Mu]/(-4 + [Mu]))]] - [Mu]^3 Log[
        1 + I Sqrt[-([Mu]/(-4 + [Mu]))]] +
        4 Log[1 - (I [Mu])/Sqrt[-(-4 + [Mu]) [Mu]]] - [Mu] Log[
        1 - (I [Mu])/Sqrt[-(-4 + [Mu]) [Mu]]] -
        4 Log[1 + (I [Mu])/Sqrt[-(-4 + [Mu]) [Mu]]] + [Mu] Log[
        1 + (I [Mu])/Sqrt[-(-4 + [Mu]) [Mu]]] -
        4 Log[(2 I - I [Mu] + Sqrt[-(-4 + [Mu]) [Mu]])/
        Sqrt[-(-4 + [Mu]) [Mu]]] -
        11 [Mu] Log[(2 I - I [Mu] + Sqrt[-(-4 + [Mu]) [Mu]])/
        Sqrt[-(-4 + [Mu]) [Mu]]] +
        7 [Mu]^2 Log[(2 I - I [Mu] + Sqrt[-(-4 + [Mu]) [Mu]])/
        Sqrt[-(-4 + [Mu]) [Mu]]] - [Mu]^3 Log[(
        2 I - I [Mu] + Sqrt[-(-4 + [Mu]) [Mu]])/
        Sqrt[-(-4 + [Mu]) [Mu]]] +
        4 Log[(-2 I + I [Mu] + Sqrt[-(-4 + [Mu]) [Mu]])/
        Sqrt[-(-4 + [Mu]) [Mu]]] +
        11 [Mu] Log[(-2 I + I [Mu] + Sqrt[-(-4 + [Mu]) [Mu]])/
        Sqrt[-(-4 + [Mu]) [Mu]]] -
        7 [Mu]^2 Log[(-2 I + I [Mu] + Sqrt[-(-4 + [Mu]) [Mu]])/
        Sqrt[-(-4 + [Mu]) [Mu]]] + [Mu]^3 Log[(-2 I + I [Mu] +
        Sqrt[-(-4 + [Mu]) [Mu]])/Sqrt[-(-4 + [Mu]) [Mu]]])




        Series[%, {[Mu], 0, 2}, Assumptions -> [Mu] > 0 && [Mu] < 1]



        $$ frac{1}{2} (-log (mu )-1)+frac{pi sqrt{mu }}{4}+frac{1}{4} mu (-2 log (mu )-5)+frac{25}{32} pi mu ^{3/2}+frac{1}{24} mu ^2 (12 log (mu )-19)+Oleft(mu ^{5/2}right)$$




        Addition. In order to complete the answer, let us consider the asymptotics for negative values of $mu$:



        Integrate[(x*(2 - x)*(1 - x))/((1 - x)^2 + [Mu]*x), {x, 0, 1}, 
        
PrincipalValue -> True, Assumptions -> [Mu] < 0 && [Mu] > -1]



        1/2 (-1 - 2 [Mu] + (-1 - [Mu] + [Mu]^2) Log[-[Mu]] +
        Sqrt[[Mu]/(-4 + [Mu])] (-1 - 3 [Mu] + [Mu]^2) Log[
        1 + Sqrt[[Mu]/(-4 + [Mu])]] +
        Sqrt[[Mu]/(-4 + [Mu])] Log[1 + [Mu]/Sqrt[(-4 + [Mu]) [Mu]]] +
        3 [Mu] Sqrt[[Mu]/(-4 + [Mu])]
        Log[1 + [Mu]/
        Sqrt[(-4 + [Mu]) [Mu]]] - [Mu]^2 Sqrt[[Mu]/(-4 + [Mu])]
        Log[1 + [Mu]/Sqrt[(-4 + [Mu]) [Mu]]] -
        Sqrt[[Mu]/(-4 + [Mu])]
        Log[(2 - [Mu] - Sqrt[(-4 + [Mu]) [Mu]])/
        Sqrt[(-4 + [Mu]) [Mu]]] -
        3 [Mu] Sqrt[[Mu]/(-4 + [Mu])]
        Log[(2 - [Mu] - Sqrt[(-4 + [Mu]) [Mu]])/
        Sqrt[(-4 + [Mu]) [Mu]]] + [Mu]^2 Sqrt[[Mu]/(-4 + [Mu])]
        Log[(2 - [Mu] - Sqrt[(-4 + [Mu]) [Mu]])/
        Sqrt[(-4 + [Mu]) [Mu]]] +
        Sqrt[[Mu]/(-4 + [Mu])]
        Log[(2 - [Mu] + Sqrt[(-4 + [Mu]) [Mu]])/
        Sqrt[(-4 + [Mu]) [Mu]]] +
        3 [Mu] Sqrt[[Mu]/(-4 + [Mu])]
        Log[(2 - [Mu] + Sqrt[(-4 + [Mu]) [Mu]])/
        Sqrt[(-4 + [Mu]) [Mu]]] - [Mu]^2 Sqrt[[Mu]/(-4 + [Mu])]
        Log[(2 - [Mu] + Sqrt[(-4 + [Mu]) [Mu]])/
        Sqrt[(-4 + [Mu]) [Mu]]])




        Series[%, {[Mu], 0, 2}, Assumptions -> [Mu] < 0 && [Mu] > -1]



        $$frac{1}{2} (-log (-mu )-1)+frac{1}{4} mu (-2 log (-mu )-5)+frac{1}{24} mu ^2 (12 log (-mu )-19)+Oleft(mu ^{5/2}right) $$








        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 1 hour ago

























        answered 2 hours ago









        user64494user64494

        3,64311122




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