Mathematica gives an unexpected answer for Integrate [closed]












4












$begingroup$


I need to integrate the following:
begin{equation}tag{1}
frac{sqrt{C + (1 - C) x^3}}{x},
end{equation}
where $0 < C < 1$ and $x$ is a positive variable (then $x^3 ge 0$). When I integrate:



Integrate[Sqrt[C + (1 - C) x^3]/x, x, Assumptions -> 0 < C < 1]


I get this answer:
begin{equation}tag{2}
frac{2}{3} sqrt{C + (1 - C), x^3} - frac{2}{3} sqrt{C} , mathrm{argtanh} Big( frac{sqrt{C + (1 - C) , x^3}}{sqrt{C}} Big).
end{equation}
This answer is wrong because the term inside parenthesis (inside the argtanh) is larger than 1, which invalidates the answer.



So how can I get the proper integration of (1), using Mathematica?




EDIT: Actually, what I need is to find the general solution of the following differential equation ($lambda$ is a positive constant):
begin{equation}tag{3}
frac{d a}{d t} = pm , frac{lambda , a}{sqrt{C + (1 - C) , a^3}},
end{equation}
i.e. the function $a(t)$ or the time $t(a)$.






differential-equations calculus-and-analysis functions trigonometry integral-equations







share|improve this question











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closed as off-topic by Szabolcs, Daniel Lichtblau, m_goldberg, Bob Hanlon, Bill Watts Jan 14 at 2:11


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – Szabolcs, Daniel Lichtblau, m_goldberg, Bob Hanlon, Bill Watts

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    When evaluating a definite integral don't the imaginary parts cancel out?
    $endgroup$
    – JimB
    Jan 13 at 16:31










  • $begingroup$
    @JimB, I need the general solution on the indefinite integration, since I don't know the limits of the variable $x$ (except that $x > 0$). It's the solution of some differential equation.
    $endgroup$
    – Cham
    Jan 13 at 16:34








  • 1




    $begingroup$
    You should change the subject line of the question from "wrong" to perhaps "unexpected" since Assuming[{0<C<1, x>=0}, Sqrt[C+(1-C) x^3]/x == D[Integrate[Sqrt[C+(1-C) x^3]/x, x], x] // Simplify] evaluates to True
    $endgroup$
    – Bob Hanlon
    Jan 13 at 20:25






  • 2




    $begingroup$
    Tha antiderivative shown is fine. Can check by differentiation: In[646]:= i0 = Sqrt[C + (1 - C) x^3]/x; i1 = Integrate[i0, x]; Simplify[D[i1, x] - i0] Out[648]= 0
    $endgroup$
    – Daniel Lichtblau
    Jan 13 at 21:59
















4












$begingroup$


I need to integrate the following:
begin{equation}tag{1}
frac{sqrt{C + (1 - C) x^3}}{x},
end{equation}
where $0 < C < 1$ and $x$ is a positive variable (then $x^3 ge 0$). When I integrate:



Integrate[Sqrt[C + (1 - C) x^3]/x, x, Assumptions -> 0 < C < 1]


I get this answer:
begin{equation}tag{2}
frac{2}{3} sqrt{C + (1 - C), x^3} - frac{2}{3} sqrt{C} , mathrm{argtanh} Big( frac{sqrt{C + (1 - C) , x^3}}{sqrt{C}} Big).
end{equation}
This answer is wrong because the term inside parenthesis (inside the argtanh) is larger than 1, which invalidates the answer.



So how can I get the proper integration of (1), using Mathematica?




EDIT: Actually, what I need is to find the general solution of the following differential equation ($lambda$ is a positive constant):
begin{equation}tag{3}
frac{d a}{d t} = pm , frac{lambda , a}{sqrt{C + (1 - C) , a^3}},
end{equation}
i.e. the function $a(t)$ or the time $t(a)$.






differential-equations calculus-and-analysis functions trigonometry integral-equations







share|improve this question











$endgroup$



closed as off-topic by Szabolcs, Daniel Lichtblau, m_goldberg, Bob Hanlon, Bill Watts Jan 14 at 2:11


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – Szabolcs, Daniel Lichtblau, m_goldberg, Bob Hanlon, Bill Watts

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    When evaluating a definite integral don't the imaginary parts cancel out?
    $endgroup$
    – JimB
    Jan 13 at 16:31










  • $begingroup$
    @JimB, I need the general solution on the indefinite integration, since I don't know the limits of the variable $x$ (except that $x > 0$). It's the solution of some differential equation.
    $endgroup$
    – Cham
    Jan 13 at 16:34








  • 1




    $begingroup$
    You should change the subject line of the question from "wrong" to perhaps "unexpected" since Assuming[{0<C<1, x>=0}, Sqrt[C+(1-C) x^3]/x == D[Integrate[Sqrt[C+(1-C) x^3]/x, x], x] // Simplify] evaluates to True
    $endgroup$
    – Bob Hanlon
    Jan 13 at 20:25






  • 2




    $begingroup$
    Tha antiderivative shown is fine. Can check by differentiation: In[646]:= i0 = Sqrt[C + (1 - C) x^3]/x; i1 = Integrate[i0, x]; Simplify[D[i1, x] - i0] Out[648]= 0
    $endgroup$
    – Daniel Lichtblau
    Jan 13 at 21:59














4












4








4


1



$begingroup$


I need to integrate the following:
begin{equation}tag{1}
frac{sqrt{C + (1 - C) x^3}}{x},
end{equation}
where $0 < C < 1$ and $x$ is a positive variable (then $x^3 ge 0$). When I integrate:



Integrate[Sqrt[C + (1 - C) x^3]/x, x, Assumptions -> 0 < C < 1]


I get this answer:
begin{equation}tag{2}
frac{2}{3} sqrt{C + (1 - C), x^3} - frac{2}{3} sqrt{C} , mathrm{argtanh} Big( frac{sqrt{C + (1 - C) , x^3}}{sqrt{C}} Big).
end{equation}
This answer is wrong because the term inside parenthesis (inside the argtanh) is larger than 1, which invalidates the answer.



So how can I get the proper integration of (1), using Mathematica?




EDIT: Actually, what I need is to find the general solution of the following differential equation ($lambda$ is a positive constant):
begin{equation}tag{3}
frac{d a}{d t} = pm , frac{lambda , a}{sqrt{C + (1 - C) , a^3}},
end{equation}
i.e. the function $a(t)$ or the time $t(a)$.






differential-equations calculus-and-analysis functions trigonometry integral-equations







share|improve this question











$endgroup$




I need to integrate the following:
begin{equation}tag{1}
frac{sqrt{C + (1 - C) x^3}}{x},
end{equation}
where $0 < C < 1$ and $x$ is a positive variable (then $x^3 ge 0$). When I integrate:



Integrate[Sqrt[C + (1 - C) x^3]/x, x, Assumptions -> 0 < C < 1]


I get this answer:
begin{equation}tag{2}
frac{2}{3} sqrt{C + (1 - C), x^3} - frac{2}{3} sqrt{C} , mathrm{argtanh} Big( frac{sqrt{C + (1 - C) , x^3}}{sqrt{C}} Big).
end{equation}
This answer is wrong because the term inside parenthesis (inside the argtanh) is larger than 1, which invalidates the answer.



So how can I get the proper integration of (1), using Mathematica?




EDIT: Actually, what I need is to find the general solution of the following differential equation ($lambda$ is a positive constant):
begin{equation}tag{3}
frac{d a}{d t} = pm , frac{lambda , a}{sqrt{C + (1 - C) , a^3}},
end{equation}
i.e. the function $a(t)$ or the time $t(a)$.






differential-equations calculus-and-analysis functions trigonometry integral-equations




differential-equations calculus-and-analysis functions trigonometry integral-equations






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Jan 13 at 22:15







Cham

















asked Jan 13 at 15:50









ChamCham

1,4421227




1,4421227




closed as off-topic by Szabolcs, Daniel Lichtblau, m_goldberg, Bob Hanlon, Bill Watts Jan 14 at 2:11


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – Szabolcs, Daniel Lichtblau, m_goldberg, Bob Hanlon, Bill Watts

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Szabolcs, Daniel Lichtblau, m_goldberg, Bob Hanlon, Bill Watts Jan 14 at 2:11


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – Szabolcs, Daniel Lichtblau, m_goldberg, Bob Hanlon, Bill Watts

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $begingroup$
    When evaluating a definite integral don't the imaginary parts cancel out?
    $endgroup$
    – JimB
    Jan 13 at 16:31










  • $begingroup$
    @JimB, I need the general solution on the indefinite integration, since I don't know the limits of the variable $x$ (except that $x > 0$). It's the solution of some differential equation.
    $endgroup$
    – Cham
    Jan 13 at 16:34








  • 1




    $begingroup$
    You should change the subject line of the question from "wrong" to perhaps "unexpected" since Assuming[{0<C<1, x>=0}, Sqrt[C+(1-C) x^3]/x == D[Integrate[Sqrt[C+(1-C) x^3]/x, x], x] // Simplify] evaluates to True
    $endgroup$
    – Bob Hanlon
    Jan 13 at 20:25






  • 2




    $begingroup$
    Tha antiderivative shown is fine. Can check by differentiation: In[646]:= i0 = Sqrt[C + (1 - C) x^3]/x; i1 = Integrate[i0, x]; Simplify[D[i1, x] - i0] Out[648]= 0
    $endgroup$
    – Daniel Lichtblau
    Jan 13 at 21:59


















  • $begingroup$
    When evaluating a definite integral don't the imaginary parts cancel out?
    $endgroup$
    – JimB
    Jan 13 at 16:31










  • $begingroup$
    @JimB, I need the general solution on the indefinite integration, since I don't know the limits of the variable $x$ (except that $x > 0$). It's the solution of some differential equation.
    $endgroup$
    – Cham
    Jan 13 at 16:34








  • 1




    $begingroup$
    You should change the subject line of the question from "wrong" to perhaps "unexpected" since Assuming[{0<C<1, x>=0}, Sqrt[C+(1-C) x^3]/x == D[Integrate[Sqrt[C+(1-C) x^3]/x, x], x] // Simplify] evaluates to True
    $endgroup$
    – Bob Hanlon
    Jan 13 at 20:25






  • 2




    $begingroup$
    Tha antiderivative shown is fine. Can check by differentiation: In[646]:= i0 = Sqrt[C + (1 - C) x^3]/x; i1 = Integrate[i0, x]; Simplify[D[i1, x] - i0] Out[648]= 0
    $endgroup$
    – Daniel Lichtblau
    Jan 13 at 21:59
















$begingroup$
When evaluating a definite integral don't the imaginary parts cancel out?
$endgroup$
– JimB
Jan 13 at 16:31




$begingroup$
When evaluating a definite integral don't the imaginary parts cancel out?
$endgroup$
– JimB
Jan 13 at 16:31












$begingroup$
@JimB, I need the general solution on the indefinite integration, since I don't know the limits of the variable $x$ (except that $x > 0$). It's the solution of some differential equation.
$endgroup$
– Cham
Jan 13 at 16:34






$begingroup$
@JimB, I need the general solution on the indefinite integration, since I don't know the limits of the variable $x$ (except that $x > 0$). It's the solution of some differential equation.
$endgroup$
– Cham
Jan 13 at 16:34






1




1




$begingroup$
You should change the subject line of the question from "wrong" to perhaps "unexpected" since Assuming[{0<C<1, x>=0}, Sqrt[C+(1-C) x^3]/x == D[Integrate[Sqrt[C+(1-C) x^3]/x, x], x] // Simplify] evaluates to True
$endgroup$
– Bob Hanlon
Jan 13 at 20:25




$begingroup$
You should change the subject line of the question from "wrong" to perhaps "unexpected" since Assuming[{0<C<1, x>=0}, Sqrt[C+(1-C) x^3]/x == D[Integrate[Sqrt[C+(1-C) x^3]/x, x], x] // Simplify] evaluates to True
$endgroup$
– Bob Hanlon
Jan 13 at 20:25




2




2




$begingroup$
Tha antiderivative shown is fine. Can check by differentiation: In[646]:= i0 = Sqrt[C + (1 - C) x^3]/x; i1 = Integrate[i0, x]; Simplify[D[i1, x] - i0] Out[648]= 0
$endgroup$
– Daniel Lichtblau
Jan 13 at 21:59




$begingroup$
Tha antiderivative shown is fine. Can check by differentiation: In[646]:= i0 = Sqrt[C + (1 - C) x^3]/x; i1 = Integrate[i0, x]; Simplify[D[i1, x] - i0] Out[648]= 0
$endgroup$
– Daniel Lichtblau
Jan 13 at 21:59










2 Answers
2






active

oldest

votes


















4












$begingroup$

$intfrac{sqrt{C + (1 - C) x^3}}{x}dx=intfrac{left(sqrt{C + (1 - C) x^3}right)x^2}{x^3}dx$



Let $C + (1 - C) x^3=u^2$



$3(1 - C) x^2dx=2udu$



$frac{2}{3}intfrac{u^2du}{u^2-C}du$



$frac{2}{3}intleft(1+frac{c}{u^2-C}right)du$



$frac{2}{3}left(u+Cintfrac{du}{u^2-sqrt{C}^2}right)$



$frac{2}{3}left(u-Cfrac{Arctanh(frac{u}{sqrt{C}})}{sqrt{C}}right)$



Substituting back gives



begin{equation}tag{2}
frac{2}{3} sqrt{C + (1 - C), x^3} - frac{2}{3} sqrt{C} , mathrm{argtanh} Big( frac{sqrt{C + (1 - C) , x^3}}{sqrt{C}} Big).
end{equation}



FunctionDomain[ArcTanh[Sqrt[C + (1 - C) x^3]/Sqrt[C]], x]



-1 < Sqrt[C + x^3 - C x^3]/Sqrt[C] < 1 && -C - x^3 + C x^3 <= 0




Or



 ArcTanh[Sqrt[C + (1 - C) x^3]/Sqrt[C]] // TrigToExp



-(1/2) Log[1 - Sqrt[C + (1 - C) x^3]/Sqrt[C]] + 1/2 Log[1 + Sqrt[C + (1 - C) x^3]/Sqrt[C]]







share|improve this answer











$endgroup$













  • $begingroup$
    There's still a problem with the argument of arctanh.
    $endgroup$
    – Cham
    Jan 13 at 17:06










  • $begingroup$
    I think it's better to split the integral of $frac{1}{u^2 - C}$ in two parts : $frac{1}{u - sqrt{C}} - frac{1}{u + sqrt{C}}$, then integration gives a $ln{dots}$.
    $endgroup$
    – Cham
    Jan 13 at 17:09












  • $begingroup$
    They are equivalent..
    $endgroup$
    – Okkes Dulgerci
    Jan 13 at 17:12










  • $begingroup$
    My solution appears to be this: begin{equation} frac{2 u}{3 sqrt{C}} + frac{1}{3} , ln{Big(frac{u - sqrt{C}}{u + sqrt{C}}Big)},end{equation} which is good since $u ge sqrt{C}$.
    $endgroup$
    – Cham
    Jan 13 at 17:12





















5












$begingroup$

Here are four five six forms for an antiderivative, the first ad1 being the same as the OP's and complex-valued over the intended domain. The other three four five are real-valued.



ad1 = Integrate[Sqrt[c + (1 - c) x^3]/x, x, Assumptions -> 0 < c < 1]

(*  2/3 Sqrt[c + x^3 - c x^3] - 2/3 Sqrt[c] ArcTanh[Sqrt[c + (1 - c) x^3]/Sqrt[c]]  *)


The imaginary part of ad is constant so we can subtract it off (which is the point of @JimB's comment under the OP):



ad2 = ad1 /. {{x -> 1}, {x -> x}} // Differences // First;



Assuming[0 < c < 1 && x > 0,

 TrigToExp@ad2 /. {Log[z_] /; Simplify[z < 0] :> Log[-1] + Log[-z]} // 

   Expand // FullSimplify

 ]

(*

2/3 (-1 + Sqrt[c + x^3 - c x^3] + 

   Sqrt[c] ArcTanh[(Sqrt[c] (x^3 - Sqrt[c + x^3 - c x^3]))/(c + x^3)])

*)


Applying the identity,
$$tanh ^{-1}left(frac{1}{z}right) - tanh ^{-1}(z)=frac{i pi}{2} ,,$$
changes ad1 by a constant, which results in another antiderivative:



ad3 = ad1 /. ArcTanh[z_] :> ArcTanh[1/z]

(*  2/3 Sqrt[c + x^3 - c x^3] - 2/3 Sqrt[c] ArcTanh[Sqrt[c]/Sqrt[c + (1 - c) x^3]] *)


Rewrite ad3 in terms of logarithms:



ad4 = TrigToExp@ad3 /. 

  a__ Log[u_] + b__ Log[v_] + w__ /; Times@a == -Times@b :> a Log@Simplify[u/v] + w

(*

  2/3 Sqrt[c + x^3 - c x^3] + 

   1/3 Sqrt[c] Log[(-Sqrt[c] + Sqrt[c + x^3 - c x^3]) /

     (Sqrt[c] + Sqrt[c + x^3 - c x^3])]

*)


Check:



D[{ad1, ad2, ad3, ad4}, x] // Simplify

(*

  {Sqrt[c + x^3 - c x^3]/x, Sqrt[c + x^3 - c x^3]/x,

   Sqrt[c + x^3 - c x^3]/x, Sqrt[c + x^3 - c x^3]/x}

*)


Addendum:
Here is a fifth, the real part of ad1:



ad5 = ComplexExpand[Re[ad1], TargetFunctions -> {Re, Im}] // 

 Simplify[#, 0 < c < 1 && x > (c/(1 - c))^(1/3)] &

(*

  1/6 (4 Sqrt[c + x^3 - c x^3] + 

     Sqrt[c] Log[(x^3 - c (-2 + x^3) - 2 Sqrt[c (c + x^3 - c x^3)])/c] -

     Sqrt[c] Log[(x^3 - c (-2 + x^3) + 2 Sqrt[c (c + x^3 - c x^3)])/c])

*)


Addendum 2:
Doing a line integral within the domain of interest will often give the desired result; however, this time it ran forever, until I turned off GenerateConditions. The antiderivative comes back in terms of HypergeometricPFQ which can be simplified with the aid of FunctionExpand.



Assuming[0 < c < 1 && x > 0 && t > 0, 

 ad6 = Simplify@ FunctionExpand@

    Integrate[Sqrt[c + (1 - c) t^3]/t, {t, 1, x}, 

     Assumptions -> 0 < c < 1 && x > 0, GenerateConditions -> False]

 ]

(*

  1/3 (-2 + 2 Sqrt[c + x^3 - c x^3] + 2 Sqrt[c] Log[1 + 1/Sqrt[c]] + 

     3 Sqrt[c] Log[x] - 2 Sqrt[c] Log[1 + Sqrt[1 + (-1 + 1/c) x^3]])

*)



(* Check *)

Simplify[D[ad6, x] - Sqrt[c + x^3 - c x^3]/x, 0 < c < 1 && x > 0]





share|improve this answer











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    $intfrac{sqrt{C + (1 - C) x^3}}{x}dx=intfrac{left(sqrt{C + (1 - C) x^3}right)x^2}{x^3}dx$



    Let $C + (1 - C) x^3=u^2$



    $3(1 - C) x^2dx=2udu$



    $frac{2}{3}intfrac{u^2du}{u^2-C}du$



    $frac{2}{3}intleft(1+frac{c}{u^2-C}right)du$



    $frac{2}{3}left(u+Cintfrac{du}{u^2-sqrt{C}^2}right)$



    $frac{2}{3}left(u-Cfrac{Arctanh(frac{u}{sqrt{C}})}{sqrt{C}}right)$



    Substituting back gives



    begin{equation}tag{2}
    frac{2}{3} sqrt{C + (1 - C), x^3} - frac{2}{3} sqrt{C} , mathrm{argtanh} Big( frac{sqrt{C + (1 - C) , x^3}}{sqrt{C}} Big).
    end{equation}



    FunctionDomain[ArcTanh[Sqrt[C + (1 - C) x^3]/Sqrt[C]], x]



    -1 < Sqrt[C + x^3 - C x^3]/Sqrt[C] < 1 && -C - x^3 + C x^3 <= 0




    Or



     ArcTanh[Sqrt[C + (1 - C) x^3]/Sqrt[C]] // TrigToExp



    -(1/2) Log[1 - Sqrt[C + (1 - C) x^3]/Sqrt[C]] + 1/2 Log[1 + Sqrt[C + (1 - C) x^3]/Sqrt[C]]







    share|improve this answer











    $endgroup$













    • $begingroup$
      There's still a problem with the argument of arctanh.
      $endgroup$
      – Cham
      Jan 13 at 17:06










    • $begingroup$
      I think it's better to split the integral of $frac{1}{u^2 - C}$ in two parts : $frac{1}{u - sqrt{C}} - frac{1}{u + sqrt{C}}$, then integration gives a $ln{dots}$.
      $endgroup$
      – Cham
      Jan 13 at 17:09












    • $begingroup$
      They are equivalent..
      $endgroup$
      – Okkes Dulgerci
      Jan 13 at 17:12










    • $begingroup$
      My solution appears to be this: begin{equation} frac{2 u}{3 sqrt{C}} + frac{1}{3} , ln{Big(frac{u - sqrt{C}}{u + sqrt{C}}Big)},end{equation} which is good since $u ge sqrt{C}$.
      $endgroup$
      – Cham
      Jan 13 at 17:12


















    4












    $begingroup$

    $intfrac{sqrt{C + (1 - C) x^3}}{x}dx=intfrac{left(sqrt{C + (1 - C) x^3}right)x^2}{x^3}dx$



    Let $C + (1 - C) x^3=u^2$



    $3(1 - C) x^2dx=2udu$



    $frac{2}{3}intfrac{u^2du}{u^2-C}du$



    $frac{2}{3}intleft(1+frac{c}{u^2-C}right)du$



    $frac{2}{3}left(u+Cintfrac{du}{u^2-sqrt{C}^2}right)$



    $frac{2}{3}left(u-Cfrac{Arctanh(frac{u}{sqrt{C}})}{sqrt{C}}right)$



    Substituting back gives



    begin{equation}tag{2}
    frac{2}{3} sqrt{C + (1 - C), x^3} - frac{2}{3} sqrt{C} , mathrm{argtanh} Big( frac{sqrt{C + (1 - C) , x^3}}{sqrt{C}} Big).
    end{equation}



    FunctionDomain[ArcTanh[Sqrt[C + (1 - C) x^3]/Sqrt[C]], x]



    -1 < Sqrt[C + x^3 - C x^3]/Sqrt[C] < 1 && -C - x^3 + C x^3 <= 0




    Or



     ArcTanh[Sqrt[C + (1 - C) x^3]/Sqrt[C]] // TrigToExp



    -(1/2) Log[1 - Sqrt[C + (1 - C) x^3]/Sqrt[C]] + 1/2 Log[1 + Sqrt[C + (1 - C) x^3]/Sqrt[C]]







    share|improve this answer











    $endgroup$













    • $begingroup$
      There's still a problem with the argument of arctanh.
      $endgroup$
      – Cham
      Jan 13 at 17:06










    • $begingroup$
      I think it's better to split the integral of $frac{1}{u^2 - C}$ in two parts : $frac{1}{u - sqrt{C}} - frac{1}{u + sqrt{C}}$, then integration gives a $ln{dots}$.
      $endgroup$
      – Cham
      Jan 13 at 17:09












    • $begingroup$
      They are equivalent..
      $endgroup$
      – Okkes Dulgerci
      Jan 13 at 17:12










    • $begingroup$
      My solution appears to be this: begin{equation} frac{2 u}{3 sqrt{C}} + frac{1}{3} , ln{Big(frac{u - sqrt{C}}{u + sqrt{C}}Big)},end{equation} which is good since $u ge sqrt{C}$.
      $endgroup$
      – Cham
      Jan 13 at 17:12
















    4












    4








    4





    $begingroup$

    $intfrac{sqrt{C + (1 - C) x^3}}{x}dx=intfrac{left(sqrt{C + (1 - C) x^3}right)x^2}{x^3}dx$



    Let $C + (1 - C) x^3=u^2$



    $3(1 - C) x^2dx=2udu$



    $frac{2}{3}intfrac{u^2du}{u^2-C}du$



    $frac{2}{3}intleft(1+frac{c}{u^2-C}right)du$



    $frac{2}{3}left(u+Cintfrac{du}{u^2-sqrt{C}^2}right)$



    $frac{2}{3}left(u-Cfrac{Arctanh(frac{u}{sqrt{C}})}{sqrt{C}}right)$



    Substituting back gives



    begin{equation}tag{2}
    frac{2}{3} sqrt{C + (1 - C), x^3} - frac{2}{3} sqrt{C} , mathrm{argtanh} Big( frac{sqrt{C + (1 - C) , x^3}}{sqrt{C}} Big).
    end{equation}



    FunctionDomain[ArcTanh[Sqrt[C + (1 - C) x^3]/Sqrt[C]], x]



    -1 < Sqrt[C + x^3 - C x^3]/Sqrt[C] < 1 && -C - x^3 + C x^3 <= 0




    Or



     ArcTanh[Sqrt[C + (1 - C) x^3]/Sqrt[C]] // TrigToExp



    -(1/2) Log[1 - Sqrt[C + (1 - C) x^3]/Sqrt[C]] + 1/2 Log[1 + Sqrt[C + (1 - C) x^3]/Sqrt[C]]







    share|improve this answer











    $endgroup$



    $intfrac{sqrt{C + (1 - C) x^3}}{x}dx=intfrac{left(sqrt{C + (1 - C) x^3}right)x^2}{x^3}dx$



    Let $C + (1 - C) x^3=u^2$



    $3(1 - C) x^2dx=2udu$



    $frac{2}{3}intfrac{u^2du}{u^2-C}du$



    $frac{2}{3}intleft(1+frac{c}{u^2-C}right)du$



    $frac{2}{3}left(u+Cintfrac{du}{u^2-sqrt{C}^2}right)$



    $frac{2}{3}left(u-Cfrac{Arctanh(frac{u}{sqrt{C}})}{sqrt{C}}right)$



    Substituting back gives



    begin{equation}tag{2}
    frac{2}{3} sqrt{C + (1 - C), x^3} - frac{2}{3} sqrt{C} , mathrm{argtanh} Big( frac{sqrt{C + (1 - C) , x^3}}{sqrt{C}} Big).
    end{equation}



    FunctionDomain[ArcTanh[Sqrt[C + (1 - C) x^3]/Sqrt[C]], x]



    -1 < Sqrt[C + x^3 - C x^3]/Sqrt[C] < 1 && -C - x^3 + C x^3 <= 0




    Or



     ArcTanh[Sqrt[C + (1 - C) x^3]/Sqrt[C]] // TrigToExp



    -(1/2) Log[1 - Sqrt[C + (1 - C) x^3]/Sqrt[C]] + 1/2 Log[1 + Sqrt[C + (1 - C) x^3]/Sqrt[C]]








    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Jan 13 at 17:10

























    answered Jan 13 at 16:47









    Okkes DulgerciOkkes Dulgerci

    4,9621817




    4,9621817












    • $begingroup$
      There's still a problem with the argument of arctanh.
      $endgroup$
      – Cham
      Jan 13 at 17:06










    • $begingroup$
      I think it's better to split the integral of $frac{1}{u^2 - C}$ in two parts : $frac{1}{u - sqrt{C}} - frac{1}{u + sqrt{C}}$, then integration gives a $ln{dots}$.
      $endgroup$
      – Cham
      Jan 13 at 17:09












    • $begingroup$
      They are equivalent..
      $endgroup$
      – Okkes Dulgerci
      Jan 13 at 17:12










    • $begingroup$
      My solution appears to be this: begin{equation} frac{2 u}{3 sqrt{C}} + frac{1}{3} , ln{Big(frac{u - sqrt{C}}{u + sqrt{C}}Big)},end{equation} which is good since $u ge sqrt{C}$.
      $endgroup$
      – Cham
      Jan 13 at 17:12




















    • $begingroup$
      There's still a problem with the argument of arctanh.
      $endgroup$
      – Cham
      Jan 13 at 17:06










    • $begingroup$
      I think it's better to split the integral of $frac{1}{u^2 - C}$ in two parts : $frac{1}{u - sqrt{C}} - frac{1}{u + sqrt{C}}$, then integration gives a $ln{dots}$.
      $endgroup$
      – Cham
      Jan 13 at 17:09












    • $begingroup$
      They are equivalent..
      $endgroup$
      – Okkes Dulgerci
      Jan 13 at 17:12










    • $begingroup$
      My solution appears to be this: begin{equation} frac{2 u}{3 sqrt{C}} + frac{1}{3} , ln{Big(frac{u - sqrt{C}}{u + sqrt{C}}Big)},end{equation} which is good since $u ge sqrt{C}$.
      $endgroup$
      – Cham
      Jan 13 at 17:12


















    $begingroup$
    There's still a problem with the argument of arctanh.
    $endgroup$
    – Cham
    Jan 13 at 17:06




    $begingroup$
    There's still a problem with the argument of arctanh.
    $endgroup$
    – Cham
    Jan 13 at 17:06












    $begingroup$
    I think it's better to split the integral of $frac{1}{u^2 - C}$ in two parts : $frac{1}{u - sqrt{C}} - frac{1}{u + sqrt{C}}$, then integration gives a $ln{dots}$.
    $endgroup$
    – Cham
    Jan 13 at 17:09






    $begingroup$
    I think it's better to split the integral of $frac{1}{u^2 - C}$ in two parts : $frac{1}{u - sqrt{C}} - frac{1}{u + sqrt{C}}$, then integration gives a $ln{dots}$.
    $endgroup$
    – Cham
    Jan 13 at 17:09














    $begingroup$
    They are equivalent..
    $endgroup$
    – Okkes Dulgerci
    Jan 13 at 17:12




    $begingroup$
    They are equivalent..
    $endgroup$
    – Okkes Dulgerci
    Jan 13 at 17:12












    $begingroup$
    My solution appears to be this: begin{equation} frac{2 u}{3 sqrt{C}} + frac{1}{3} , ln{Big(frac{u - sqrt{C}}{u + sqrt{C}}Big)},end{equation} which is good since $u ge sqrt{C}$.
    $endgroup$
    – Cham
    Jan 13 at 17:12






    $begingroup$
    My solution appears to be this: begin{equation} frac{2 u}{3 sqrt{C}} + frac{1}{3} , ln{Big(frac{u - sqrt{C}}{u + sqrt{C}}Big)},end{equation} which is good since $u ge sqrt{C}$.
    $endgroup$
    – Cham
    Jan 13 at 17:12













    5












    $begingroup$

    Here are four five six forms for an antiderivative, the first ad1 being the same as the OP's and complex-valued over the intended domain. The other three four five are real-valued.



    ad1 = Integrate[Sqrt[c + (1 - c) x^3]/x, x, Assumptions -> 0 < c < 1]
    
(*  2/3 Sqrt[c + x^3 - c x^3] - 2/3 Sqrt[c] ArcTanh[Sqrt[c + (1 - c) x^3]/Sqrt[c]]  *)


    The imaginary part of ad is constant so we can subtract it off (which is the point of @JimB's comment under the OP):



    ad2 = ad1 /. {{x -> 1}, {x -> x}} // Differences // First;
    

    
Assuming[0 < c < 1 && x > 0,
    
 TrigToExp@ad2 /. {Log[z_] /; Simplify[z < 0] :> Log[-1] + Log[-z]} // 
    
   Expand // FullSimplify
    
 ]
    
(*
    
2/3 (-1 + Sqrt[c + x^3 - c x^3] + 
    
   Sqrt[c] ArcTanh[(Sqrt[c] (x^3 - Sqrt[c + x^3 - c x^3]))/(c + x^3)])
    
*)


    Applying the identity,
    $$tanh ^{-1}left(frac{1}{z}right) - tanh ^{-1}(z)=frac{i pi}{2} ,,$$
    changes ad1 by a constant, which results in another antiderivative:



    ad3 = ad1 /. ArcTanh[z_] :> ArcTanh[1/z]
    
(*  2/3 Sqrt[c + x^3 - c x^3] - 2/3 Sqrt[c] ArcTanh[Sqrt[c]/Sqrt[c + (1 - c) x^3]] *)


    Rewrite ad3 in terms of logarithms:



    ad4 = TrigToExp@ad3 /. 
    
  a__ Log[u_] + b__ Log[v_] + w__ /; Times@a == -Times@b :> a Log@Simplify[u/v] + w
    
(*
    
  2/3 Sqrt[c + x^3 - c x^3] + 
    
   1/3 Sqrt[c] Log[(-Sqrt[c] + Sqrt[c + x^3 - c x^3]) /
    
     (Sqrt[c] + Sqrt[c + x^3 - c x^3])]
    
*)


    Check:



    D[{ad1, ad2, ad3, ad4}, x] // Simplify
    
(*
    
  {Sqrt[c + x^3 - c x^3]/x, Sqrt[c + x^3 - c x^3]/x,
    
   Sqrt[c + x^3 - c x^3]/x, Sqrt[c + x^3 - c x^3]/x}
    
*)


    Addendum:
    Here is a fifth, the real part of ad1:



    ad5 = ComplexExpand[Re[ad1], TargetFunctions -> {Re, Im}] // 
    
 Simplify[#, 0 < c < 1 && x > (c/(1 - c))^(1/3)] &
    
(*
    
  1/6 (4 Sqrt[c + x^3 - c x^3] + 
    
     Sqrt[c] Log[(x^3 - c (-2 + x^3) - 2 Sqrt[c (c + x^3 - c x^3)])/c] -
    
     Sqrt[c] Log[(x^3 - c (-2 + x^3) + 2 Sqrt[c (c + x^3 - c x^3)])/c])
    
*)


    Addendum 2:
    Doing a line integral within the domain of interest will often give the desired result; however, this time it ran forever, until I turned off GenerateConditions. The antiderivative comes back in terms of HypergeometricPFQ which can be simplified with the aid of FunctionExpand.



    Assuming[0 < c < 1 && x > 0 && t > 0, 
    
 ad6 = Simplify@ FunctionExpand@
    
    Integrate[Sqrt[c + (1 - c) t^3]/t, {t, 1, x}, 
    
     Assumptions -> 0 < c < 1 && x > 0, GenerateConditions -> False]
    
 ]
    
(*
    
  1/3 (-2 + 2 Sqrt[c + x^3 - c x^3] + 2 Sqrt[c] Log[1 + 1/Sqrt[c]] + 
    
     3 Sqrt[c] Log[x] - 2 Sqrt[c] Log[1 + Sqrt[1 + (-1 + 1/c) x^3]])
    
*)
    

    
(* Check *)
    
Simplify[D[ad6, x] - Sqrt[c + x^3 - c x^3]/x, 0 < c < 1 && x > 0]





    share|improve this answer











    $endgroup$


















      5












      $begingroup$

      Here are four five six forms for an antiderivative, the first ad1 being the same as the OP's and complex-valued over the intended domain. The other three four five are real-valued.



      ad1 = Integrate[Sqrt[c + (1 - c) x^3]/x, x, Assumptions -> 0 < c < 1]
      
(*  2/3 Sqrt[c + x^3 - c x^3] - 2/3 Sqrt[c] ArcTanh[Sqrt[c + (1 - c) x^3]/Sqrt[c]]  *)


      The imaginary part of ad is constant so we can subtract it off (which is the point of @JimB's comment under the OP):



      ad2 = ad1 /. {{x -> 1}, {x -> x}} // Differences // First;
      

      
Assuming[0 < c < 1 && x > 0,
      
 TrigToExp@ad2 /. {Log[z_] /; Simplify[z < 0] :> Log[-1] + Log[-z]} // 
      
   Expand // FullSimplify
      
 ]
      
(*
      
2/3 (-1 + Sqrt[c + x^3 - c x^3] + 
      
   Sqrt[c] ArcTanh[(Sqrt[c] (x^3 - Sqrt[c + x^3 - c x^3]))/(c + x^3)])
      
*)


      Applying the identity,
      $$tanh ^{-1}left(frac{1}{z}right) - tanh ^{-1}(z)=frac{i pi}{2} ,,$$
      changes ad1 by a constant, which results in another antiderivative:



      ad3 = ad1 /. ArcTanh[z_] :> ArcTanh[1/z]
      
(*  2/3 Sqrt[c + x^3 - c x^3] - 2/3 Sqrt[c] ArcTanh[Sqrt[c]/Sqrt[c + (1 - c) x^3]] *)


      Rewrite ad3 in terms of logarithms:



      ad4 = TrigToExp@ad3 /. 
      
  a__ Log[u_] + b__ Log[v_] + w__ /; Times@a == -Times@b :> a Log@Simplify[u/v] + w
      
(*
      
  2/3 Sqrt[c + x^3 - c x^3] + 
      
   1/3 Sqrt[c] Log[(-Sqrt[c] + Sqrt[c + x^3 - c x^3]) /
      
     (Sqrt[c] + Sqrt[c + x^3 - c x^3])]
      
*)


      Check:



      D[{ad1, ad2, ad3, ad4}, x] // Simplify
      
(*
      
  {Sqrt[c + x^3 - c x^3]/x, Sqrt[c + x^3 - c x^3]/x,
      
   Sqrt[c + x^3 - c x^3]/x, Sqrt[c + x^3 - c x^3]/x}
      
*)


      Addendum:
      Here is a fifth, the real part of ad1:



      ad5 = ComplexExpand[Re[ad1], TargetFunctions -> {Re, Im}] // 
      
 Simplify[#, 0 < c < 1 && x > (c/(1 - c))^(1/3)] &
      
(*
      
  1/6 (4 Sqrt[c + x^3 - c x^3] + 
      
     Sqrt[c] Log[(x^3 - c (-2 + x^3) - 2 Sqrt[c (c + x^3 - c x^3)])/c] -
      
     Sqrt[c] Log[(x^3 - c (-2 + x^3) + 2 Sqrt[c (c + x^3 - c x^3)])/c])
      
*)


      Addendum 2:
      Doing a line integral within the domain of interest will often give the desired result; however, this time it ran forever, until I turned off GenerateConditions. The antiderivative comes back in terms of HypergeometricPFQ which can be simplified with the aid of FunctionExpand.



      Assuming[0 < c < 1 && x > 0 && t > 0, 
      
 ad6 = Simplify@ FunctionExpand@
      
    Integrate[Sqrt[c + (1 - c) t^3]/t, {t, 1, x}, 
      
     Assumptions -> 0 < c < 1 && x > 0, GenerateConditions -> False]
      
 ]
      
(*
      
  1/3 (-2 + 2 Sqrt[c + x^3 - c x^3] + 2 Sqrt[c] Log[1 + 1/Sqrt[c]] + 
      
     3 Sqrt[c] Log[x] - 2 Sqrt[c] Log[1 + Sqrt[1 + (-1 + 1/c) x^3]])
      
*)
      

      
(* Check *)
      
Simplify[D[ad6, x] - Sqrt[c + x^3 - c x^3]/x, 0 < c < 1 && x > 0]





      share|improve this answer











      $endgroup$
















        5












        5








        5





        $begingroup$

        Here are four five six forms for an antiderivative, the first ad1 being the same as the OP's and complex-valued over the intended domain. The other three four five are real-valued.



        ad1 = Integrate[Sqrt[c + (1 - c) x^3]/x, x, Assumptions -> 0 < c < 1]
        
(*  2/3 Sqrt[c + x^3 - c x^3] - 2/3 Sqrt[c] ArcTanh[Sqrt[c + (1 - c) x^3]/Sqrt[c]]  *)


        The imaginary part of ad is constant so we can subtract it off (which is the point of @JimB's comment under the OP):



        ad2 = ad1 /. {{x -> 1}, {x -> x}} // Differences // First;
        

        
Assuming[0 < c < 1 && x > 0,
        
 TrigToExp@ad2 /. {Log[z_] /; Simplify[z < 0] :> Log[-1] + Log[-z]} // 
        
   Expand // FullSimplify
        
 ]
        
(*
        
2/3 (-1 + Sqrt[c + x^3 - c x^3] + 
        
   Sqrt[c] ArcTanh[(Sqrt[c] (x^3 - Sqrt[c + x^3 - c x^3]))/(c + x^3)])
        
*)


        Applying the identity,
        $$tanh ^{-1}left(frac{1}{z}right) - tanh ^{-1}(z)=frac{i pi}{2} ,,$$
        changes ad1 by a constant, which results in another antiderivative:



        ad3 = ad1 /. ArcTanh[z_] :> ArcTanh[1/z]
        
(*  2/3 Sqrt[c + x^3 - c x^3] - 2/3 Sqrt[c] ArcTanh[Sqrt[c]/Sqrt[c + (1 - c) x^3]] *)


        Rewrite ad3 in terms of logarithms:



        ad4 = TrigToExp@ad3 /. 
        
  a__ Log[u_] + b__ Log[v_] + w__ /; Times@a == -Times@b :> a Log@Simplify[u/v] + w
        
(*
        
  2/3 Sqrt[c + x^3 - c x^3] + 
        
   1/3 Sqrt[c] Log[(-Sqrt[c] + Sqrt[c + x^3 - c x^3]) /
        
     (Sqrt[c] + Sqrt[c + x^3 - c x^3])]
        
*)


        Check:



        D[{ad1, ad2, ad3, ad4}, x] // Simplify
        
(*
        
  {Sqrt[c + x^3 - c x^3]/x, Sqrt[c + x^3 - c x^3]/x,
        
   Sqrt[c + x^3 - c x^3]/x, Sqrt[c + x^3 - c x^3]/x}
        
*)


        Addendum:
        Here is a fifth, the real part of ad1:



        ad5 = ComplexExpand[Re[ad1], TargetFunctions -> {Re, Im}] // 
        
 Simplify[#, 0 < c < 1 && x > (c/(1 - c))^(1/3)] &
        
(*
        
  1/6 (4 Sqrt[c + x^3 - c x^3] + 
        
     Sqrt[c] Log[(x^3 - c (-2 + x^3) - 2 Sqrt[c (c + x^3 - c x^3)])/c] -
        
     Sqrt[c] Log[(x^3 - c (-2 + x^3) + 2 Sqrt[c (c + x^3 - c x^3)])/c])
        
*)


        Addendum 2:
        Doing a line integral within the domain of interest will often give the desired result; however, this time it ran forever, until I turned off GenerateConditions. The antiderivative comes back in terms of HypergeometricPFQ which can be simplified with the aid of FunctionExpand.



        Assuming[0 < c < 1 && x > 0 && t > 0, 
        
 ad6 = Simplify@ FunctionExpand@
        
    Integrate[Sqrt[c + (1 - c) t^3]/t, {t, 1, x}, 
        
     Assumptions -> 0 < c < 1 && x > 0, GenerateConditions -> False]
        
 ]
        
(*
        
  1/3 (-2 + 2 Sqrt[c + x^3 - c x^3] + 2 Sqrt[c] Log[1 + 1/Sqrt[c]] + 
        
     3 Sqrt[c] Log[x] - 2 Sqrt[c] Log[1 + Sqrt[1 + (-1 + 1/c) x^3]])
        
*)
        

        
(* Check *)
        
Simplify[D[ad6, x] - Sqrt[c + x^3 - c x^3]/x, 0 < c < 1 && x > 0]





        share|improve this answer











        $endgroup$



        Here are four five six forms for an antiderivative, the first ad1 being the same as the OP's and complex-valued over the intended domain. The other three four five are real-valued.



        ad1 = Integrate[Sqrt[c + (1 - c) x^3]/x, x, Assumptions -> 0 < c < 1]
        
(*  2/3 Sqrt[c + x^3 - c x^3] - 2/3 Sqrt[c] ArcTanh[Sqrt[c + (1 - c) x^3]/Sqrt[c]]  *)


        The imaginary part of ad is constant so we can subtract it off (which is the point of @JimB's comment under the OP):



        ad2 = ad1 /. {{x -> 1}, {x -> x}} // Differences // First;
        

        
Assuming[0 < c < 1 && x > 0,
        
 TrigToExp@ad2 /. {Log[z_] /; Simplify[z < 0] :> Log[-1] + Log[-z]} // 
        
   Expand // FullSimplify
        
 ]
        
(*
        
2/3 (-1 + Sqrt[c + x^3 - c x^3] + 
        
   Sqrt[c] ArcTanh[(Sqrt[c] (x^3 - Sqrt[c + x^3 - c x^3]))/(c + x^3)])
        
*)


        Applying the identity,
        $$tanh ^{-1}left(frac{1}{z}right) - tanh ^{-1}(z)=frac{i pi}{2} ,,$$
        changes ad1 by a constant, which results in another antiderivative:



        ad3 = ad1 /. ArcTanh[z_] :> ArcTanh[1/z]
        
(*  2/3 Sqrt[c + x^3 - c x^3] - 2/3 Sqrt[c] ArcTanh[Sqrt[c]/Sqrt[c + (1 - c) x^3]] *)


        Rewrite ad3 in terms of logarithms:



        ad4 = TrigToExp@ad3 /. 
        
  a__ Log[u_] + b__ Log[v_] + w__ /; Times@a == -Times@b :> a Log@Simplify[u/v] + w
        
(*
        
  2/3 Sqrt[c + x^3 - c x^3] + 
        
   1/3 Sqrt[c] Log[(-Sqrt[c] + Sqrt[c + x^3 - c x^3]) /
        
     (Sqrt[c] + Sqrt[c + x^3 - c x^3])]
        
*)


        Check:



        D[{ad1, ad2, ad3, ad4}, x] // Simplify
        
(*
        
  {Sqrt[c + x^3 - c x^3]/x, Sqrt[c + x^3 - c x^3]/x,
        
   Sqrt[c + x^3 - c x^3]/x, Sqrt[c + x^3 - c x^3]/x}
        
*)


        Addendum:
        Here is a fifth, the real part of ad1:



        ad5 = ComplexExpand[Re[ad1], TargetFunctions -> {Re, Im}] // 
        
 Simplify[#, 0 < c < 1 && x > (c/(1 - c))^(1/3)] &
        
(*
        
  1/6 (4 Sqrt[c + x^3 - c x^3] + 
        
     Sqrt[c] Log[(x^3 - c (-2 + x^3) - 2 Sqrt[c (c + x^3 - c x^3)])/c] -
        
     Sqrt[c] Log[(x^3 - c (-2 + x^3) + 2 Sqrt[c (c + x^3 - c x^3)])/c])
        
*)


        Addendum 2:
        Doing a line integral within the domain of interest will often give the desired result; however, this time it ran forever, until I turned off GenerateConditions. The antiderivative comes back in terms of HypergeometricPFQ which can be simplified with the aid of FunctionExpand.



        Assuming[0 < c < 1 && x > 0 && t > 0, 
        
 ad6 = Simplify@ FunctionExpand@
        
    Integrate[Sqrt[c + (1 - c) t^3]/t, {t, 1, x}, 
        
     Assumptions -> 0 < c < 1 && x > 0, GenerateConditions -> False]
        
 ]
        
(*
        
  1/3 (-2 + 2 Sqrt[c + x^3 - c x^3] + 2 Sqrt[c] Log[1 + 1/Sqrt[c]] + 
        
     3 Sqrt[c] Log[x] - 2 Sqrt[c] Log[1 + Sqrt[1 + (-1 + 1/c) x^3]])
        
*)
        

        
(* Check *)
        
Simplify[D[ad6, x] - Sqrt[c + x^3 - c x^3]/x, 0 < c < 1 && x > 0]






        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Jan 13 at 22:29

























        answered Jan 13 at 18:26









        Michael E2Michael E2

        147k12197473




        147k12197473















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            A Topological Invariant for $pi_3(U(n))$

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            3 3 $begingroup$ Recently, I saw a construction of topological invariant for $pi_3(U(n))$ with $ngeq 2$ : $$ N=frac{1}{24pi^2}int_{S^3} d^3x epsilon^{ijk} Tr[(U^{-1}partial_{x_i}U)(U^{-1}partial_{x_j}U)(U^{-1}partial_{x_k}U)] , $$ where $Uin U(n)$ depends on $boldsymbol{x}=(x_1,x_2,x_3)in S^3$ , $epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$ , and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why? Update 02/02/2019 I think I got an argument for $n=2$ . In this case, $U=e^{i varphi} q$ with $qin SU(2)$ . Due to the trace and the Levi-Civita symbol in $N$ , $varphi$ does not contribute to $N$ . As $Tr[q^{dagger}partial_i q]=0$ and $(q^{dagger}partial_i q)^{dagger}=-q^{dagger}partial_i q$ , $q^{dagger}partial_i q$ in geneeral has the form $q^{dagger}partia
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