Approximating integral with small parameter Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?How to do symbolic definite integral without copy and paste the intermediate results?Forcing an integral to be solved in separate termsSeries approximation to integralComputation (or estimation) of an trigonometric Integral with a parameterHow to Mellin transform a complicated Log integrand?Evaluate an Exponential involving an Integral OperatorAnalytic result for integral?How to compute my integralapproximations and limitsTrouble with numerical evaluation of a four-fold integral
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Approximating integral with small parameter
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?How to do symbolic definite integral without copy and paste the intermediate results?Forcing an integral to be solved in separate termsSeries approximation to integralComputation (or estimation) of an trigonometric Integral with a parameterHow to Mellin transform a complicated Log integrand?Evaluate an Exponential involving an Integral OperatorAnalytic result for integral?How to compute my integralapproximations and limitsTrouble with numerical evaluation of a four-fold integral
$begingroup$
I want to approximately compute integral $$I =int_0^1 dx fracx(2-x)(1-x)(1-x)^2+mu x$$ 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, $int_0^1 dx fracx(2-x)(1-x)(1-x)^2$, is logarithmically divergent on the upper limit). Therefore, the whole effect of this term can be accounted by shifting the upper limit:
$$I approx int_0^1-mu dx fracx(2-x)(1-x)(1-x)^2$$
Now Mathematica can compute this integral easily
Integrate [(2 - x) (1 - x) x/(1 - x)^2, x, 0, 1 - A, Assumptions -> A > 0]
giving $I = -frac12(1+ln(mu^2)-mu^2)$, which can be approximated as $I approx -frac12(1+ln(mu^2))$
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
asked 3 hours ago
Kolya TerzievKolya Terziev
778
$endgroup$
add a comment |
$begingroup$
I want to approximately compute integral $$I =int_0^1 dx fracx(2-x)(1-x)(1-x)^2+mu x$$ 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, $int_0^1 dx fracx(2-x)(1-x)(1-x)^2$, is logarithmically divergent on the upper limit). Therefore, the whole effect of this term can be accounted by shifting the upper limit:
$$I approx int_0^1-mu dx fracx(2-x)(1-x)(1-x)^2$$
Now Mathematica can compute this integral easily
Integrate [(2 - x) (1 - x) x/(1 - x)^2, x, 0, 1 - A, Assumptions -> A > 0]
giving $I = -frac12(1+ln(mu^2)-mu^2)$, which can be approximated as $I approx -frac12(1+ln(mu^2))$
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
asked 3 hours ago
Kolya TerzievKolya Terziev
778
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
I want to approximately compute integral $$I =int_0^1 dx fracx(2-x)(1-x)(1-x)^2+mu x$$ 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, $int_0^1 dx fracx(2-x)(1-x)(1-x)^2$, is logarithmically divergent on the upper limit). Therefore, the whole effect of this term can be accounted by shifting the upper limit:
$$I approx int_0^1-mu dx fracx(2-x)(1-x)(1-x)^2$$
Now Mathematica can compute this integral easily
Integrate [(2 - x) (1 - x) x/(1 - x)^2, x, 0, 1 - A, Assumptions -> A > 0]
giving $I = -frac12(1+ln(mu^2)-mu^2)$, which can be approximated as $I approx -frac12(1+ln(mu^2))$
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
asked 3 hours ago
Kolya TerzievKolya Terziev
778
$endgroup$
I want to approximately compute integral $$I =int_0^1 dx fracx(2-x)(1-x)(1-x)^2+mu x$$ 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, $int_0^1 dx fracx(2-x)(1-x)(1-x)^2$, is logarithmically divergent on the upper limit). Therefore, the whole effect of this term can be accounted by shifting the upper limit:
$$I approx int_0^1-mu dx fracx(2-x)(1-x)(1-x)^2$$
Now Mathematica can compute this integral easily
Integrate [(2 - x) (1 - x) x/(1 - x)^2, x, 0, 1 - A, Assumptions -> A > 0]
giving $I = -frac12(1+ln(mu^2)-mu^2)$, which can be approximated as $I approx -frac12(1+ln(mu^2))$
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
asked 3 hours ago
Kolya TerzievKolya Terziev
778
asked 3 hours ago
Kolya TerzievKolya Terziev
778
edited 2 hours ago
Kolya Terziev
asked 3 hours ago
Kolya TerzievKolya Terziev
778
asked 3 hours ago
Kolya TerzievKolya Terziev
778
asked 3 hours ago
Kolya TerzievKolya Terziev
778
778
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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
answered 3 hours ago
Carl WollCarl Woll
75.7k3100197
$endgroup$
$begingroup$
Sorry, but this is a surrogate, not the true answer.
$endgroup$
– user64494
2 hours ago
add a comment |
$begingroup$
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]
$$ frac12 (-log (mu )-1)+fracpi sqrtmu 4+frac14 mu (-2 log (mu )-5)+frac2532 pi mu ^3/2+frac124 mu ^2 (12 log (mu )-19)+Oleft(mu ^5/2right)$$
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]
$$frac12 (-log (-mu )-1)+frac14 mu (-2 log (-mu )-5)+frac124 mu ^2 (12 log (-mu )-19)+Oleft(mu ^5/2right) $$
answered 2 hours ago
user64494user64494
3,65311122
$endgroup$
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
answered 3 hours ago
Carl WollCarl Woll
75.7k3100197
$endgroup$
$begingroup$
Sorry, but this is a surrogate, not the true answer.
$endgroup$
– user64494
2 hours ago
add a comment |
$begingroup$
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
answered 3 hours ago
Carl WollCarl Woll
75.7k3100197
$endgroup$
$begingroup$
Sorry, but this is a surrogate, not the true answer.
$endgroup$
– user64494
2 hours ago
add a comment |
$begingroup$
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
answered 3 hours ago
Carl WollCarl Woll
75.7k3100197
$endgroup$
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
answered 3 hours ago
Carl WollCarl Woll
75.7k3100197
answered 3 hours ago
Carl WollCarl Woll
75.7k3100197
answered 3 hours ago
Carl WollCarl Woll
75.7k3100197
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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]
$$ frac12 (-log (mu )-1)+fracpi sqrtmu 4+frac14 mu (-2 log (mu )-5)+frac2532 pi mu ^3/2+frac124 mu ^2 (12 log (mu )-19)+Oleft(mu ^5/2right)$$
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]
$$frac12 (-log (-mu )-1)+frac14 mu (-2 log (-mu )-5)+frac124 mu ^2 (12 log (-mu )-19)+Oleft(mu ^5/2right) $$
answered 2 hours ago
user64494user64494
3,65311122
$endgroup$
add a comment |
$begingroup$
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]
$$ frac12 (-log (mu )-1)+fracpi sqrtmu 4+frac14 mu (-2 log (mu )-5)+frac2532 pi mu ^3/2+frac124 mu ^2 (12 log (mu )-19)+Oleft(mu ^5/2right)$$
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]
$$frac12 (-log (-mu )-1)+frac14 mu (-2 log (-mu )-5)+frac124 mu ^2 (12 log (-mu )-19)+Oleft(mu ^5/2right) $$
answered 2 hours ago
user64494user64494
3,65311122
$endgroup$
add a comment |
$begingroup$
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]
$$ frac12 (-log (mu )-1)+fracpi sqrtmu 4+frac14 mu (-2 log (mu )-5)+frac2532 pi mu ^3/2+frac124 mu ^2 (12 log (mu )-19)+Oleft(mu ^5/2right)$$
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]
$$frac12 (-log (-mu )-1)+frac14 mu (-2 log (-mu )-5)+frac124 mu ^2 (12 log (-mu )-19)+Oleft(mu ^5/2right) $$
answered 2 hours ago
user64494user64494
3,65311122
$endgroup$
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]
$$ frac12 (-log (mu )-1)+fracpi sqrtmu 4+frac14 mu (-2 log (mu )-5)+frac2532 pi mu ^3/2+frac124 mu ^2 (12 log (mu )-19)+Oleft(mu ^5/2right)$$
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]
$$frac12 (-log (-mu )-1)+frac14 mu (-2 log (-mu )-5)+frac124 mu ^2 (12 log (-mu )-19)+Oleft(mu ^5/2right) $$
answered 2 hours ago
user64494user64494
3,65311122
edited 1 hour ago
answered 2 hours ago
user64494user64494
3,65311122
answered 2 hours ago
user64494user64494
3,65311122
answered 2 hours ago
user64494user64494
3,65311122
3,65311122
add a comment |
add a comment |
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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