Approximating integral with small parameterHow to do symbolic definite integral without copy and paste the...
Are there physical dangers to preparing a prepared piano?
What happens in the secondary winding if there's no spark plug connected?
How could Tony Stark make this in Endgame?
Two field separators (colon and space) in awk
Why did C use the -> operator instead of reusing the . operator?
How to have a sharp product image?
Which big number is bigger?
Minor Revision with suggestion of an alternative proof by reviewer
bldc motor, esc and battery draw, nominal vs peak
How to pronounce 'c++' in Spanish
How come there are so many candidates for the 2020 Democratic party presidential nomination?
Why did some of my point & shoot film photos come back with one third light white or orange?
"Whatever a Russian does, they end up making the Kalashnikov gun"? Are there any similar proverbs in English?
What happened to Captain America in Endgame?
Why was the Spitfire's elliptical wing almost uncopied by other aircraft of World War 2?
Why does nature favour the Laplacian?
Function pointer with named arguments?
Aligning equation numbers vertically
Rivers without rain
"Hidden" theta-term in Hamiltonian formulation of Yang-Mills theory
Converting a sprinkler system's 24V AC outputs to 3.3V DC logic inputs
Overlay of two functions leaves gaps
How can I practically buy stocks?
Why does Mind Blank stop the Feeblemind spell?
Approximating integral with small parameter
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 $$I =int_0^1 dx frac{x(2-x)(1-x)}{(1-x)^2+mu x}$$ 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
asked yesterday
Kolya TerzievKolya Terziev
779
$endgroup$
add a comment |
$begingroup$
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
asked yesterday
Kolya TerzievKolya Terziev
779
$endgroup$
add a comment |
$begingroup$
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
asked yesterday
Kolya TerzievKolya Terziev
779
$endgroup$
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
asked yesterday
Kolya TerzievKolya Terziev
779
asked yesterday
Kolya TerzievKolya Terziev
779
edited yesterday
Kolya Terziev
asked yesterday
Kolya TerzievKolya Terziev
779
asked yesterday
Kolya TerzievKolya Terziev
779
asked yesterday
Kolya TerzievKolya Terziev
779
779
add a comment |
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
Addendum
AsymptoticIntegrate also works when using $mu x$, but is much slower, and needs to use a higher order than 1 to get a correct answer:
asymp = AsymptoticIntegrate[(x(2-x)(1-x))/((1-x)^2+μ x), {x,0,1}, {μ,0,2}]; //AbsoluteTiming
asymp //TeXForm
{127.263, Null}
$-frac{(mu -1) left(mu ^2-4 mu right) tanh ^{-1}left(frac{mu -2}{sqrt{mu -4}
sqrt{mu }}right)}{2 sqrt{mu -4} sqrt{mu }}+frac{(mu -1) left(mu ^2-4 mu
right) tanh ^{-1}left(frac{sqrt{mu }}{sqrt{mu -4}}right)}{2 sqrt{mu -4}
sqrt{mu }}-mu +frac{1}{4} (mu -2) (mu -1) log (mu )+(mu -1) log (mu
)-frac{2 (mu -3) tanh ^{-1}left(frac{mu -2}{sqrt{mu -4} sqrt{mu
}}right)}{sqrt{frac{mu -4}{mu }}}-frac{(mu -3) (mu -2) tanh
^{-1}left(frac{mu -2}{sqrt{mu -4} sqrt{mu }}right)}{2 sqrt{frac{mu -4}{mu
}}}+frac{2 (mu -3) tanh ^{-1}left(frac{sqrt{mu }}{sqrt{mu
-4}}right)}{sqrt{frac{mu -4}{mu }}}+frac{sqrt{mu -4} (mu -3) sqrt{mu }
left(frac{log (mu )}{2}+frac{(mu -2) tanh ^{-1}left(frac{sqrt{mu
}}{sqrt{mu -4}}right)}{sqrt{mu -4} sqrt{mu }}right)}{2 sqrt{frac{mu
-4}{mu }}}-frac{1}{2}$
The Series expansion of asymp reproduces user64494's result:
Series[asymp, {μ, 0, 2}, Assumptions->μ>0] //TeXForm
$left(-frac{log (mu )}{2}-frac{1}{2}right)+frac{pi sqrt{mu }}{4}+mu
left(-frac{log (mu )}{2}-frac{5}{4}right)+frac{25}{32} pi mu ^{3/2}+mu ^2
left(frac{log (mu )}{2}-frac{19}{24}right)+Oleft(mu ^{5/2}right)$
answered yesterday
Carl WollCarl Woll
75.9k3100198
$endgroup$
$begingroup$
Maybe interesting to someone: using the full μx term here, leadsAsymptoticIntegrateto return the fully correct analytic answer, equivalent to the answer ofIntegratein this case!
$endgroup$
– Thies Heidecke
yesterday
$begingroup$
Sorry, but this is a surrogate, not the true answer. Please don't delete my comment as before.
$endgroup$
– user64494
yesterday
$begingroup$
BTW, the AsymptoticIntegrate command produces an incorrect result for the integral from the question. The report [CASE:4251479] was submitted by me.
$endgroup$
– user64494
yesterday
$begingroup$
@user64494 I did not delete your comment. AsymptoticIntegrate at lowest order produces an incorrect result, which can be fixed by raising the order. Of course, if Mathematica can evaluate the integral symbolically, then there is no reason to use AsymptoticIntegrate. Still, the OP asked for such a method.
$endgroup$
– Carl Woll
yesterday
$begingroup$
Unfortunately, the AsymptoticIntegrate command of order 2 without the assumption $mu>0$ or with the assumption $mu<0$ produces an incorrect answer.
$endgroup$
– user64494
yesterday
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]
$$ frac{1}{2} (-log (mu )-1)+frac{pi sqrt{mu }}{4}+frac{1}{4} mu (-2 log (mu )-5)+frac{25}{32} pi mu ^{3/2}+frac{1}{24} mu ^2 (12 log (mu )-19)+Oleft(mu ^{5/2}right)$$
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]
$$frac{1}{2} (-log (-mu )-1)+frac{1}{4} mu (-2 log (-mu )-5)+frac{1}{24} mu ^2 (12 log (-mu )-19)+Oleft(mu ^{5/2}right) $$
answered yesterday
user64494user64494
3,65311122
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "387"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f196985%2fapproximating-integral-with-small-parameter%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
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
Addendum
AsymptoticIntegrate also works when using $mu x$, but is much slower, and needs to use a higher order than 1 to get a correct answer:
asymp = AsymptoticIntegrate[(x(2-x)(1-x))/((1-x)^2+μ x), {x,0,1}, {μ,0,2}]; //AbsoluteTiming
asymp //TeXForm
{127.263, Null}
$-frac{(mu -1) left(mu ^2-4 mu right) tanh ^{-1}left(frac{mu -2}{sqrt{mu -4}
sqrt{mu }}right)}{2 sqrt{mu -4} sqrt{mu }}+frac{(mu -1) left(mu ^2-4 mu
right) tanh ^{-1}left(frac{sqrt{mu }}{sqrt{mu -4}}right)}{2 sqrt{mu -4}
sqrt{mu }}-mu +frac{1}{4} (mu -2) (mu -1) log (mu )+(mu -1) log (mu
)-frac{2 (mu -3) tanh ^{-1}left(frac{mu -2}{sqrt{mu -4} sqrt{mu
}}right)}{sqrt{frac{mu -4}{mu }}}-frac{(mu -3) (mu -2) tanh
^{-1}left(frac{mu -2}{sqrt{mu -4} sqrt{mu }}right)}{2 sqrt{frac{mu -4}{mu
}}}+frac{2 (mu -3) tanh ^{-1}left(frac{sqrt{mu }}{sqrt{mu
-4}}right)}{sqrt{frac{mu -4}{mu }}}+frac{sqrt{mu -4} (mu -3) sqrt{mu }
left(frac{log (mu )}{2}+frac{(mu -2) tanh ^{-1}left(frac{sqrt{mu
}}{sqrt{mu -4}}right)}{sqrt{mu -4} sqrt{mu }}right)}{2 sqrt{frac{mu
-4}{mu }}}-frac{1}{2}$
The Series expansion of asymp reproduces user64494's result:
Series[asymp, {μ, 0, 2}, Assumptions->μ>0] //TeXForm
$left(-frac{log (mu )}{2}-frac{1}{2}right)+frac{pi sqrt{mu }}{4}+mu
left(-frac{log (mu )}{2}-frac{5}{4}right)+frac{25}{32} pi mu ^{3/2}+mu ^2
left(frac{log (mu )}{2}-frac{19}{24}right)+Oleft(mu ^{5/2}right)$
answered yesterday
Carl WollCarl Woll
75.9k3100198
$endgroup$
$begingroup$
Maybe interesting to someone: using the full μx term here, leadsAsymptoticIntegrateto return the fully correct analytic answer, equivalent to the answer ofIntegratein this case!
$endgroup$
– Thies Heidecke
yesterday
$begingroup$
Sorry, but this is a surrogate, not the true answer. Please don't delete my comment as before.
$endgroup$
– user64494
yesterday
$begingroup$
BTW, the AsymptoticIntegrate command produces an incorrect result for the integral from the question. The report [CASE:4251479] was submitted by me.
$endgroup$
– user64494
yesterday
$begingroup$
@user64494 I did not delete your comment. AsymptoticIntegrate at lowest order produces an incorrect result, which can be fixed by raising the order. Of course, if Mathematica can evaluate the integral symbolically, then there is no reason to use AsymptoticIntegrate. Still, the OP asked for such a method.
$endgroup$
– Carl Woll
yesterday
$begingroup$
Unfortunately, the AsymptoticIntegrate command of order 2 without the assumption $mu>0$ or with the assumption $mu<0$ produces an incorrect answer.
$endgroup$
– user64494
yesterday
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
Addendum
AsymptoticIntegrate also works when using $mu x$, but is much slower, and needs to use a higher order than 1 to get a correct answer:
asymp = AsymptoticIntegrate[(x(2-x)(1-x))/((1-x)^2+μ x), {x,0,1}, {μ,0,2}]; //AbsoluteTiming
asymp //TeXForm
{127.263, Null}
$-frac{(mu -1) left(mu ^2-4 mu right) tanh ^{-1}left(frac{mu -2}{sqrt{mu -4}
sqrt{mu }}right)}{2 sqrt{mu -4} sqrt{mu }}+frac{(mu -1) left(mu ^2-4 mu
right) tanh ^{-1}left(frac{sqrt{mu }}{sqrt{mu -4}}right)}{2 sqrt{mu -4}
sqrt{mu }}-mu +frac{1}{4} (mu -2) (mu -1) log (mu )+(mu -1) log (mu
)-frac{2 (mu -3) tanh ^{-1}left(frac{mu -2}{sqrt{mu -4} sqrt{mu
}}right)}{sqrt{frac{mu -4}{mu }}}-frac{(mu -3) (mu -2) tanh
^{-1}left(frac{mu -2}{sqrt{mu -4} sqrt{mu }}right)}{2 sqrt{frac{mu -4}{mu
}}}+frac{2 (mu -3) tanh ^{-1}left(frac{sqrt{mu }}{sqrt{mu
-4}}right)}{sqrt{frac{mu -4}{mu }}}+frac{sqrt{mu -4} (mu -3) sqrt{mu }
left(frac{log (mu )}{2}+frac{(mu -2) tanh ^{-1}left(frac{sqrt{mu
}}{sqrt{mu -4}}right)}{sqrt{mu -4} sqrt{mu }}right)}{2 sqrt{frac{mu
-4}{mu }}}-frac{1}{2}$
The Series expansion of asymp reproduces user64494's result:
Series[asymp, {μ, 0, 2}, Assumptions->μ>0] //TeXForm
$left(-frac{log (mu )}{2}-frac{1}{2}right)+frac{pi sqrt{mu }}{4}+mu
left(-frac{log (mu )}{2}-frac{5}{4}right)+frac{25}{32} pi mu ^{3/2}+mu ^2
left(frac{log (mu )}{2}-frac{19}{24}right)+Oleft(mu ^{5/2}right)$
answered yesterday
Carl WollCarl Woll
75.9k3100198
$endgroup$
$begingroup$
Maybe interesting to someone: using the full μx term here, leadsAsymptoticIntegrateto return the fully correct analytic answer, equivalent to the answer ofIntegratein this case!
$endgroup$
– Thies Heidecke
yesterday
$begingroup$
Sorry, but this is a surrogate, not the true answer. Please don't delete my comment as before.
$endgroup$
– user64494
yesterday
$begingroup$
BTW, the AsymptoticIntegrate command produces an incorrect result for the integral from the question. The report [CASE:4251479] was submitted by me.
$endgroup$
– user64494
yesterday
$begingroup$
@user64494 I did not delete your comment. AsymptoticIntegrate at lowest order produces an incorrect result, which can be fixed by raising the order. Of course, if Mathematica can evaluate the integral symbolically, then there is no reason to use AsymptoticIntegrate. Still, the OP asked for such a method.
$endgroup$
– Carl Woll
yesterday
$begingroup$
Unfortunately, the AsymptoticIntegrate command of order 2 without the assumption $mu>0$ or with the assumption $mu<0$ produces an incorrect answer.
$endgroup$
– user64494
yesterday
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
Addendum
AsymptoticIntegrate also works when using $mu x$, but is much slower, and needs to use a higher order than 1 to get a correct answer:
asymp = AsymptoticIntegrate[(x(2-x)(1-x))/((1-x)^2+μ x), {x,0,1}, {μ,0,2}]; //AbsoluteTiming
asymp //TeXForm
{127.263, Null}
$-frac{(mu -1) left(mu ^2-4 mu right) tanh ^{-1}left(frac{mu -2}{sqrt{mu -4}
sqrt{mu }}right)}{2 sqrt{mu -4} sqrt{mu }}+frac{(mu -1) left(mu ^2-4 mu
right) tanh ^{-1}left(frac{sqrt{mu }}{sqrt{mu -4}}right)}{2 sqrt{mu -4}
sqrt{mu }}-mu +frac{1}{4} (mu -2) (mu -1) log (mu )+(mu -1) log (mu
)-frac{2 (mu -3) tanh ^{-1}left(frac{mu -2}{sqrt{mu -4} sqrt{mu
}}right)}{sqrt{frac{mu -4}{mu }}}-frac{(mu -3) (mu -2) tanh
^{-1}left(frac{mu -2}{sqrt{mu -4} sqrt{mu }}right)}{2 sqrt{frac{mu -4}{mu
}}}+frac{2 (mu -3) tanh ^{-1}left(frac{sqrt{mu }}{sqrt{mu
-4}}right)}{sqrt{frac{mu -4}{mu }}}+frac{sqrt{mu -4} (mu -3) sqrt{mu }
left(frac{log (mu )}{2}+frac{(mu -2) tanh ^{-1}left(frac{sqrt{mu
}}{sqrt{mu -4}}right)}{sqrt{mu -4} sqrt{mu }}right)}{2 sqrt{frac{mu
-4}{mu }}}-frac{1}{2}$
The Series expansion of asymp reproduces user64494's result:
Series[asymp, {μ, 0, 2}, Assumptions->μ>0] //TeXForm
$left(-frac{log (mu )}{2}-frac{1}{2}right)+frac{pi sqrt{mu }}{4}+mu
left(-frac{log (mu )}{2}-frac{5}{4}right)+frac{25}{32} pi mu ^{3/2}+mu ^2
left(frac{log (mu )}{2}-frac{19}{24}right)+Oleft(mu ^{5/2}right)$
answered yesterday
Carl WollCarl Woll
75.9k3100198
$endgroup$
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
Addendum
AsymptoticIntegrate also works when using $mu x$, but is much slower, and needs to use a higher order than 1 to get a correct answer:
asymp = AsymptoticIntegrate[(x(2-x)(1-x))/((1-x)^2+μ x), {x,0,1}, {μ,0,2}]; //AbsoluteTiming
asymp //TeXForm
{127.263, Null}
$-frac{(mu -1) left(mu ^2-4 mu right) tanh ^{-1}left(frac{mu -2}{sqrt{mu -4}
sqrt{mu }}right)}{2 sqrt{mu -4} sqrt{mu }}+frac{(mu -1) left(mu ^2-4 mu
right) tanh ^{-1}left(frac{sqrt{mu }}{sqrt{mu -4}}right)}{2 sqrt{mu -4}
sqrt{mu }}-mu +frac{1}{4} (mu -2) (mu -1) log (mu )+(mu -1) log (mu
)-frac{2 (mu -3) tanh ^{-1}left(frac{mu -2}{sqrt{mu -4} sqrt{mu
}}right)}{sqrt{frac{mu -4}{mu }}}-frac{(mu -3) (mu -2) tanh
^{-1}left(frac{mu -2}{sqrt{mu -4} sqrt{mu }}right)}{2 sqrt{frac{mu -4}{mu
}}}+frac{2 (mu -3) tanh ^{-1}left(frac{sqrt{mu }}{sqrt{mu
-4}}right)}{sqrt{frac{mu -4}{mu }}}+frac{sqrt{mu -4} (mu -3) sqrt{mu }
left(frac{log (mu )}{2}+frac{(mu -2) tanh ^{-1}left(frac{sqrt{mu
}}{sqrt{mu -4}}right)}{sqrt{mu -4} sqrt{mu }}right)}{2 sqrt{frac{mu
-4}{mu }}}-frac{1}{2}$
The Series expansion of asymp reproduces user64494's result:
Series[asymp, {μ, 0, 2}, Assumptions->μ>0] //TeXForm
$left(-frac{log (mu )}{2}-frac{1}{2}right)+frac{pi sqrt{mu }}{4}+mu
left(-frac{log (mu )}{2}-frac{5}{4}right)+frac{25}{32} pi mu ^{3/2}+mu ^2
left(frac{log (mu )}{2}-frac{19}{24}right)+Oleft(mu ^{5/2}right)$
answered yesterday
Carl WollCarl Woll
75.9k3100198
edited yesterday
answered yesterday
Carl WollCarl Woll
75.9k3100198
answered yesterday
Carl WollCarl Woll
75.9k3100198
answered yesterday
Carl WollCarl Woll
75.9k3100198
75.9k3100198
$begingroup$
Maybe interesting to someone: using the full μx term here, leadsAsymptoticIntegrateto return the fully correct analytic answer, equivalent to the answer ofIntegratein this case!
$endgroup$
– Thies Heidecke
yesterday
$begingroup$
Sorry, but this is a surrogate, not the true answer. Please don't delete my comment as before.
$endgroup$
– user64494
yesterday
$begingroup$
BTW, the AsymptoticIntegrate command produces an incorrect result for the integral from the question. The report [CASE:4251479] was submitted by me.
$endgroup$
– user64494
yesterday
$begingroup$
@user64494 I did not delete your comment. AsymptoticIntegrate at lowest order produces an incorrect result, which can be fixed by raising the order. Of course, if Mathematica can evaluate the integral symbolically, then there is no reason to use AsymptoticIntegrate. Still, the OP asked for such a method.
$endgroup$
– Carl Woll
yesterday
$begingroup$
Unfortunately, the AsymptoticIntegrate command of order 2 without the assumption $mu>0$ or with the assumption $mu<0$ produces an incorrect answer.
$endgroup$
– user64494
yesterday
add a comment |
$begingroup$
Maybe interesting to someone: using the full μx term here, leadsAsymptoticIntegrateto return the fully correct analytic answer, equivalent to the answer ofIntegratein this case!
$endgroup$
– Thies Heidecke
yesterday
$begingroup$
Sorry, but this is a surrogate, not the true answer. Please don't delete my comment as before.
$endgroup$
– user64494
yesterday
$begingroup$
BTW, the AsymptoticIntegrate command produces an incorrect result for the integral from the question. The report [CASE:4251479] was submitted by me.
$endgroup$
– user64494
yesterday
$begingroup$
@user64494 I did not delete your comment. AsymptoticIntegrate at lowest order produces an incorrect result, which can be fixed by raising the order. Of course, if Mathematica can evaluate the integral symbolically, then there is no reason to use AsymptoticIntegrate. Still, the OP asked for such a method.
$endgroup$
– Carl Woll
yesterday
$begingroup$
Unfortunately, the AsymptoticIntegrate command of order 2 without the assumption $mu>0$ or with the assumption $mu<0$ produces an incorrect answer.
$endgroup$
– user64494
yesterday
$begingroup$
Maybe interesting to someone: using the full μx term here, leads
AsymptoticIntegrate to return the fully correct analytic answer, equivalent to the answer of Integrate in this case!$endgroup$
– Thies Heidecke
yesterday
$begingroup$
Maybe interesting to someone: using the full μx term here, leads
AsymptoticIntegrate to return the fully correct analytic answer, equivalent to the answer of Integrate in this case!$endgroup$
– Thies Heidecke
yesterday
$begingroup$
Sorry, but this is a surrogate, not the true answer. Please don't delete my comment as before.
$endgroup$
– user64494
yesterday
$begingroup$
Sorry, but this is a surrogate, not the true answer. Please don't delete my comment as before.
$endgroup$
– user64494
yesterday
$begingroup$
BTW, the AsymptoticIntegrate command produces an incorrect result for the integral from the question. The report [CASE:4251479] was submitted by me.
$endgroup$
– user64494
yesterday
$begingroup$
BTW, the AsymptoticIntegrate command produces an incorrect result for the integral from the question. The report [CASE:4251479] was submitted by me.
$endgroup$
– user64494
yesterday
$begingroup$
@user64494 I did not delete your comment. AsymptoticIntegrate at lowest order produces an incorrect result, which can be fixed by raising the order. Of course, if Mathematica can evaluate the integral symbolically, then there is no reason to use AsymptoticIntegrate. Still, the OP asked for such a method.
$endgroup$
– Carl Woll
yesterday
$begingroup$
@user64494 I did not delete your comment. AsymptoticIntegrate at lowest order produces an incorrect result, which can be fixed by raising the order. Of course, if Mathematica can evaluate the integral symbolically, then there is no reason to use AsymptoticIntegrate. Still, the OP asked for such a method.
$endgroup$
– Carl Woll
yesterday
$begingroup$
Unfortunately, the AsymptoticIntegrate command of order 2 without the assumption $mu>0$ or with the assumption $mu<0$ produces an incorrect answer.
$endgroup$
– user64494
yesterday
$begingroup$
Unfortunately, the AsymptoticIntegrate command of order 2 without the assumption $mu>0$ or with the assumption $mu<0$ produces an incorrect answer.
$endgroup$
– user64494
yesterday
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]
$$ frac{1}{2} (-log (mu )-1)+frac{pi sqrt{mu }}{4}+frac{1}{4} mu (-2 log (mu )-5)+frac{25}{32} pi mu ^{3/2}+frac{1}{24} mu ^2 (12 log (mu )-19)+Oleft(mu ^{5/2}right)$$
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]
$$frac{1}{2} (-log (-mu )-1)+frac{1}{4} mu (-2 log (-mu )-5)+frac{1}{24} mu ^2 (12 log (-mu )-19)+Oleft(mu ^{5/2}right) $$
answered yesterday
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]
$$ frac{1}{2} (-log (mu )-1)+frac{pi sqrt{mu }}{4}+frac{1}{4} mu (-2 log (mu )-5)+frac{25}{32} pi mu ^{3/2}+frac{1}{24} mu ^2 (12 log (mu )-19)+Oleft(mu ^{5/2}right)$$
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]
$$frac{1}{2} (-log (-mu )-1)+frac{1}{4} mu (-2 log (-mu )-5)+frac{1}{24} mu ^2 (12 log (-mu )-19)+Oleft(mu ^{5/2}right) $$
answered yesterday
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]
$$ frac{1}{2} (-log (mu )-1)+frac{pi sqrt{mu }}{4}+frac{1}{4} mu (-2 log (mu )-5)+frac{25}{32} pi mu ^{3/2}+frac{1}{24} mu ^2 (12 log (mu )-19)+Oleft(mu ^{5/2}right)$$
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]
$$frac{1}{2} (-log (-mu )-1)+frac{1}{4} mu (-2 log (-mu )-5)+frac{1}{24} mu ^2 (12 log (-mu )-19)+Oleft(mu ^{5/2}right) $$
answered yesterday
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]
$$ frac{1}{2} (-log (mu )-1)+frac{pi sqrt{mu }}{4}+frac{1}{4} mu (-2 log (mu )-5)+frac{25}{32} pi mu ^{3/2}+frac{1}{24} mu ^2 (12 log (mu )-19)+Oleft(mu ^{5/2}right)$$
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]
$$frac{1}{2} (-log (-mu )-1)+frac{1}{4} mu (-2 log (-mu )-5)+frac{1}{24} mu ^2 (12 log (-mu )-19)+Oleft(mu ^{5/2}right) $$
answered yesterday
user64494user64494
3,65311122
edited yesterday
answered yesterday
user64494user64494
3,65311122
answered yesterday
user64494user64494
3,65311122
answered yesterday
user64494user64494
3,65311122
3,65311122
add a comment |
add a comment |
Thanks for contributing an answer to Mathematica Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f196985%2fapproximating-integral-with-small-parameter%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
