Approximating integral with small parameter

Question

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.

Answer 1

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

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

Answer 2

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]

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

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]

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

Answer 3

If you are only trying to find the convergent piece of the integral, you can set GenerateConditions to be False:

Integrate[((1-x) (2-x) x)/(1-x)^2,{x,0,1},GenerateConditions->False] gives $-1/2$.

This is not an isolated example, I have observed this behavior for many divergent integrals; setting GenerateConditions to be false seems to give the convergent part each case, convergent in the sense that the leading divergent piece is assumed to be cancelled by a regulator.

Another such an example is

Integrate[1/(x + a), {x, 0, \[Infinity]}, GenerateConditions -> False] Log[1/a] which matches the convergent piece of cut-off regularization: Series[Normal[Integrate[1/(x+a),{x,0,L},Assumptions->L>0]],{L,\[Infinity],0}] Log[L/a]+O[1/L]^1

I would appreciate if anyone with a deeper knowledge of this behaviour can enlighten us.