Multivalued functions and branch points: Evaluate the following integral

Question

I have this integral :

$$ {J}=\int_{-1}^{+1}\frac{(1-x)^{-\frac{2}{3}}(1+x)^{-\frac{1}{3}}}{4+x^2}\,dx$$

There are three branch points for this integral : $1,-i,+i$ and two pole points $-2i$ and $+2i$ which lie outside. So can we say that there are no pole points for this integral and how to solve it.

Answer 1

The trick, when dealing with tricky contours, is to make first the standard trick work. In our case the standard way would be to substitute, so that we see only polynomials, then get a result for them. Explicitly:

$$ \begin{aligned} J &=\int_{-1}^{+1} \frac{(1-x)^{-2/3}(1+x)^{-1/3}}{4+x^2}\,dx \\ &=\int_{-1}^{+1} \frac{\displaystyle\left(\frac{1+x}{1-x}\right)^{2/3}} {(1+x)(4+x^2)}\,dx \\ &\qquad\text{Substitution: }y^3=\frac{1+x}{1-x}\ ,\ x=\frac{y^3-1}{y^3+1}=1-\frac2{y^3+1}\ ,\ dx =\frac{2\cdot 3y^2}{(y^3+1)^2}\; dy\ , \\ &= \int_0^\infty \frac {y^2} {\displaystyle\frac{2y^3}{y^3+1}\cdot\frac {5y^6+6y^3+5}{(y^3+1)^2}} \cdot \frac{2\cdot 3y^2}{(y^3+1)^2}\; dy \\ &= 3\int_0^\infty \frac {y(y^3+1)} {5y^6+6y^3+5}\; dy \ . \end{aligned} $$ Now, the contour to be considered is completely clear:

and passing to the limit with $R\to\infty$, $\epsilon\to0$,

• the integral on the contour $C_1$ converges to $J$,
• the integral on $C_2$ converges to zero because the degree in the denominator is two more the one in the numerator, so we integrate an expression in $O(R^{-2})$ against a contour which is in $O(R)$, so the whole contribution is in $O(R^{-1}$,
• and the integral on the contour $C_3$ converges to $-e^{2\cdot2\pi i/3}J=u^2J$, since all $y^3$ occurrences become real, and the lonely factor $y$ in the numerator brings the rotation factor with it.

Here, $u=e^{2\pi i/3}$ is the primitive third root on unity in the second quadrant. We also need $v=\sqrt u= e^{2\pi i/6}$ in the sequel.

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Let us apply the Residue Theorem, and pass to the limits, getting thus: $$ \begin{aligned} J + 0 -u^2J &= 2\pi i\sum_{a\text{ residue inside }C_1\cup C_2\cup C_3} \operatorname{Res}_{z=a} \frac {3z(z^3+1)} {5z^6+6z^3+5} \\ &= 2\pi i\cdot \frac 3{20} \sum_a \operatorname{Res}_{z=a} z \left[ \frac{2+i}{z^3+\frac 15(3+4i)} + \frac{2-i}{z^3+\frac 15(3-4i)} \right] \ . \end{aligned} $$ Let now $a,b$ be values in the first quadrant with $a^3=\frac 15(-3-4i)$, $b^3=\frac 15(-3+4i)$. So $$ \begin{aligned} a &= v\cdot\underbrace{\sqrt[3]{\frac 15(3+4i)}}_{:=c}=vc\ ,\qquad v^3 = -1\ ,\\ b &= u\bar a =\frac ua =\frac v{\displaystyle\sqrt[3]{\frac 15(3+4i)}}=\frac vc \ . \end{aligned} $$ Then $a$ is residual point for the first fraction in the bracket, $b$ for the second one, we also denote by $u=\exp\frac {2\pi i}3$ the one primitive root of unity related to the contour, and the only involved residues are $$ \begin{aligned} \operatorname{Res}_{z=a} \frac{(2+i)z}{z^3+\frac 15(3+4i)} &= \operatorname{Res}_{z=a} \frac{(2+i)z}{z^3-a^3} \\ &= \operatorname{Res}_{z=a} \frac{(2+i)z}{(z-a)(z-ua)(z-u^2a)} \\ &= \operatorname{Value}_{z=a} \frac{(2+i)z}{(z-ua)(z-u^2a)} = \operatorname{Value}_{z=a} \frac{(2+i)z}{(z^2+az+a^2)} \\ &=\frac{2+i}{3a}\ , \\[2mm] \operatorname{Res}_{z=b} \frac{(2-i)z}{z^3+\frac 15(3-4i)} &=\operatorname{Res}_{z=b} \frac{(2-i)z}{z^3-b^3} =\dots \\ &=\frac{2-i}{3b}\ , \end{aligned} $$ Putting all together $$ \begin{aligned} J&= \frac 1{1-u^2}\cdot 2\pi i\cdot \frac 1{20}\cdot \left[\frac {2+i}{vc}+\frac {2-i}{v/c}\right] \\ &= 2\pi\cdot \frac i{v(1-u^2)} \cdot \frac 1{20}\cdot \left[\frac {2+i}{\sqrt[3]{\frac 15(3+4i)}} +\frac {2-i}{\sqrt[3]{\frac 15(3-4i)}}\right] \\ &= 2\pi\cdot \frac 1{\sqrt 3} \cdot \frac 1{20}\cdot 2\cdot\text{Real}\frac {2+i}{\sqrt[3]{\frac 15(3+4i)}} \ . \end{aligned} $$

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To obtain a proof in terms of the given function under the integral transform the contour first using $y\to y^3$, then a Möbius transformation in the complex plane.

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Computer check, we use pari/gp:

? f(x) = (1-x)^(-2/3) * (1+x)^(-1/3) / (x^2+4);
? J = intnum( x=-1, 1, f(x));
? v = ( 1+ sqrt(-3) ) / 2;
? u = v^2;
? c = ( (3+4*I)/5 )^(1/3)
%130 = 0.952608221822056413578700298090 + 0.304199894340908282509646316476*I
? 
? 2*Pi / sqrt(3) / 10 * real( (2+I)/c )
%131 = 0.801487589833185015198191025384
? J
%132 = 0.801487589833185015171411244041

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We cannot expect a total simplification, since the minimal polynomial of some of the numbers that appear is still complicated.

sage: s = (2+i) / ( (3+4*i)/5 )^(1/3)
sage: s.minpoly()
x^6 - 20*x^3 + 125
sage: K.<S> = NumberField(s.minpoly())
sage: s.minpoly().base_extend(K).factor()
(x - S) * (x + 1/25*S^5 - 4/5*S^2) * (x^2 + S*x + S^2) * (x^2 + (-1/25*S^5 + 4/5*S^2)*x - 1/5*S^4 + 4*S)
sage: 5/S
-1/25*S^5 + 4/5*S^2
sage: (S+5/S).minpoly()
x^3 - 15*x - 20

The best we can obtain is to calculate the roots of the last polynomial,

sage: _.roots(ring=RR, multiplicities=False)
[-2.80560283257759, -1.61322984339245, 4.41883267597004]

we then take the last root and...

sage: root1, root2, root3 = ( S + 5/S ).minpoly().roots(ring=RR, multiplicities=False)
sage: root3
4.41883267597004
sage: ( 2*pi / sqrt(3) / 20 * root3 ).n()
0.801487589833185
sage: numerical_integral( lambda x: 1 / (1-x)^(2/3) / (1+x)^(1/3) / (4+x^2), (-1,1) )[0]
0.801485689398172

Answer 2

OK, here is a solution that works with the integral directly, branch points and all. First, consider the complex integral

$$\oint_C dz \frac{\left ( z-1 \right )^{-2/3} \left ( z+1 \right )^{-1/3}}{z^2+4} $$

where $C$ is the following contour

The outer circle has radius $R$ and the inner arcs have radius $\epsilon$, and there are branch points at $z=\pm 1$. Note that because $C$ is a closed contour, the poles at $z=\pm i 2$ are inside the contour and we compute their residues.

(As an aside, I find the way people perform these integrations, using a "dogbone" contour and considering the residues at infinity, to be utterly confusing.)

Note that the contour $C$ is detoured around the branch points. The argument of $z$ just above the cut is $\pi$ while below the cut it is $-\pi$. Thus, the integrand is restricted to have arguments between $[-\pi,\pi]$. This is significant.

The complex integral above may now be written as follows:

$$e^{i \pi} \int_R^{1+\epsilon} dx \frac{(x+1)^{-2/3} e^{-i 2 \pi/3} (x-1)^{-1/3} e^{-i \pi/3}}{x^2+4} + \int_{-1+\epsilon}^{1-\epsilon} dx \frac{(1-x)^{-2/3} e^{-i 2 \pi/3} (1+x)^{-1/3}}{x^2+4} \\ + \int_{1-\epsilon}^{-1+\epsilon} dx \frac{(1-x)^{-2/3} e^{+i 2 \pi/3} (1+x)^{-1/3}}{x^2+4} + e^{-i \pi} \int_{1+\epsilon}^R dx \frac{(x+1)^{-2/3} e^{i 2 \pi/3} (x-1)^{-1/3} e^{i \pi/3}}{x^2+4} \\ + i R \int_{-\pi}^{\pi} d\theta \, e^{i \theta} \frac{\left ( R e^{i \theta}-1 \right )^{-2/3} \left ( R e^{i \theta}+1 \right )^{-1/3}}{R^2 e^{i 2 \theta} + 4} $$

A few notes: the factors of $e^{\pm i 2 \pi/3}$, etc. in the integrals come from conversion of minus signs to their proper arguments, and then taken to the given powers. Also note that I left out the integrals over the small arcs about the branch points, as they go to zero as $\epsilon \to 0$.

That said, it should be clear that the fifth integral over $\theta$ goes to zero as $R \to \infty$. Further, the first and fourth integrals cancel, as the exponents sum to $-1$. Thus, we are left with, as the contour integral:

$$\left (e^{-i 2 \pi/3} - e^{i 2 \pi/3} \right ) \int_{-1}^{1} dx \frac{(1-x)^{-2/3} (1+x)^{-1/3}}{x^2+4} $$

The residue theorem states that the contour integral is also equal to $i 2 \pi$ times the sum of the residues of the poles inside $C$, i.e., $z=\pm i 2$. Thus, we have

$$-i 2 \sin{\frac{2 \pi}{3}} \int_{-1}^{1} dx \frac{(1-x)^{-2/3} (1+x)^{-1/3}}{x^2+4} = i 2 \pi \frac1{i 4} \left [(-1+i 2)^{-2/3} (1+i 2)^{-1/3} - (-1-i 2)^{-2/3} (1-i 2)^{-1/3} \right ] $$

Are we done? No! We cannot leave the above in this form because it is not clear how to compute it. What are the arguments of each term? Well, as I stated above, the choice of branch we used is significant: all arguments must be between $-\pi$ and $\pi$. Under this constraint, we may write the RHS in polar form where it simplifies amazingly:

$$\frac{\pi}{2 \sqrt{5}} \left (e^{-i 2 \pi/3} e^{i \frac13 \arctan{2}} - e^{i 2 \pi/3} e^{-i \frac13 \arctan{2}} \right ) $$

Simplifying everything, we get our final result:

$$\int_{-1}^{1} dx \frac{(1-x)^{-2/3} (1+x)^{-1/3}}{x^2+4} = \frac{\pi}{\sqrt{15}} \sin{\left ( \frac{2 \pi}{3} - \frac13 \arctan{2} \right ) }$$

Numerically, this is about $0.801488$, which agrees with Dan's answer and Mathematica. Mathematica's analytical answer, however, is a mess of hypergeometrics, etc., so pen and paper still have the advantage.