How can we prove ∫10x2/3(1−x)−1/31−x+x2dx=2π3√3?
Thought 1
It cannot be solved by using contour integration directly. If we replace −1/3 with −2/3 or 1/3 or something else, we can use contour integration directly to solve it.
Thought 2
I have tried substitution x=t3 and x=1−t. None of them worked. But I noticed that the form of 1−x+x2 does not change while applying x=1−t.
Thought 3
Recall the integral representation of 2F1 function, I was able to convert it into a formula with 2F1(2/3,1;4/3;eπi/3) involved. But I think it will only make the integral more "complex". Moreover, I prefer a elementary approach. (But I also appreciate hypergeometric approach)
Answer
The solution heavily exploits symmetry of the integrand.
Let I=∫10x2/3(1−x)−1/31−x+x2dx
Replace x by 1−x and sum up gives
2I=∫10x2/3(1−x)−1/3+(1−x)2/3x−1/31−x+x2dx=∫10x−1/3(1−x)−1/31−x+x2dx
Let ln1 be complex logarithm with branch cut at positive real axis, while ln2 be the one whose cut is at negative real axis. Denote
f(z)=23ln1(x)−13ln2(1−x)
Then f(z) is discontinuous along the positive axis, but have different jump in arg across intervals [0,1] and [1,∞).
Now integrate g(z)=ef(z)/(1−z+z2) using keyhole contour. Let γ1 be path slightly above [0,1], γ4 below. γ2 be path slightly above [1,∞), γ3 below. It is easily checked that
∫γ1g(z)dz=I∫γ4g(z)dz=Ie4πi/3
∫γ2g(z)dz=eπi/3∫∞1x2/3(x−1)−1/31−x+x2dx⏟J∫γ3g(z)dz=eπiJ
If we perform x↦1/x on J, we get ∫10x−1/3(1−x)−1/3/(1−x+x2)dx, thus J=2I by (1).
Therefore I(1−e4πi/3)+2I(eπi/3−eπi)=2πi×Sum of residues of g(z) at e±2πi/3
From which I believe you can work out the value of I.
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