Friday, December 27, 2019

calculus - Evaluating int10fracx2/3(1x)1/31x+x2dx





How can we prove 10x2/3(1x)1/31x+x2dx=2π33?




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=1t. None of them worked. But I noticed that the form of 1x+x2 does not change while applying x=1t.
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(1x)1/31x+x2dx
Replace x by 1x and sum up gives

2I=10x2/3(1x)1/3+(1x)2/3x1/31x+x2dx=10x1/3(1x)1/31x+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(1x)
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)/(1z+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=Iγ4g(z)dz=Ie4πi/3

γ2g(z)dz=eπi/31x2/3(x1)1/31x+x2dxJγ3g(z)dz=eπiJ



If we perform x1/x on J, we get 10x1/3(1x)1/3/(1x+x2)dx, thus J=2I by (1).



Therefore I(1e4πi/3)+2I(eπi/3eπi)=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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