Integral ∫1−11x√1+x1−xlog((r−1)x2+sx+1(r−1)x2−sx+1)dx\int_{-1}^{1} \frac{1}{x}\sqrt{\frac{1+x}{1-x}} \log \left( \frac{(r-1)x^{2} + sx + 1}{(r-1)x^{2} – sx + 1} \right) \, \mathrm dx

Regarding this problem, I conjectured that

I(r,s)=111x1+x1xlog((r1)x2+sx+1(r1)x2sx+1)dx=4πarccot2r+2r2s2s21.

Though we may try the same technique as in the previous problem, now I’m curious if this generality leads us to a different (and possible a more elegant) proof.

Indeed, I observed that I(r,0)=0 and

Is(r,s)=0{2y(rs)y2+2(2r)y+(r+s)+2y(r+s)y2+2(2r)y+(rs)}dy,

which can be evaluated using standard contour integration technique. But simplifying the residue and integrating them seems still daunting.


EDIT. By applying a series of change of variables, I noticed that the problem is equivalent to prove that

˜I(α,s):=111x1+x1xlog(1+2sxsinα+(s2cos2α)x212sxsinα+(s2cos2α)x2)dx=4πα

for π2<α<π2 and s>1. (This is equivalent to the condition that the expression inside the logarithm is positive for all xR.)

Another simple observation. once you prove that ˜I(α,s) does not depend on the variable s for s>1, then by suitable limiting process it follows that

˜I(α,s)=log(1+2xsinα+x212xsinα+x2)dxx,

which (I guess) can be calculated by hand. The following graph may also help us understand the behavior of this integral.

enter image description here

Answer

So, following the procedure I outlined here, I get for the transformed integral:

I(r,s)=0dv4s(v21)(v4(4r6)v2+1)v8+4(2rs21)v6+2(8r28r4s2+3)v4+4(2rs21)v2+1logv

Note that this reduces to the integral in the original problem when r=3 and s=2. Then we see that the roots of the denominator satisfy the same symmetries as before, so we need only find one root of the form ρeiθ where

ρ=r+r2s22+r+r2s221

and

θ=arctan2(r+r2s2)s21

Using the same methodology I derived, I am able to confirm your conjecture.

Attribution
Source : Link , Question Author : Sangchul Lee , Answer Author : Community

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