Regarding this problem, I conjectured that

I(r,s)=∫1−11x√1+x1−xlog((r−1)x2+sx+1(r−1)x2−sx+1)dx=4πarccot√2r+2√r2−s2s2−1.

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

∂I∂s(r,s)=∫∞0{2√y(r−s)y2+2(2−r)y+(r+s)+2√y(r+s)y2+2(2−r)y+(r−s)}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):=∫1−11x√1+x1−xlog(1+2sxsinα+(s2−cos2α)x21−2sxsinα+(s2−cos2α)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 x∈R.)

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+x21−2xsinα+x2)dxx,

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

**Answer**

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

I(r,s)=∫∞0dv4s(v2−1)(v4−(4r−6)v2+1)v8+4(2r−s2−1)v6+2(8r2−8r−4s2+3)v4+4(2r−s2−1)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+√r2−s22+√r+√r2−s22−1

and

θ=arctan√2(r+√r2−s2)s2−1

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

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