Integral ∫10arctan2x√1−x2dx\int_0^1\frac{\arctan^2x}{\sqrt{1-x^2}}\mathrm dx

Is it possible to evaluate this integral in a closed form?
I=10arctan2x1x2dx
It also can be represented as
I=π/40ϕ2cosϕcos2ϕdϕ

Answer

Okay, finally I was able to prove it.

Step 0. Observations. In view of the following identity

π20arctan(rsinθ)dθ=2χ2(1+r21r),

Vladimir’s result suggests that there may exists a general formula connecting

I(r,s)=π20arctan(rsinθ)arctan(ssinθ)dθ

and the Legendre chi function χ2. Indeed, inspired by Vladimir’s result, I conjectured that

I(r,s)=πχ2(1+r21r1+s21s).

I succeeded in proving this identity, so I post a solution here.

Step 1. Proof of the identity (1). It is easy to check that the following identity holds:

arctan(ab)=1/badxa2+x2.

So it follows that

I(r,s)=1/r1/sπ20sin2θ(x2+sin2θ)(y2+sin2θ)dθdydx=1/r1/sπ201x2y2(x2x2+sin2θy2y2+sin2θ)dθdydx=π21/r1/s1x2y2(xx2+1yy2+1)dydx.

For the convenience of notation, we put

α=r2+11randβ=s2+11s.

Then it is easy to check that arsinh(1/r)=logα and likewise for s and β. Thus with the substitution xsinhx and ysinhy, we have

I(r,s)=π2logαlogβsinhxcoshysinhycoshxsinh2xsinh2ydydx.

Applying the substitution exx and eyy, it follows that

I(r,s)=πα0β0dydx1x2y2=πn=0(α0x2ndx)(β0y2ndx)=π0(αβ)2n+1(2n+1)2=πχ2(αβ)

as desired, proving the identity (1).


EDIT. I found a much simpler and intuitive proof of (1). We first observe that (1) is equivalent to the following identity

π20arctan(2rsinθ1r2)arctan(2ssinθ1s2)dθ=πχ2(rs).

Now we first observe that from the addition formula for the hyperbolic tangent, we obtain the following formula

artanhxartanhy=artanh(xy1xy)

which holds for sufficiently small x,y. Thus

arctan(2rsinθ1r2)=1iartanh(2irsinθ1r2)=artanh(reiθ)artanh(reiθ)i=2artanh(reiθ)=2n=0sin(2n+1)θ2n+1r2n+1.

We readily check this holds for any |r|<1. Therefore

π20arctan(2rsinθ1r2)arctan(2ssinθ1s2)dθ=4m=0n=0r2m+1s2n+1(2m+1)(2n+1)π20sin(2m+1)θsin(2n+1)θdθ=2m=0n=0r2m+1s2n+1(2m+1)(2n+1)π20{cos(2m2n)θcos(2m+2n+2)θ}dθ=πm=0n=0r2m+1s2n+1(2m+1)(2n+1)δm,n=πχ2(rs).

Attribution
Source : Link , Question Author : Zakharia Stanley , Answer Author : Sangchul Lee

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