Ramanujan log-trigonometric integrals

I discovered the following conjectured identity numerically while studying a family of related integrals.

Let’s set
R+:=2ππ/208x2+ln2cosx12+1212+12ln2cosxx2+ln2cosxdx,
R:=2ππ/2018x2+ln2cosx12+1212+12ln2cosxx2+ln2cosxdx.
We may numerically observe with at least 500 digits of precision that
R+R?=1R+?=4ln2R?=14ln2.

How can we prove it?

A version of this has been sent to Eric Weisstein, these integrals are on Mathworld as Ramanujan log-trigonometric integrals.

Answer

Here is an approach.

Theorem.
Let s be a real number such that 1<s<1. Then

π20cos(sarctan(xlncosx))(x2+ln2cosx)s2dx=π21lns2

Proof. First assume that 0<s<1. Then we may write
π20cos(sarctan(xlncosx))(x2+ln2cosx)s/2dx=1Γ(s)π20+0us1cos(ux)eulncosxdudx=1Γ(s)+0us1π20cosuxcos(ux)dxdu=1Γ(s)+0us1π2u+1du=π21Γ(s)+0us1euln2du=π21Γ(s)Γ(s)lns2=π21lns2
where we have used Fubini's theorem and the classic results (here and there)
+0us1cos(au)ebudu=Γ(s)cos(sarctan(ab))(a2+b2)s/2,((s)>0,b>0,a>0)π20cosuxcos(ux)dx=π2u+1,u>1.
We may extend identity (7) by analytic continuation to obtain (1).

Example 1. We have

π/20x2+ln2cosxlncosxdx=π22ln2

and

π/201x2+ln2cosxx2+ln2cosxlncosxdx=π2ln2

Proof. Let 0<x<π2 and set t:=arctan(xlncosx). Observe that 0<t<π2 and
cost=cos(arctan(xlncosx))=lncosxx2+ln2cosx,
cos(t2)=12+12cost
then put successively s=12, s=12 in (1) to obtain (10) and (11).

Example 2.

2ππ/208x2+ln2cosx12+1212+12ln2cosxx2+ln2cosxdx=4ln2

and

2ππ/2018x2+ln2cosx12+1212+12ln2cosxx2+ln2cosxdx=14ln2

Proof. Let 0<x<π2 and set t:=arctan(xlncosx). Observe that 0<t<π2 and
cost=cos(arctan(xlncosx))=ln2cosxx2+ln2cosx,
cos(t4)=12+1212+12cost, then put successively s=14, s=14 in (1) to obtain (12) and (13).

Example n.

Set

R+n:=2ππ/202nx2+ln2cosx12+1212++1212+12ln2cosxx2+ln2cosxdx

and

Rn:=2ππ/2012nx2+ln2cosx12+1212++1212+12ln2cosxx2+ln2cosxdx.

Then

R+n=2nln2

and

Rn=12nln2.

Proof. Let 0<x<π2 and set t:=arctan(xlncosx). Observe that 0<t<π2 and
cost=ln2cosxx2+ln2cosx,
cos(t2n)=12+1212+1212++1212+12cost, then put successively s=12n, s=12n, n1, in (1) to obtain (14) and (15).

Attribution
Source : Link , Question Author : Olivier Oloa , Answer Author : Venus

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