I discovered the following conjectured identity numerically while studying a family of related integrals.
Let’s set
R+:=2π∫π/208√x2+ln2cosx√12+12√12+12√ln2cosxx2+ln2cosxdx,
R−:=2π∫π/2018√x2+ln2cosx√12+12√12+12√ln2cosxx2+ln2cosxdx.
We may numerically observe with at least 500 digits of precision that
R+R−?=1R+?=4√ln2R−?=14√ln2.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∫+∞0us−1cos(ux)eulncosxdudx=1Γ(s)∫+∞0us−1∫π20cosuxcos(ux)dxdu=1Γ(s)∫+∞0us−1π2u+1du=π21Γ(s)∫+∞0us−1e−uln2du=π21Γ(s)Γ(s)lns2=π21lns2
where we have used Fubini's theorem and the classic results (here and there)
∫+∞0us−1cos(au)e−budu=Γ(s)cos(sarctan(ab))(a2+b2)s/2,(ℜ(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/20√√x2+ln2cosx−lncosxdx=π2√2ln2
and
∫π/201√x2+ln2cosx√√x2+ln2cosx−lncosxdx=π√2ln2
Proof. Let 0<x<π2 and set t:=arctan(−xlncosx). Observe that 0<t<π2 and
cost=cos(arctan(−xlncosx))=−lncosx√x2+ln2cosx,
cos(t2)=√12+12cost
then put successively s=−12, s=12 in (1) to obtain (10) and (11).
Example 2.
2π∫π/208√x2+ln2cosx√12+12√12+12√ln2cosxx2+ln2cosxdx=4√ln2
and
2π∫π/2018√x2+ln2cosx√12+12√12+12√ln2cosxx2+ln2cosxdx=14√ln2
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+12√12+12cost, then put successively s=−14, s=14 in (1) to obtain (12) and (13).
Example n.
Set
R+n:=2π∫π/202n√x2+ln2cosx√12+12√12+⋯+12√12+12√ln2cosxx2+ln2cosxdx
and
R−n:=2π∫π/2012n√x2+ln2cosx√12+12√12+⋯+12√12+12√ln2cosxx2+ln2cosxdx.
Then
R+n=2n√ln2
and
R−n=12n√ln2.
Proof. Let 0<x<π2 and set t:=arctan(−xlncosx). Observe that 0<t<π2 and
cost=√ln2cosxx2+ln2cosx,
cos(t2n)=√12+12√12+12√12+⋯+12√12+12cost, then put successively s=−12n, s=12n, n≥1, in (1) to obtain (14) and (15).
Attribution
Source : Link , Question Author : Olivier Oloa , Answer Author : Venus