# Ramanujan log-trigonometric integrals

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

Let’s set

We may numerically observe with at least 500 digits of precision that

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.

Here is an approach.

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

Proof. First assume that $0 Then we may write

where we have used Fubini's theorem and the classic results (here and there)

We may extend identity $(7)$ by analytic continuation to obtain $(1)$.

Example 1. We have

and

Proof. Let $0 and set $t:=\arctan \left(-\frac{x}{\ln \cos x}\right)$. Observe that $0 < t < \frac{\pi}{2}$ and

then put successively $\displaystyle s=-\frac{1}{2}$, $\displaystyle s=\frac{1}{2}$ in $(1)$ to obtain $(10)$ and $(11)$.

Example 2.

and

Proof. Let $0 and set $t:=\arctan \left(-\frac{x}{\ln \cos x}\right)$. Observe that $0 < t < \frac{\pi}{2}$ and

then put successively $\displaystyle s=-\frac{1}{4}$, $\displaystyle s=\frac{1}{4}$ in $(1)$ to obtain $(12)$ and $(13)$.

Example n.

Set

and

Then

and

Proof. Let $0 and set $t:=\arctan \left(-\frac{x}{\ln \cos x}\right)$. Observe that $0 < t < \frac{\pi}{2}$ and

then put successively $\displaystyle s=-\frac{1}{2^n}$, $\displaystyle s=\frac{1}{2^n}$, $n\geq 1,$ in $(1)$ to obtain $(14)$ and $(15)$.