Ramanujan log-trigonometric integrals

Question

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

Let's set $$ R^{+}:= \frac{2}{\pi}\int_{0}^{\pi/2}\sqrt[\normalsize{8}]{x^2 + \ln^2\!\cos x} \sqrt{ \frac{1}{2}+\frac{1}{2}\sqrt{ \frac{1}{2}+ \frac{1}{2} \sqrt{ \frac{\ln^{2}\!\cos x}{ x^2 + \ln^2\! \cos x}}}}\,\mathrm{d}x, \tag1 $$ $$ R^{-}:= \frac{2}{\pi}\int_{0}^{\pi/2}\frac{1}{\sqrt[\normalsize{8}]{x^2 + \ln^2\!\cos x}} \sqrt{ \frac{1}{2}+\frac{1}{2}\sqrt{ \frac{1}{2}+ \frac{1}{2} \sqrt{ \frac{\ln^{2}\!\cos x}{ x^2 + \ln^2\! \cos x}}}}\,\mathrm{d}x. \tag2 $$ We may numerically observe with at least 500 digits of precision that $$ \begin{align} & R^{+}R^{-} \stackrel{?}{=}1 \tag 3 \\\\ & R^{+} \stackrel{?}{=} \sqrt[\normalsize{4}]{\ln 2} \tag4 \\\\ & R^{-} \stackrel{?}{=}\frac{1}{\sqrt[\normalsize{4}]{\ln 2}}. \tag5 \end{align} $$

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 1

Here is an approach.

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

\begin{equation}{\Large\int_{0}^{\!\Large \frac{\pi}{2}}} \frac{\cos \left(\! s \arctan \left(-\frac{x}{\ln \cos x}\right)\right)}{(x^2+\ln^2\! \cos x)^{\Large\frac{s}{2}}}\, \mathrm{d}x = \frac{\pi}{2}\frac{1}{\ln^{\Large s}\!2} \tag1\end{equation}

Proof. First assume that $0<s<1.$ Then we may write $$ \begin{align} \int_{0}^{\!\Large \frac{\pi}{2}} \frac{\cos \left(\! s \arctan \left(-\frac{x}{\ln \cos x}\right)\right)}{(x^2+\ln^2\! \cos x)^{s/2}} \mathrm{d}x & = \frac{1}{\Gamma(s)}\int_{0}^{\!\Large \frac{\pi}{2}}\!\!\int_{0}^{+\infty} u^{s-1} \cos (u x) \:e^{u \ln \cos x}\:\mathrm{d}u \:\mathrm{d}x \tag2\\\\ & = \frac{1}{\Gamma(s)}\int_{0}^{+\infty}u^{s-1}\!\!\int_{0}^{\!\Large \frac{\pi}{2}} \cos^u\! x \cos (u x)\:\mathrm{d}x \:\mathrm{d}u \tag3\\\\ & = \frac{1}{\Gamma(s)}\int_{0}^{+\infty}u^{s-1} \frac{\pi}{2^{u+1}}\:\mathrm{d}u \tag4\\\\ & = \frac{\pi}{2} \frac{1}{\Gamma(s)}\int_{0}^{+\infty}u^{s-1} e^{-u\ln 2}\:\mathrm{d}u \tag5\\\\ & = \frac{\pi}{2}\frac{1}{\Gamma(s)}\frac{\Gamma(s)}{\ln^{s}\!2} \tag6\\\\ & = \frac{\pi}{2}\frac{1}{\ln^{s}\!2} \tag7 \end{align} $$ where we have used Fubini's theorem and the classic results (here and there) $$ \begin{align} & \int_{0}^{+\infty} u^{s-1} \cos (a u) \:e^{-b u}\:\mathrm{d}u = \Gamma (s)\frac{\cos \left(\! s \arctan \left(\frac{a}{b}\right)\right)}{(a^2+b^2)^{s/2}}, \, \left(\Re(s)>0, b>0, a>0 \right) \tag8 \\ & \int_{0}^{\!\Large \frac{\pi}{2}} \cos^u\! x \cos (u x)\:\mathrm{d}x = \frac{\pi}{2^{u+1}}, \quad u>-1. \tag9 \end{align} $$ We may extend identity $(7)$ by analytic continuation to obtain $(1)$.

Example 1. We have

$$ \int_{0}^{\pi/2}\displaystyle \sqrt{\sqrt{x^2 + \ln^2\!\cos x}-\ln\! \cos x}\,\,\mathrm{d}x = \frac{\pi}{2}\sqrt{2\ln 2} \tag{10} $$

and

$$ \int_{0}^{\pi/2}\displaystyle \frac{1}{\sqrt{x^2 + \ln^2\!\cos x}} \sqrt{\sqrt{x^2 + \ln^2\!\cos x}-\ln\! \cos x}\,\,\mathrm{d}x = \frac{\pi}{\sqrt{2\ln 2}} \tag{11} $$

Proof. Let $0<x<\frac{\pi}{2}$ and set $t:=\arctan \left(-\frac{x}{\ln \cos x}\right)$. Observe that $0 < t < \frac{\pi}{2}$ and $$ \cos t=\cos \left(\!\arctan \left(-\frac{x}{\ln \cos x}\right)\right)= -\frac{\ln \cos x}{\sqrt{x^2 + \ln^2\!\cos x}}, $$ $$ \cos \left(\frac{t}{2}\right) = \sqrt{ \frac{1}{2}+ \frac{1}{2} \cos t} $$ then put successively $\displaystyle s=-\frac{1}{2}$, $\displaystyle s=\frac{1}{2}$ in $(1)$ to obtain $(10)$ and $(11)$.

Example 2.

$$ \frac{2}{\pi}\int_{0}^{\pi/2}\sqrt[\normalsize{8}]{x^2 + \ln^2\!\cos x} \sqrt{ \frac{1}{2}+\frac{1}{2}\sqrt{ \frac{1}{2}+ \frac{1}{2} \sqrt{ \frac{\ln^{2}\!\cos x}{ x^2 + \ln^2\! \cos x}}}}\,\mathrm{d}x = \sqrt[\normalsize{4}]{\ln 2} \tag{12} $$

and

$$ \frac{2}{\pi}\int_{0}^{\pi/2}\frac{1}{\sqrt[\normalsize{8}]{x^2 + \ln^2\!\cos x}} \sqrt{ \frac{1}{2}+\frac{1}{2}\sqrt{ \frac{1}{2}+ \frac{1}{2} \sqrt{ \frac{\ln^{2}\!\cos x}{ x^2 + \ln^2\! \cos x}}}}\,\mathrm{d}x=\frac{1}{\sqrt[\normalsize{4}]{\ln 2}} \tag{13} $$

Proof. Let $0<x<\frac{\pi}{2}$ and set $t:=\arctan \left(-\frac{x}{\ln \cos x}\right)$. Observe that $0 < t < \frac{\pi}{2}$ and $$ \cos t=\cos \left(\!\arctan \left(-\frac{x}{\ln \cos x}\right)\right)=\sqrt{ \frac{\ln^{2}\!\cos x}{x^2 + \ln^2\! \cos x}}, $$ $$ \cos \left(\frac{t}{4}\right) = \sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{ \frac{1}{2}+ \frac{1}{2} \cos t}}, $$ then put successively $\displaystyle s=-\frac{1}{4}$, $\displaystyle s=\frac{1}{4}$ in $(1)$ to obtain $(12)$ and $(13)$.

Example n.

Set

$$ R_n^{+}:=\frac{2}{\pi}\int_{0}^{\pi/2}\sqrt[\Large {2^n}]{x^2 + \ln^2\!\cos x} \sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\cdots+\frac{1}{2}\sqrt{ \frac{1}{2}+ \frac{1}{2}\sqrt{ \frac{\ln^{2}\!\cos x}{ x^2 + \ln^2\! \cos x}}}}}\,\mathrm{d}x $$

and

$$ R_n^{-}:=\frac{2}{\pi}\int_{0}^{\pi/2}\frac{1}{\sqrt[\Large {2^n}]{x^2 + \ln^2\!\cos x}} \sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\cdots+\frac{1}{2}\sqrt{ \frac{1}{2}+ \frac{1}{2}\sqrt{ \frac{\ln^{2}\!\cos x}{ x^2 + \ln^2\! \cos x}}}}}\,\mathrm{d}x. $$

Then

$$ R_n^{+}= \sqrt[\Large {2^n}]{\ln 2} \tag{14} $$

and

$$ R_n^{-}= \frac{1}{\sqrt[\Large {2^n}]{\ln 2}} \tag{15}. $$

Proof. Let $0<x<\frac{\pi}{2}$ and set $t:=\arctan \left(-\frac{x}{\ln \cos x}\right)$. Observe that $0 < t < \frac{\pi}{2}$ and $$ \cos t=\sqrt{ \frac{\ln^{2}\!\cos x}{x^2 + \ln^2\! \cos x}}, $$ $$ \cos \left(\frac{t}{2^n}\right) = \sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\cdots+\frac{1}{2}\sqrt{ \frac{1}{2}+ \frac{1}{2} \cos t}}}}, $$ then put successively $\displaystyle s=-\frac{1}{2^n}$, $\displaystyle s=\frac{1}{2^n}$, $n\geq 1,$ in $(1)$ to obtain $(14)$ and $(15)$.

Answer 2

Oloa was right, I was on the right track last night (it makes me wonder if he has already solved it and just didn't tell us?). The final trick is to realize $(ix-\ln(\cos x))^{1/4}=\ln(1+i\tan(x))^{1/4}$ and then make the substitution $u=1+i\tan x$. This gives $R^+=\frac{2}{\pi}\text{Re}\lbrace\frac{i}{2}\Gamma(5/4) \text{PolyLog}(5/4, 2)\rbrace=(\ln 2)^{1/4}$. Details to follow in an edit in the next few minutes.

Edit: details follow So, this may not be the best way but at least it's a way: To get to the compact form I posted in the above comment, let $\theta(x)$ be such that $$\cos(\theta(x))=\frac{-\ln(\cos x)}{\sqrt{x^2+\ln^2(\cos x)}}.$$ Considering the right triangle with legs $x$ and $-\ln(\cos x)$ we notice that the hypotenuse is given by $h(x)=\sqrt{x^2+\ln^2(\cos x)}$ and further that $\sin(\theta(x))=\frac{x}{h(x)}$. This gives, using the correct identity $\sqrt{1/2+1/2\cos x}=\cos(x/2)$: $$R^+:=\frac{2}{\pi}\int_0^{\pi/2}\left(\frac{x}{\sin(\theta[x])}\right )^{1/4}\cos\left (\frac{\theta[x]}{4}\right )dx\\R^-:=\frac{2}{\pi}\int_0^{\pi/2}\left(\frac{x}{\sin(\theta[x])}\right )^{-1/4}\cos\left (\frac{\theta[x]}{4}\right )dx. $$

The trick is now to turn that $\theta/4$ into some kind of 4th root. We use the exponential definition of cosine: $\cos(\theta/4)=\frac{1}{2}\left ( e^{i\theta/4}+e^{-i\theta/4} \right )=\frac{1}{2}\left ( \left [\cos\theta+i\sin\theta \right ]^{1/4}+\left [\cos\theta-i\sin\theta \right ]^{1/4} \right )$. Then we have $$\left(\frac{x}{\sin(\theta[x])}\right )^{1/4}\cos\left (\frac{\theta[x]}{4}\right )=\frac{1}{2}\left ( [x\cot\theta+ix]^{1/4}+[x\cot\theta-ix]^{1/4} \right ).$$ Considering the triangle again this gives $$\left(\frac{x}{\sin(\theta[x])}\right )^{1/4}\cos\left (\frac{\theta[x]}{4}\right )=\frac{1}{2}\left ( [-\ln(\cos x)+ix]^{1/4}+[-\ln(\cos x)-ix]^{1/4} \right )\\=\text{Re}\left \lbrace [-\ln(\cos x)+ix]^{1/4} \right \rbrace. $$ Similar methods show $$\left(\frac{x}{\sin(\theta[x])}\right )^{-1/4}\cos\left (\frac{\theta[x]}{4}\right )=\text{Re}\left \lbrace [-\ln(\cos x)-ix]^{-1/4} \right \rbrace.$$ Now we use the addition property of logarithms and the exponental form of cosine again:$$-\ln(\cos x)+ix=-\ln(e^{ix}/2+e^{-ix}/2)+\ln(e^{ix})=\ln\left ( \frac{2e^{ix}}{e^{ix}+e^{-ix}} \right ) \\=\ln(1+i\tan x).$$ Similarly, $-\ln(\cos x)-ix=\ln(1-i\tan x).$ Okay, putting this together gives, using the fact that the integral over a real part is the real part of an integral: $$R^+=\frac{2}{\pi}\text{Re}\left \lbrace \int_0^{\pi/2}\ln(1+i\tan x)^{1/4}dx \right \rbrace \\ R^-=\frac{2}{\pi}\text{Re}\left \lbrace \int_0^{\pi/2}\ln(1-i\tan x)^{-1/4}dx \right \rbrace.$$ It seems that Mathematica still couldn't solve this, but with a simple substitution it works. For $R^+$ let $u=1+i\tan x$. Then $$ R^+=\frac{2}{\pi}\text{Re}\left \lbrace i\int_1^{1+i\infty}\frac{\ln(u)^{1/4}}{u^2-2u}du \right \rbrace\\=\frac{2}{\pi}\text{Re}\lbrace\frac{i}{2}\Gamma(5/4) \text{PolyLog}(5/4, 2)\rbrace\\=(\ln 2)^{1/4}.$$ I assume someone skilled in contour integration can do whatever Mathematica did to get this answer. For $R^-$ let $u=1-i\tan x$. Then $$ R^-=\frac{2}{\pi}\text{Re}\left \lbrace i\int_1^{1-i\infty}\frac{\ln(u)^{-1/4}}{u^2-2u}du \right \rbrace\\=\frac{2}{\pi}\text{Re}\lbrace\frac{i}{2}\Gamma(3/4) \text{PolyLog}(3/4, 2)\rbrace\\=(\ln 2)^{-1/4}.$$ I had to fudge a minus sign on this one, probably something with direction of contour integration or something. Anyway, conjecture confirmed. I assume similar methods can be used to confirm the nth order Ramanujan Log-Trigonometric Integral as mentioned by Oloa. Specifically, it seems we have $$R_n^+=\frac{2}{\pi}\text{Re}\left \lbrace \int_0^{\pi/2}\ln(1+i\tan x)^{1/(2^n)}dx \right \rbrace \\ R_n^-=\frac{2}{\pi}\text{Re}\left \lbrace \int_0^{\pi/2}\ln(1-i\tan x)^{-1/(2^n)}dx \right \rbrace.$$

Edit: It was silly for me to choose different representations for the real part. Better to chose the same one in which case we have: $$R_n^\pm=\frac{2}{\pi}\text{Re}\left \lbrace \int_0^{\pi/2}\ln(1+i\tan x)^{\pm 1/(2^n)}dx \right \rbrace.$$