74
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ My immediate (albeit likely overly optimistic) thought is that their product could be viewed as a double integral, and this "somehow" shown to be equal to unity. $\endgroup$ – Semiclassical Jul 31 '14 at 20:10
  • $\begingroup$ @semiclassical The double integral seems frightening :) $\endgroup$ – Olivier Oloa Jul 31 '14 at 20:17
  • $\begingroup$ Oh, yes. That somehow is a bit like suggesting that it would "somehow" be possible for me to climb Mount Everest---it may be true, but it's still not bloody likely! On a more topical note, it also vaguely reminds me of certain properties of 'functional determinants' in field theory. (But that doesn't help much either!) $\endgroup$ – Semiclassical Jul 31 '14 at 20:18
  • 5
    $\begingroup$ The identity $$\sqrt{(1/2)+(1/2)\cos\theta}=\sin(\theta/2)$$ suggests making the substitution $$\cos\theta=\sqrt{\log^2\cos x/(x^2+\log^2\cos x)}$$ $\endgroup$ – Gerry Myerson Aug 17 '14 at 1:29
  • 5
    $\begingroup$ The integrals can be rewritten a bit more compactly in the following way: $$R^+:=\frac{2}{\pi}\int_0^{\pi/2}\text{Re}\left [ (ix-\ln(\cos x))^{1/4} \right ]dx\\R^-:=\frac{2}{\pi}\int_0^{\pi/2}\text{Re}\left [ \left (-ix-\ln(\cos x)\right )^{-1/4} \right ]dx$$ $\endgroup$ – SDiv Sep 22 '14 at 23:04
40
+100
$\begingroup$

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 $(15)$.

$\endgroup$
  • $\begingroup$ Very brilliant approach & great answer, but how to prove equation (8) & (9)? The cited links don't contain the proofs. +1 anyway $\endgroup$ – Anastasiya-Romanova 秀 Sep 24 '14 at 9:19
  • 8
    $\begingroup$ Much more professional than my answer! I knew you were holding out on us. $\endgroup$ – SDiv Sep 24 '14 at 17:55
  • 2
    $\begingroup$ OK fine, I am able to prove equation (8) & (9) by myself $\endgroup$ – Anastasiya-Romanova 秀 Sep 25 '14 at 9:53
  • 1
    $\begingroup$ @Anastasiya-Romanova Because the positive square root of $\log^{2} (\cos x)$ for $0 \le x \le \frac{\pi}{2}$ is $-\log \cos x$. Perhaps $s$ needs to be further restricted. $\endgroup$ – Random Variable Sep 25 '14 at 16:47
  • 1
    $\begingroup$ Wow, this is very impressive, but the only thing I haven't understood yet is how did you get the equality $(2)$, may I have miss something in your reasonning or it is obvious? :) $\endgroup$ – Hexacoordinate-C Oct 29 '15 at 20:07
15
$\begingroup$

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: $$R_n^\pm=\frac{2}{\pi}\text{Re}\left \lbrace \int_0^{\pi/2}\ln(1+i\tan x)^{\pm 1/(2^n)}dx \right \rbrace.$$

$\endgroup$
  • 6
    $\begingroup$ Congratulations! They are some details to clarify, but you did it! "Mille bravos !" Thank you for your time! $\endgroup$ – Olivier Oloa Sep 23 '14 at 16:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.