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How to prove $$ \int_{0}^{\pi/2}\ln\left(\,x^{2} + \ln^{2}\left(\,\cos\left(\,x\,\right)\,\right) \,\right)\,{\rm d}x\ =\ \pi\ln\left(\,\ln\left(\, 2\,\right)\,\right) $$

I don't know how to answer it.

When I asked this integral to my brother, after less than half hours he said it has a nice closed-form involving $\pi$ and $\ln\left(2\right)$ but, as always, he didn't tell me the closed-form and how to obtain it ( I didn't believe him and I think he tried to mess around with me ).

I have also searched the similar question here but it looks like nothing is similar or related.

Could anyone here please help me to obtain the closed form of the integral preferably with elementary ways ( high school methods )?. Any help would be greatly appreciated. Thank you.

Edit:

He is being a little bit nice to me today, he said the closed form is $\pi\ln\ln2$ and it's numerically correct.

This is not a duplicate problem, I am looking for a proof without using complex analysis.

GNUSupporter 8964民主女神 地下教會
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Anastasiya-Romanova 秀
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  • I have no idea, but perhaps you can use the identity $\Re(\ln(x+iy))=\dfrac12\ln(x^2+y^2)$ where $x,y\in\mathbb R$. – Akiva Weinberger Aug 31 '14 at 19:51
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    The indefinite integral can be written as $$x\log\left(x^2+\log^2\cos x\right) - \int\dfrac{2x\left(x-\tan (x)\log\cos x\right)}{x^2+\log^2\cos x},dx$$ Note that if $f(x)$ is the denominator then the numerator is $xf'(x)$ – Alice Ryhl Aug 31 '14 at 20:10
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    This looks simpler $$\int_{0}^{\pi/2} \ln \left(x^{2} + \ln^{2}\cos x \right) , dx=2 \Re \left[\int_{0}^{\pi/2} \ln \ln \left( \frac{1 + e^{2ix}}{2} \right) , dx\right] $$ – Anastasiya-Romanova 秀 Aug 31 '14 at 20:16
  • @imranfat I got stuck trying to solve for $x$ in $t=x^2+\log^2\cos x$ – Alice Ryhl Aug 31 '14 at 20:19
  • @Darksonn But that first part you don't need to integrate that. I assume that's what you meant? That's plug in work, and since it is an improper integral, you may have to work with L'Hospital or something. For the second one, if the num is the derivative of the denom, that would become an ln term? – imranfat Aug 31 '14 at 20:24
  • Wolframalpha doesn't even return an exact value or give a closed form for the indefinite integral. So calculus level is pretty much impossible, and there is some definite chain pulling going on here. – zibadawa timmy Aug 31 '14 at 20:25
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    @zibadawatimmy While I don't want to dispute that generally, I have come across integrals that were doable with normal integration but the computersystems couldn't do it. If (and only IF) Darksonn's approach works, that should be the way...That would also explain the double ln in the answer. – imranfat Aug 31 '14 at 20:28
  • @imranfat True. And in any case I'm sure one of the resident integral masters will crack it and get many upvotes. – zibadawa timmy Aug 31 '14 at 20:30
  • @Darksonn That factor $x$ in the num creates a problem here... – imranfat Aug 31 '14 at 20:33
  • @imranfat If $f(x)=x^2+\log^2\cos x$ and $t=x^2+\log^2\cos x$ then the integral is equal to $$x\log(x^2+\log^2\cos x)-\int\dfrac{f^{-1}(t)}{t},dt$$ where $f^{-1}(x)$ is the inverse of $f(x)$ – Alice Ryhl Aug 31 '14 at 20:37
  • I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. – GNUSupporter 8964民主女神 地下教會 Mar 12 '18 at 16:53

3 Answers3

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Here is a real-analytic method.

We have

$$ \int_{0}^{\Large \frac{\pi}{2}} \ln \left(x^{2} + \ln^2\cos x\right) \, {\rm d}x=\pi\ln(\ln2) \tag1 $$

Proof. Let $s$ be a real number such that $-1<s<1$. One may use the following theorem (proved here) $$ \int_{0}^{\Large \frac{\pi}{2}} \frac{\cos \left( s \arctan \left(\frac{x}{-\ln \cos x}\right)\right)}{(x^2+\ln^2\! \cos x)^{s/2}} \mathrm{d}x = \frac{\pi}{2}\frac{1}{\ln^{s}\!2}. \tag2 $$

We are then allowed to differentiate both sides of $(2)$ $$ \begin{align} \partial_s \left. \left( \frac{\cos \left( s \arctan \left(\frac{x}{-\ln \cos x}\right)\right)}{(x^2+\ln^2\! \cos x)^{s/2}}\right) \right|_{s=0} &=-\frac 12 \ln \left(x^{2} + \ln^2\cos x\right) \\\\ \partial_s \left. \left( \frac{\pi}{2}\frac{1}{\ln^{s}\!2}\right) \right|_{s=0} &=-\frac{\pi}{2}\ln(\ln2) \end{align} $$ which gives the result $(1)$.

Larry
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Olivier Oloa
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16

As noted in the comments, the integral is: $$2\Re\int_0^{\pi/2} \ln\ln\left(\frac{1+e^{2ix}}{2}\right)\,dx=2\Re\int_0^{\pi/2} \ln\left(\ln\left(1+e^{2ix}\right)-\ln 2\right)\,dx$$ Consider $$f(x)=\ln(\ln(1+x)-\ln2)$$ Around $x=0$, the taylor expansion can be written as: $$f(x)=f(0)+f'(0)x+f''(0)\frac{x^2}{2!}+f'''(0)\frac{x^3}{3!}+....$$ Replace $x$ with $e^{2ix}$. Notice that integrating the powers of $e^{2ix}$ would result in either zero or a purely imaginary number and since the derivatives of $f(x)$ at $0$ are real, we need to consider only the constant term i.e $f(0)$. Since $f(0)=\ln(-\ln 2)=\ln\ln 2+i\pi$, hence, $$2\Re\int_0^{\pi/2} \ln\left(\ln\left(1+e^{2ix}\right)-\ln 2\right)\,dx=2\int_0^{\pi/2} \ln\ln 2\,dx=\boxed{\pi\ln\ln 2}$$

$\blacksquare$

Pranav Arora
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  • Very nice Pranav! This is brilliant answer!! d(ˆ▽ˆ)b – Anastasiya-Romanova 秀 Sep 03 '14 at 10:48
  • Summon @VladimirReshetnikov! （‐＾▽＾‐） – Anastasiya-Romanova 秀 Sep 03 '14 at 10:49
  • Wait!? I'm missing something here. You said $f(x)=\ln(\ln(1+e^{\large2ix})-\ln2)$, right? But, then $f(0)\to-\infty$. – Anastasiya-Romanova 秀 Sep 03 '14 at 11:41
  • Nope, look at f(x) again. – Pranav Arora Sep 03 '14 at 11:42
  • Does the series converge ?. I'm worry about it because $\left\vert,{\rm e}^{2{\rm i}x},\right\vert = 1$. – Felix Marin Sep 03 '14 at 21:47
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    Indeed, I agree with Mr. @FelixMarin. I think the correct form for Taylor expansion should be $$f(x)=f(x)\bigg|{x=0}+f'(x)\bigg|{x=0}x+ f''(x)\bigg|{x=0} \frac{x^2}{2!} + \cdots$$ Hence $$f(e^{2ix})=f(e^{2ix})\bigg|{x=0}+f'(e^{2ix})\bigg|{x=0}(e^{2ix}) + f''(e^{2ix})\bigg|{x=0} \frac{(e^{2ix})^2}{2!} + \cdots$$ – Anastasiya-Romanova 秀 Sep 04 '14 at 12:59
  • @PranavArora Thank you for the compliment. Concerning your answer, it would be nice to tell which determination you considered for $\log(x)$ and it would be nice to justify why you are allowed to switch $\int$ and $\sum$. – Olivier Oloa Sep 24 '14 at 17:33
  • @OlivierOloa: Sorry, I do not know what does "determination for $\log(x)$" mean. About switching $\int$ and $\sum$, I did not do that, I integrate term by term. I integrated the first few powers to see that they result in either zero or an imaginary number so I kept only the first term and took the real part. Can you please answer the questions raised by OP and Felix? I do not have much knowledge about convergence of series and expansions, thanks! – Pranav Arora Sep 24 '14 at 18:27
  • @PranavArora You said "I integrated the first few powers to see that they result in either zero or an imaginary number so I kept only the first term and took the real part." This is the point: is the integral of the rest of your series really tending to zero? I think yes it is, but I won't prove it now. Your reasoning is similar to the one I had given in my answer to your question about $\int_0^{\pi} \frac{x}{x^2+\ln^2(2\sin x)},dx,$ that's why I had dwelt with convergence issue there. Thanks! – Olivier Oloa Sep 24 '14 at 19:01
3

A solution using complex analysis is given here by sos440.

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