Prove $\int_{0}^{\pi/2} \ln \left(x^{2} + (\ln\cos x)^2 \right) \, dx=\pi\ln\ln2 $

Question

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.

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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.

Answer 1

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 $(1)$.

Answer 2

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, $$2\Re\int_0^{\pi/2} \ln\left(\ln\left(1+e^{2ix}\right)-\ln 2\right)\,dx=2\int_0^{\pi/2} \ln\ln 2\,dx=\boxed{\pi\ln\ln 2}$$

$\blacksquare$

Answer 3

A solution using complex analysis is given here by sos440.

Answer 4

Feynman’s trick

Using the fact that $$ \int_0^{\frac{\pi}{2}} \ln \left(x^2+\ln ^2(\cos x)\right) d x=I(1) $$ where $$I(a)=\int_0^{\frac{\pi}{2}} \ln \left(x^2+\ln ^2(a\cos x)\right) d x $$ for $0<a<2$, whose derivative w.r.t. $a$ is $$ I^{\prime}(a)=\int_0^{\frac{\pi}{2}} \frac{2 \ln (a \cos x)}{a\left(x^2+\ln ^2(a \cos x)\right)} d x . $$ Replacing $t$ by $-\ln(a\cos x)$ in $$ \int_0^{\infty} e^{-t y} \cos (x y) d y=\frac{t}{x^2+t^2}, $$ we can express the integral $I^{\prime}(a) $ as $$ \begin{aligned}I^{\prime}(a) & =-\frac{2}{a} \int_0^{\frac{\pi}{2}} \int_0^{\infty} e^{y \ln (a \cos x)} \cos (x y) d y d x \\ & =-\frac{2}{a} \int_0^{\infty} \int_0^{\frac{\pi}{2}}(a \cos x)^y \cos (x y) d x d y \\ & =-\frac{2}{a} \int_0^{\infty} a^y \int_0^{\frac{\pi}{2}} \cos ^y x \cos (x y) d x d y \end{aligned} $$ By the post,

$$\int_0^{\frac{\pi}{2}} \cos ^y x \cos (x y) dx=\frac{\pi}{2^{y+1}} $$

we have $$ I^{\prime}(a)=-\frac{2}{a} \int_0^{\infty} \frac{a^y \pi}{2^{y+1}} d y=-\frac{\pi}{a} \int_0^{\infty}\left(\frac{a}{2}\right)^y d y=-\frac{\pi}{a \ln\left(\frac{a}{2}\right)} $$

Integrating back w.r.t $a$ yields $$ \int_0^{\frac{\pi}{2}} \ln \left(x^2+\ln ^2(a\cos x)\right) d x =\pi \ln (\ln 2-\ln a)=\pi \ln \ln \left(\frac{2}{a}\right)+C $$ Now we are going to find the value of the constant $C$ by adding $-\int_0^{\frac{\pi}{2}} \ln \left(\ln ^2 a\right) d x=-\frac{\pi}{2} \ln \left(\ln ^2a \right) $ on both sides.

$$ \int_0^{\frac{\pi}{2}}\ln \left(\frac{x^2+\ln ^2(a \cos x)}{\ln ^2 a}\right) d x=\frac{\pi}{2} \ln \left(\ln^2 \frac{2}{a}\right)-\frac{\pi}{2} \ln \left(\ln ^2 a\right)+C $$

By L’Hospital Rule, we can show that $$ \lim _{a \rightarrow 0} \frac{x^2+\ln ^2(a \cos x)}{\ln ^2 a}=\lim _{a \rightarrow 0}\left(\frac{\ln \frac{2}{a}}{\ln a}\right)^2=1 \Rightarrow C=0 $$

Hence we can conclude that in general, $$\boxed{ \int_0^{\frac{\pi}{2}} \ln \left(x^2+\ln ^2(a\cos x)\right) d x=\pi \ln (\ln \frac 2a)} $$

In particular, $$\boxed{ \int_0^{\frac{\pi}{2}} \ln \left(x^2+\ln ^2(\cos x)\right) d x=I(1)=\pi \ln (\ln 2)} $$