Prove that \begin{equation} \int_0^{\pi/2}\ln^2(\cos x)\,dx=\frac{\pi}{2}\ln^2 2+\frac{\pi^3}{24} \end{equation}

I tried to use by parts method and ended with \begin{equation} \int \ln^2(\cos x)\,dx=x\ln^2(\cos x)+2\int x\ln(\cos x)\tan x\,dx \end{equation} The latter integral seems hard to evaluate. Could anyone here please help me to prove it preferably with elementary ways (high school methods)? Any help would be greatly appreciated. Thank you.


I also found this nice closed-form \begin{equation} -\int_0^{\pi/2}\ln^3(\cos x)\,dx=\frac{\pi}{2}\ln^3 2+\frac{\pi^3}{8}\ln 2 +\frac{3\pi}{4}\zeta(3) \end{equation} I hope someone here also help me to prove it. (>‿◠)✌

  • $\begingroup$ @GTXOC It looks a good idea, I'll try it. Thanks. (ô‿ô) $\endgroup$ – Anastasiya-Romanova 秀 Sep 1 '14 at 16:33
  • $\begingroup$ please put the Addendum in a separate question, I think is better. $\endgroup$ – Jack D'Aurizio Sep 1 '14 at 16:43
  • $\begingroup$ @JackD'Aurizio Aye Sir! (✿◠‿◠) $\endgroup$ – Anastasiya-Romanova 秀 Sep 1 '14 at 16:45
  • $\begingroup$ @V-Moy: See here. $\endgroup$ – Mhenni Benghorbal Sep 1 '14 at 22:43

Have a look at this other question. We have that $\log(2\cos x)$ has a nice Fourier series:

$$ \log(2\cos x) = \sum_{n=1}^{+\infty}\frac{(-1)^{n+1}}{n}\cos(2n x)\tag{1}$$ and since: $$ \int_{0}^{\pi/2}\cos(2nx)\cos(2mx)\,dx = \frac{\pi}{4}\delta_{m,n}\tag{2}$$ it follows that: $$ \int_{0}^{\pi/2}\log^2(2\cos x)\,dx = \frac{\pi}{4}\sum_{n=1}^{+\infty}\frac{1}{n^2}=\frac{\pi}{4}\zeta(2) = \frac{\pi^3}{24},\tag{3}$$ while $$\int_{0}^{\pi/2}\log(\cos x)\,dx = -\frac{\pi}{2}\log 2\tag{4}$$ is a well-known result. $(3)$ and $(4)$ proves your claim: $$\int_0^{\pi/2}\ln^2(\cos x)\,dx=\frac{\pi}{2}\ln^2 2+\frac{\pi^3}{24}.$$

Since $$\log(2\sin x)=-\sum_{n=1}^{+\infty}\frac{\cos(2nx)}{n}\tag{5}$$ and $$\int_{0}^{\pi/2}\cos(2n_1 x)\cos(2n_2 x)\cos(2n_3 x)\,dx = \frac{\pi}{8}\delta_{2\cdot\max n_i=(n_1+n_2+n_3)}\tag{6}$$ it follows that $$\int_{0}^{\pi/2}\log^3(2\sin x)\,dx = -\frac{3\pi}{4}\sum_{n=1}^{+\infty}\frac{H_{n-1}}{n^2}=-\frac{3\pi}{4}\zeta(3),\tag{7}$$ from which it is easy to prove your second claim, too.

  • 1
    $\begingroup$ Wow!? It looks easy after I study it but I never know the Fourier series, no wonder I can answer it. Since you already answered the second integral, I think no need to post it to another OP. (✿◠‿◠) $\endgroup$ – Anastasiya-Romanova 秀 Sep 1 '14 at 16:59
  • $\begingroup$ @Jack - I know I am a little late to the party, but I am having trouble interpreting what $\delta_{2\cdot\max n_i=(n_1+n_2+n_3)}$ you give in your log cube solution is. Any explanation would be greatly appreciated. $\endgroup$ – omegadot Oct 8 '19 at 7:49
  • $\begingroup$ @omegadot that delta is just a placeholder for the function which equals 1 if twice the maximum variable equals the sum of the variables, zero otherwise. $\endgroup$ – Jack D'Aurizio Oct 8 '19 at 11:04

Let's get all powers of $\ln(\cos(x))$ at once, using an exponential generating function:

$$ G(z) = \int_0^{\pi/2} \sum_{j=0}^\infty \dfrac{\ln^j(\cos(x)) z^j}{j!}\; dx = \int_0^{\pi/2} \exp(z \ln(\cos(x)))\; dx$$

Change variables: $\cos(x) = t^{1/2}$, and using the Beta function:

$$\eqalign{\dfrac{1}{2} &\int_0^1 \dfrac{t^{(z-1)/2}}{\sqrt{1-t}}\; dt = \dfrac{1}{2} B\left(\dfrac{1}{2}, \dfrac{z+1}{2}\right) = \dfrac{\Gamma(1/2)\Gamma((z+1)/2)}{2\; \Gamma(1+z/2)}\cr &= {\frac {\pi }{2}}-{\frac {\pi \,\ln \left( 2 \right) }{2}}z+ \left( {\frac {{\pi }^{3}}{48}}+{\frac {\pi \, \left( \ln \left( 2 \right) \right) ^{2}}{4}} \right) {z}^{2}-{\frac {\pi \, \left( 4\, \left( \ln \left( 2 \right) \right) ^{3}+{\pi }^{2}\ln \left( 2 \right) +6 \,\zeta \left( 3 \right) \right) }{48}}{z}^{3}+\ldots } $$

  • $\begingroup$ Wait!? I don't get it. Could you please elaborate a bit more Prof? Thanks. $\endgroup$ – Anastasiya-Romanova 秀 Sep 1 '14 at 16:48
  • $\begingroup$ OK, I added some details. $\endgroup$ – Robert Israel Sep 1 '14 at 17:01
  • $\begingroup$ I understand how to evaluate the right expression into a beta function but how to evaluate the left term? It doesn't ring a bell yet to me. $\endgroup$ – Anastasiya-Romanova 秀 Sep 1 '14 at 17:09
  • $\begingroup$ Thanks Prof. At least, your answer give me an idea to evaluate the integral (ô‿ô) $\endgroup$ – Anastasiya-Romanova 秀 Sep 1 '14 at 17:38
  • $\begingroup$ Aha! I understand now your solution Prof. Thanks you so much. Forgive me for noticing it so late. $\endgroup$ – Anastasiya-Romanova 秀 Sep 2 '14 at 9:00

I think I get an idea from Prof. Robert Israel's answer. Let $\cos x=\sqrt{t}$ and $dx=\dfrac{dt}{2\sqrt{t}\sqrt{1-t}}$, then \begin{align} \int_0^{\pi/2}\ln^2(\cos x)\,dx&=\frac{1}{8}\int_0^1\dfrac{\ln^2t}{\sqrt{t}\sqrt{1-t}}dt\\ &=\frac{1}{8}\lim_{x\to\frac{1}{2}}\lim_{y\to\frac{1}{2}}\frac{\partial^2}{\partial x^2}\int_0^1 t^{x-1}(1-t)^{y-1}dt\\ &=\frac{1}{8}\lim_{x\to\frac{1}{2}}\lim_{y\to\frac{1}{2}}\operatorname{B}(x,y)\left[(\Psi(x)-\Psi(x+y))^2+\Psi_1(x)-\Psi_1(x+y)\right] \end{align} where $\operatorname{B}(x,y)$ is beta function and $\Psi_k(z)$ is polygamma function. The same approach also works for $$ \int_0^{\pi/2}\ln^3(\cos x)\,dx $$ but for this one, we use third derivative of beta function.

  • $\begingroup$ There is a typo in the ${\rm d}x$ expression. It needs an additional minus sign. $\endgroup$ – Felix Marin Sep 4 '14 at 6:22

We are all familiar with the famous Wallis integrals, $W_n=\displaystyle\int_0^\frac\pi2\sin^{n-1}x~dx=\int_0^\frac\pi2\cos^{n-1}x~dx$

$=\dfrac12B\bigg(\dfrac12,~\dfrac n2\bigg)$, see beta function for more details. It follows then that our integral is nothing

other than $W''(1)$, which can be evaluated in terms of the digamma and trigamma functions, for

arguments $\dfrac12$ and $1$. The former can be expressed as $\psi_{_0}(k+1)=H_k-\gamma$, where $\gamma\approx\dfrac1{\sqrt3}~$ is the

Euler-Mascheroni constant, and $H_k=\displaystyle\int_0^1\frac{1-x^k}{1-x~~}dx=-\int_0^1\ln\Big(1-\sqrt[^k]x\Big)~dx$ is the harmonic

. Thus, that $H_1=1$ is self-evident, and that $H_\frac12=2~(1-\ln2)$ can be shown by a simple

substitution. The values are also found here. For the latter we have $~\psi_{_1}(x)=\displaystyle\sum_{k=0}^\infty\frac1{(k+x)^2}$, which

implies $\psi_{_1}(1)=\zeta(2)=\dfrac{\pi^2}6$, and $\psi_{_1}\bigg(\dfrac12\bigg)=4\bigg(1-\dfrac14\bigg)\zeta(2)=\dfrac{\pi^2}2$. See Basel problem for more


  • $\begingroup$ Thanks for your answer Mr. Lucian, we have a same approach. +1 (✿◠‿◠) $\endgroup$ – Anastasiya-Romanova 秀 Sep 1 '14 at 17:48

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