A closed form for $\int_{0}^{\pi/2}\frac{\ln\cos x}{x}\mathrm{d}x$?

Question

The following integrals are classic, initiated by L. Euler.

\begin{align} \displaystyle \int_{0}^{\pi/2} x^3 \ln\cos x\:\mathrm{d}x & = -\frac{\pi^4}{64} \ln 2-\frac{3\pi^2}{16} \zeta(3)+\frac{93}{128} \zeta(5), \\ \int_{0}^{\pi/2} x^2 \ln\cos x\:\mathrm{d}x & = -\frac{\pi^3}{24} \ln 2-\frac{\pi}{4} \zeta(3), \\ \int_{0}^{\pi/2} x^1 \ln\cos x\:\mathrm{d}x & = -\frac{\pi^2}{8} \ln 2-\frac{7}{16} \zeta(3), \\ \int_{0}^{\pi/2} x^0 \ln\cos x\:\mathrm{d}x & = -\frac{\pi}{2}\ln 2. \end{align}

We may logically consider the case when the first factor of the integrand is $\displaystyle x^{-1} = \frac 1x $ leading to the following non classic convergent integral.

$$ \int_{0}^{\pi/2} \frac{\ln\cos x}{x}\:\mathrm{d}x \qquad (*)$$

I do not have a closed form for this integral.

My question is does someone have some references/results about $(*)$?

Answer 1

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\color{#66f}{\large% \int_{0}^{\pi/2}{\ln\pars{\cos\pars{x}} \over x}\,\dd x} =\int_{0}^{\pi/2}{\ln\pars{\sin\pars{x}} \over \pi/2 - x}\,\dd x \end{align}

We'll use the expansion

$$ \ln\pars{\sin\pars{x}} =-\ln\pars{2} - \sum_{k\ = 1}^{\infty}{\cos\pars{2kx} \over k} =\sum_{k\ = 1}^{\infty}{\pars{-1}^{k} - \cos\pars{2kx} \over k} $$

Then,

\begin{align}&\color{#66f}{\large% \int_{0}^{\pi/2}{\ln\pars{\cos\pars{x}} \over x}\,\dd x} =\sum_{k\ =\ 1}^{\infty}{1 \over k} \int_{0}^{\pi/2}{\pars{-1}^{k} - \cos\pars{2kx} \over \pi/2 - x}\,\dd x \\[5mm]&=\sum_{k\ =\ 1}^{\infty}{1 \over k} \int_{0}^{\pi/2}{\pars{-1}^{k} - \cos\pars{k\pi - 2kx} \over x}\,\dd x =\sum_{k\ =\ 1}^{\infty}{\pars{-1}^{k} \over k} \int_{0}^{\pi/2}{1 - \cos\pars{2kx} \over x}\,\dd x \end{align}

The integral in the RHS is given by

$$ \int_{0}^{\pi/2}{1 - \cos\pars{2kx} \over x}\,\dd x =\gamma + \ln\pars{\pi} + \ln\pars{k} - \,{\rm Ci}\pars{\pi k} $$

where $\ds{\gamma}$ is the Euler-Mascheroni Constant and $\ds{\,{\rm Ci}}$ is the Cosine Integral function. Moreover, $\ds{\sum_{k\ =\ 1}^{\infty}\pars{-1}^{k}\,{\ln\pars{k} \over k} =\gamma\ln\pars{2} - \half\,\ln^{2}\pars{2}}$ such that

\begin{align}&\color{#66f}{\large% \int_{0}^{\pi/2}{\ln\pars{\cos\pars{x}} \over x}\,\dd x} \\[5mm]&=-\bracks{\gamma + \ln\pars{\pi}}\ln\pars{2} +\bracks{\gamma\ln\pars{2} - \half\,\ln^{2}\pars{2}} -\sum_{k\ =\ 1}^{\infty}\pars{-1}^{k}\,{\,{\rm Ci}\pars{\pi k} \over k} \\[5mm]&=\color{#66f}{\large% -\ln\pars{\pi}\ln\pars{2} - \half\,\ln^{2}\pars{2} -\sum_{k\ =\ 1}^{\infty}\pars{-1}^{k}\,{\,{\rm Ci}\pars{\pi k} \over k}} \end{align}

Answer 2

We have $$ I = \int_{0}^{\pi/4}\frac{\log(1-2\sin^2\theta)}{\theta}\,d\theta = \int_{0}^{\pi/4}\frac{\log(\sin(2\theta))}{\pi/4-\theta}\,d\theta,\tag{1}$$ hence by considering the Taylor series of $\log(1-x)$ we end with a series of CosIntegral values, not so appealing. An interesting approach may be to represent both $$\frac{\log(1-2\sin^2\theta)}{\sin\theta}\qquad\text{and}\qquad\frac{\sin\theta}{\theta}$$ or $$\frac{\log(1-2\sin^2\theta)}{\sin2\theta}\qquad\text{and}\qquad\frac{\sin2\theta}{\theta}$$ as Fouries sine series (or the first function as a Fourier sine series and the second one as a Fourier transform), then integrate their product through the orthogonality relations.

This may lead to a re-writing of $I$ as a well-known series.

Also notice that $(1)$ gives: $$ I = \sum_{j=0}^{+\infty}\left(\frac{2}{\pi}\right)^{j+1}\int_{0}^{\pi/2}x^j\log(\sin x)\,dx = \sum_{j=0}^{+\infty}\left(\frac{2}{\pi}\right)^{j+1}\int_{0}^{\pi/2}(\pi/2-x)^j\log(\cos x)\,dx,$$ hence $I$ can be written as a series of powers of $\frac{2}{\pi}$ times binomial coefficients times the values of the Euler integrals.

[Continues]

Answer 3

I've established some related explicit formulae.

Theorem 1. Let $n$ be any positive integer.

Set $$ I_{2n}:=\int_{0}^{\pi/2}\! \! x^{2n} \ln \cos x \: \mathrm{d}x $$ Then $$ I_{2n} = - \frac{\pi^{2n+1}\ln 2}{2^{2n+1}(2n+1)} - (-1)^{n}\frac{(2n)!}{2^{2n+1}}\sum_{p=1}^{n} \frac{(-1)^p}{(2p-1)!}\pi^{2p-1}\zeta(2n-2p+3) \tag1 $$ Set $$ I_{2n+1}:=\int_{0}^{\pi/2}\! \! x^{2n+1} \ln \cos x \:\mathrm{d}x $$ Then $$ \begin{align} I_{2n+1}=- \frac{\pi^{2n+2}\ln 2}{2^{2n+2}(2n+2)} - (-1)^n\left(1-\frac{1}{2^{2n+2}}\right)\frac{(2n+1)!}{2^{2n+2}}\zeta(2n+3) \\\\ - (-1)^n\frac{(2n+1)!}{2^{2n+2}} \sum_{p=0}^{n} \frac{(-1)^p}{(2p)!}\pi^{2p}\zeta(2n-2p+3) \tag2 \end{align} $$

[To be continued]

Answer 4

Other series expansions can be found by noticing that:

$$\int\limits_{0}^{\frac{\pi}{2}} \frac{\log{\left(\cos{\left(x \right)} \right)}}{x}\, dx = -\frac{\pi}{2}\int\limits_{0}^{1}\int\limits_{0}^{1} \tan{\left(\frac{\pi x y}{2} \right)}\, dx\, dy \tag{1}$$

which enables us to use the series expansion of the tangent for $-\pi/2<x<\pi/2$:

$$\tan{\left(x \right)} = \sum_{n=1}^{\infty} \frac{\left(-1\right)^{n - 1} \cdot 2^{2 n} x^{2 n - 1} \left(2^{2 n} - 1\right) B_{2 n}}{\left(2 n\right)!} \tag{2}$$

to obtain:

$$\begin{aligned} \int\limits_{0}^{\frac{\pi}{2}} \frac{\log{\left(\cos{\left(x \right)} \right)}}{x}\, dx &= \sum_{n=1}^{\infty} \frac{\left(- \pi^{2}\right)^{n} \left(4^{n} - 1\right) B_{2 n} \int\limits_{0}^{1}\int\limits_{0}^{1} \left(x y\right)^{2 n - 1}\, dx\, dy}{\left(2 n\right)!}\\ &=\sum_{n=1}^{\infty} \frac{\left(- \pi^{2}\right)^{n} \left(4^{n} - 1\right) B_{2 n}}{4 n^{2} \left(2 n\right)!}\\ &=\sum_{n=1}^{\infty} \frac{\left(- 1 + \frac{1}{2^{2 n}}\right) \zeta\left(2 n\right)}{2 n^{2}}\\ &=\sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \frac{\left(- 1 + \frac{1}{2^{2 n}}\right)}{2 n^{2}k^{2 n}}\\ &=\frac{1}{2}\sum_{k=1}^{\infty} \left(\operatorname{Li}_{2}\left(\frac{1}{4 k^{2}}\right) - \operatorname{Li}_{2}\left(\frac{1}{k^{2}}\right)\right)\\ &=-\frac{1}{2}\sum_{k=1}^{\infty} \operatorname{Li}_{2}\left(\frac{1}{\left(2 k - 1\right)^{2}}\right) \end{aligned} \tag{3}$$

where $B_{n}$ is a Bernoulli number, $\zeta$ is the Riemann zeta function, and $\operatorname{Li}_{2}$ is a polylogarithm of order 2 (aka dilogarithm or Spence's function). Unlike this answer from the op, on this occasion we do not obtain poly-Stieltjes constants but rather a higher order generalisation of sorts to the dilogarithm. The final expression in terms of the dilogarithm can be calculated to arbitrary precision in Python's Sympy package and agrees with the numerical evaluation of the integral $−0.941237674287746$ at $1e20$ terms.

More generally, following the above method it can be shown that:

$$\begin{aligned} \int\limits_{0}^{1} x^{- s} \log{\left(\cos{\left(\frac{\pi x}{2} \right)} \right)}\, dx &= - \sum_{n=1}^{\infty} \frac{\left(1 - 4^{- n}\right) \zeta\left(2 n\right)}{n \left(2 n - s + 1\right)}\\ &=- \sum_{n=1}^{\infty}\sum_{k=1}^{\infty} \frac{1}{n \left(2 n - s + 1\right)\left(2 k - 1\right)^{2 n}} \end{aligned} \tag{4}$$

and in particular, by using:

$$\sum_{n=1}^{\infty} \frac{z^{2 n}}{n \left(2 n - 1\right)} = \left(1 - z\right) \log{\left(1 - z \right)} + \left(z + 1\right) \log{\left(z + 1 \right)} \tag{5}$$

the special case with $s=2$ can be expressed in closed form in terms of the poly-Stieltjes constants:

$$\int\limits_{0}^{1} x^{- 2} \log{\left(\cos{\left(\frac{\pi x}{2} \right)} \right)}\, dx = - \sum_{n=1}^{\infty} \frac{1}{n \left(2 n - 1\right)} \\- \sum_{k=1}^{\infty} \left(\left(1 - \frac{1}{2 k + 1}\right) \log{\left(1 - \frac{1}{2 k + 1} \right)} + \left(1 + \frac{1}{2 k + 1}\right) \log{\left(1 + \frac{1}{2 k + 1} \right)}\right) \tag{6}$$

$$\int\limits_{0}^{1} x^{- 2} \log{\left(\cos{\left(\frac{\pi x}{2} \right)} \right)}\, dx = - \log{\left(\pi \right)} +\frac{\gamma_1(0,\frac{1}{2})}{2} - \frac{\gamma_1(0,-\frac{1}{2})}{2} \tag{7}$$

where the limit as $z$ tends to 1 in $\left(5\right)$ was used to derive:

$$\sum_{n=1}^{\infty} \frac{1}{n \left(2 n - 1\right)} = 2 \log{\left(2 \right)} \tag{8}$$ the product formula for the cosine was used to derive: $$\log{\left(\frac{\pi}{4}\right)}=\sum_{k=1}^{\infty} \log{\left(1 - \frac{1}{\left(2 k + 1\right)^{2}} \right)} \tag{9}$$ and the poly-Stieltjes constants are defined as: $$\gamma_k(a,b) = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log^k (n+a)}{n+b}-\frac{\log^{k+1} \!N}{k+1}\right) \tag{10}$$