Evaluating $\int_0^{\frac\pi2}\frac{\ln{(\sin x)}\ \ln{(\cos x})}{\tan x}\ dx$

Question

I need to solve $$ \int_0^{\Large\frac\pi2}\frac{\ln{(\sin x)}\ \ln{(\cos x})}{\tan x}\ dx $$

I tried to use symmetric properties of the trigonometric functions as is commonly used to compute $$ \int_0^{\Large\frac\pi2}\ln\sin x\ dx = -\frac{\pi}{2}\ln2 $$ but never succeeded. (see this for example)

Answer 1

Let's start out with the substitution $ \displaystyle \ln(\sin x) = u $ and get: $$\ln(\cos x)=\frac{\ln(1-e^{2u})}{2}$$ $$\displaystyle\frac{1}{\tan x} \ dx =du$$ that further yields $$\int_0^{\pi/2}\frac{(\ln{\sin x})(\ln{\cos x})}{\tan x}dx= \frac{1}{2} \int_{-\infty}^{0} \ln(1-e^{2u}) u \ du$$

According to Taylor expansion we have $$\ln(1-e^{2u})= \sum_{k=1}^{\infty} (-1)^{2k+1} \frac{e^{2 k u}}{k}$$ then $$\frac{1}{2} \int_{-\infty}^{0} \ln(1-e^{2u}) u \ du=$$ $$\frac{1}{2} \sum_{k=1}^{\infty} \frac{(-1)^{2k+1}}{k} \int_{-\infty}^{0} u e^{2ku} \ du =$$ $$\frac{1}{2} \sum_{k=1}^{\infty} \frac{(-1)^{2k+1}}{k} \frac{-1}{4k^2} = \frac{1}{8} \sum_{k=1}^{\infty} \frac{1}{k^3}=\frac{1}{8} \zeta(3).$$

Remark: the value of $\zeta(3)\approx1.2020569$ is called Apéry's Constant - see here.

Q.E.D. (Chris)

Answer 2

Let $u = \sin^2(x)$. Then $\frac{\mathrm{d}x}{\tan(x)} = \frac{\cos(x)}{\sin(x)} \mathrm{d}x = \frac{d\sin(x)}{\sin(x)} = \frac{1}{2}\frac{\mathrm{d}u}{u}$, $\ln(\sin(x)) = \frac{1}{2}\ln(u)$ and $\ln(\cos(x)) = \frac{1}{2} \ln(1-u)$: $$ \int_0^{\pi/2} \frac{\ln(\sin(x)) \ln(\cos(x))}{\tan(x)} \mathrm{d} x = \frac{1}{8} \int_0^{1} \frac{\ln u}{u} \cdot \ln(1-u) \mathrm{d} u = \left.\frac{1}{8} \frac{\mathrm{d}^2}{\mathrm{d} s \mathrm{d} t} \int_0^1 u^{s-1} (1-u)^{t-1} \mathrm{d}u \right|_{s\to 0^+,t=1} = \left.\frac{1}{8} \frac{\mathrm{d}^2}{\mathrm{d} s \mathrm{d} t} \frac{\Gamma(s) \Gamma(t)}{\Gamma(s+t)} \right|_{s\to 0^+,t=1} $$ First differentiate with respect to $t$ and substitute $t=1$: $$ \left.\frac{1}{8} \frac{\mathrm{d}}{\mathrm{d} s} \frac{\Gamma(s)}{\Gamma(s+1)}\left( \psi(1) - \psi(s+1)\right) \right|_{s\to 0^+} = \left.\frac{1}{8} \frac{\mathrm{d}}{\mathrm{d} s} \frac{\left( \psi(1) - \psi(s+1)\right) }{s} \right|_{s\to 0^+} $$ Using Taylor series expansion for the digamma function $\psi(s)$ we have: $$ \frac{\left( \psi(1) - \psi(s+1)\right) }{s} = -\zeta(2) + \zeta(3) s + \mathcal{o}(s) $$ Hence the value of the integral is: $$ \int_0^{\pi/2} \frac{\ln(\sin(x)) \ln(\cos(x))}{\tan(x)} \mathrm{d} x = \frac{1}{8} \zeta(3) $$

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Alternatively you could use $$\frac{\ln(1-u)}{u} = -\sum_{k=0}^\infty \frac{u^k}{k+1}$$ and integrate term-wise: $$\int_0^1 u^k \ln(u) \mathrm{d} u \stackrel{u=\exp(-t)}{=} \int_0^\infty t \exp(-t(k+1)) \mathrm{d} t = -\frac{1}{(k+1)^2}$$ which yields the result immediately.

Answer 3

$$\begin{align*} I &= \int_0^{\frac\pi2} \frac{\log(\cos(x)) \log(\sin(x))}{\tan(x)} \, dx \\[1ex] &= \int_0^\infty \frac{\log\left(\frac1{\sqrt{x^2+1}}\right) \log\left(\frac x{\sqrt{x^2+1}}\right)}{x} \, \frac{dx}{1+x^2} \tag{1} \\[1ex] &= \frac14 \int_0^\infty \frac{\log^2(x^2+1)-\log(x^2)\log(x^2+1)}{x} \, \frac{dx}{1+x^2} \\[1ex] &= \frac18 \int_0^\infty \frac{\log^2(x+1)-\log(x)\log(x+1)}{x(1+x)} \, dx \tag{2} \\[1ex] &= \frac18 \int_0^1 \left(\frac{\log^2(x+1)}{x} - \frac{\log(x)\log(x+1)}x\right) \, dx \tag{3} \\[1ex] &= \frac14 \int_0^1 \frac{\log(x)\log(x+1)}x \, dx - \frac12 \int_0^1 \frac{\log(x+1)\log(x)}{x+1} \, dx \tag{4} \\[1ex] &= -\frac14 \sum_{n=1}^\infty \frac{(-1)^n}n \int_0^1 x^{n-1} \log(x) \, dx + \frac12 \sum_{n=1}^\infty (-1)^n H_n \int_0^1 x^n \log(x) \, dx \tag{5} \\[1ex] &= -\frac14 \sum_{n=1}^\infty \frac{(-1)^n}{n^3} + \frac12 \sum_{n=1}^\infty \frac{(-1)^n H_n}{(n+1)^2} \tag{4} \\[1ex] &= \boxed{\frac{\zeta(3)}8} \tag{6} \end{align*}$$

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• $(1)$ : substitute $x\mapsto\arctan(x)$
• $(2)$ : substitute $x\mapsto\sqrt x$
• $(3)$ : split up the integral at $x=1$; substitute $x\mapsto\frac1x$ in the integral over $[1,\infty)$; partial fractions
• $(4)$ : integrate by parts
• $(5)$ : recall the power series $-\log(1-x)=\sum\limits_{n\ge1}\frac{x^n}n$ and $-\frac{\log(1-x)}{1-x}=\sum\limits_{n\ge1} H_n x^n$, where $H_n$ is the $n^{\rm th}$ harmonic number
• $(6)$ : see e.g. section $7$ for a contour-integration method to evaluating the Euler sum; I'm sure there are more efficient means