# Evaluating $\int_0^{\pi/2}\frac{(\ln{\sin x})(\ln{\cos x})}{\tan x}dx$

I need to solve $$\int_0^{\pi/2}\frac{(\ln{\sin x})(\ln{\cos x})}{\tan x}dx$$

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

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Could help that $\displaystyle\frac{d}{dx}(\ln(\sin x)) = \frac{1}{\tan x}$ ? – Chris's wise sister Aug 30 '12 at 19:41
With the substitution $\ln(\sin x)=u$ you get $\displaystyle \frac{1}{2} \int_{-oo}^{0} \ln(1-e^{2u}) u \ du$. Then you may use Taylor expansion for $\ln(1-e^{2u})$ and integrate term by term. Does it help? – Chris's wise sister Aug 30 '12 at 20:21
Finally, you get $\frac{1}{8} \zeta(3)$. – Chris's wise sister Aug 30 '12 at 20:28

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)

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 +1 I love this answer – dot dot Aug 31 '12 at 9:50 It is a nice answer. – Mhenni Benghorbal Aug 31 '12 at 13:37

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: $$\int_0^{\pi/2} \frac{\ln(\sin(x)) \ln(\cos(x))}{\tan(x)} \mathrm{d} x = \frac{1}{8} \zeta(3)$$

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.

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 It looks better the alternative with the series. – Chris's wise sister Aug 30 '12 at 21:07 I do not disagree, but I left them in the order they occurred to me. – Sasha Aug 30 '12 at 21:11 Nice solution. Good technique. – Mhenni Benghorbal Aug 31 '12 at 13:34