Integral with Logarithms

Question

$$\displaystyle \int _{ 0 }^{ \pi /2 }{ \log(\cos(x))\log(\sin(x)) \ dx } $$

One solution: Consider : $\displaystyle F(m,n)=\int _{ 0 }^{ \pi /2 }{ \sin ^{ 2m-1 }{ x } \cos ^{ 2n-1 }{ x } dx } $

To solve this put $\sin^{2}x = t $ to get our integral as :

$ \displaystyle F(m,n) =\frac{1}{2} \int _{ 0 }^{ 1 }{ { t }^{ m-1 }{ (1-t) }^{ n-1 }dt }=\frac{\beta (m,n)}{2} $

Where $\beta(m,n)$ is the beta function.

$\displaystyle F(m,n) = \frac { \Gamma (m)\Gamma (n) }{2 \Gamma (m+n) } $

Hence we have :

$\displaystyle \frac { \Gamma (m)\Gamma (n) }{ \Gamma (m+n) } = 2\int _{ 0 }^{ \pi /2 }{ \sin ^{ 2m-1 }{ x } \cos ^{ 2n-1 }{ x } dx }$

First differentiating both sides with respect to $m$ we have :

$$\displaystyle \frac { \Gamma (n) }{ ({ \Gamma (m+n)) }^{ 2 } } (\Gamma '(m)\Gamma (m+n)-\Gamma (m)\Gamma '(m+n)) = 4\int _{ 0 }^{ \pi /2 }{ log(sin(x))\sin ^{ 2m-1 }{ x } \cos ^{ 2n-1 }{ x } dx } $$

Better written as :

$\displaystyle \frac { \Gamma (m)\Gamma (n) }{ \Gamma (m+n) } (\psi (m)-\psi (m+n))=4\int _{ 0 }^{ \pi /2 }{ \log(\sin(x))\sin ^{ 2m-1 }{ x } \cos ^{ 2n-1 }{ x } dx }$

where $\psi(x)$ is the digamma function.

Now differentiate with respect to $n$ both sides we get :

$\displaystyle \frac { \Gamma (m)\Gamma (n) }{ \Gamma (m+n) } (((\psi (m)-\psi (m+n))(\psi (n)-\psi (m+n))-\psi '(m+n))$

$\displaystyle =8\int _{ 0 }^{ \pi /2 }{ \log(\sin(x))\log(\cos(x))\sin ^{ 2m-1 }{ x } \cos ^{ 2n-1 }{ x } dx } $

Put $m=n=\dfrac{1}{2}$ to get :

$\displaystyle \frac { { \Gamma }^{ 2 }(1/2) }{ \Gamma (1) } ({ (\psi (1/2)-\psi (1)) }^{ 2 }-\psi '(1))$

$\displaystyle =8\int _{ 0 }^{ \pi /2 }{ \log(\cos(x))\log(\sin(x))dx } $

Now $\displaystyle \Gamma (1/2)=\sqrt { \pi } ,\Gamma (1)=1,\psi (1/2)=-\gamma -\log(4),\psi (1)=-\gamma ,\psi '(1)=\frac { { \pi }^{ 2 } }{ 6 } $

Could somebody please help me with this Integral?

possible duplicate of Computing the integral of $\log(\sin x)$ — Chappers, May 18 '15 at 20:05
Related: http://math.stackexchange.com/questions/292468/fourier-series-of-log-sine-and-log-cos — Jack D'Aurizio, May 18 '15 at 20:14
@BetterWorld: just use the orthogonality condition for the sine and cosine function. — Jack D'Aurizio, May 18 '15 at 20:16
you may use this: http://math.stackexchange.com/questions/1146362/a-challenging-log-trigonometric-integral/1146411#1146411 — tired, May 18 '15 at 20:17

score 5 · Accepted Answer · edited Apr 13 '17 at 12:21

Given the two identities: $$ -\log(\sin(x))=\sum_{k=1}^\infty\frac{\cos(2kx)}{k}+\log(2)\tag{1} $$ and $$ -\log(\cos(x))=\sum_{k=1}^\infty(-1)^k\frac{\cos(2kx)}{k}+\log(2)\tag{2} $$ proved here, and the orthogonality relation: $$ \int_{0}^{\pi/2}\cos(2k_1 x)\cos(2k_2 x)\,dx = \frac{\pi}{4}\,\delta(k_1,k_2)\tag{3} $$ it follows that $\int_{0}^{\pi/2}\log(\sin x)\,dx=\int_{0}^{\pi/2}\log(\cos x)\,dx=-\frac{\pi}{2}\log 2$ and:

$$ \int_{0}^{\pi/2}\log(\sin x)\log(\cos x)\,dx = \frac{\pi}{2}\log^2 2+\frac{\pi}{4}\sum_{k=1}^{+\infty}\frac{(-1)^k}{k^2} = \color{red}{\frac{\pi}{2}\log^2 2-\frac{\pi^3}{48}}.\tag{4}$$

Integral with Logarithms

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