Logarithm, Trignometry, Integrals

The answer is :

$= \dfrac{\pi{log}^{2}(2)}{2} - \dfrac{ {\pi}^{3}}{48}$

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