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A friend of mine posed the following improper integral:

$\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{\ln^{2}(\sin(x))\ln^{2}(\cos(x))}{\sin(x)\cos(x)}dx=\frac{1}{2}\zeta(5)-\frac{1}{4}\zeta(2)\zeta(3)$.

This checks numerically. But, does anyone know of a nice place to begin?.

I tried the classic series $\displaystyle \ln(\sin(x))=-\sum_{k=1}^{\infty}\frac{\cos(2kx)}{k}-\ln(2)$ and

$\displaystyle \ln(\cos(x))=-\sum_{k=1}^{\infty}\frac{(-1)^{k}\cos(2kx)}{k}-\ln(2)$

Then, tried squaring and so forth. It gets rather tedious, but perhaps it'll work given the effort. Then again, the $\sin(x)\cos(x)$ in the denominator must be dealt with.

With all of those zetas in the solution, one would think that series may be the way to go.

Does anyone know of a nice approach?.


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marked as duplicate by Marvis, Sasha, Asaf Karagila, Micah, Cameron Buie Feb 18 '13 at 5:41

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Sorry for double posting. I did not realize. Feel free to delete. – Cody Feb 2 '13 at 0:40

Answer on log-sine and log-cos integral with fun result

       Huizeng Qin

       Institute of Applied Mathematics,

       Shandong University of Technology,

     Zibo, Shandong, P. R. China(

Using \begin{equation} \int_{0}^{\frac{\pi }{2}}\sin ^{2x-1}u\cos ^{2y-1}u\ln ^{p}\sin u\ln ^{q}\cos udu=2^{-p-q-1}B_{p,q}(x,y) \tag*{(1)} \end{equation} where $B(x,y)$ is Beta function and $B_{p,q}(x,y)=\frac{\partial ^{p+q}}{ \partial x^{p}\partial y^{q}}B(x,y),$ we have \begin{equation} I(p,q)=\int_{0}^{\frac{\pi }{2}}\frac{\ln ^{p}\sin u\ln ^{q}\cos u}{\sin u\cos u}du=2^{-p-q-1}B_{p,q}(0,0). \tag*{(2)} \end{equation} By the partial derivative of the identity \begin{equation} xyB(x,y)=(x+y)(x+y+1)B(x+1,y+1) \tag*{(3)} \end{equation} on $x,y$ we have \begin{equation} \begin{array}{c} xyB_{p,q}(x,y)+pyB_{p-1,q}(x,y)+qxB_{p,q-1}(x,y)+qpB_{p-1,q-1}(x,y) \\ =(x+y)(x+y+1)B_{p,q}(x+1,y+1)+q(2x+2y+1)B_{p,q-1}(x+1,y+1) \\ +p(2x+2y+1)B_{p-1,q}(x+1,y+1)+q(q-1)B_{p,q-2}(x+1,y+1) \\ +2qpB_{p-1,q-1}(x+1,y+1)+p(p-1)B_{p-2,q}(x+1,y+1) \end{array} \tag*{(4)} \end{equation} In (4) let $x\rightarrow 0,y\rightarrow 0$, there is \begin{equation} \begin{array}{c} B_{p-1,q-1}(0,0)=\frac{1}{p}B_{p,q-1}(1,1)+\frac{1}{q}B_{p-1,q}(1,1) \\ +\frac{q-1}{p}B_{p,q-2}(1,1)+2B_{p-1,q-1}(1,1)+\frac{p-1}{q}B_{p-2,q}(1,1) \end{array} \tag*{(5)} \end{equation} and \begin{equation} \begin{array}{c} B_{p,q}(0,0)=\frac{1}{p+1}B_{p+1,q}(1,1)+\frac{1}{q+1}B_{p,q+1}(1,1) \\ +\frac{q}{p+1}B_{p+1,q-1}(1,1)+2B_{p,q}(1,1)+\frac{p}{q+1}B_{p-1,q+1}(1,1) \end{array} \tag*{(6)} \end{equation} Thus we get \begin{equation} \begin{array}{c} I(p,q)=2^{-p-q-1}( \frac{B_{p,q+1}(1,1)}{q+1}+\frac{B_{p+1,q}(1,1)}{p+1} \\ +2B_{p,q}(1,1)+\frac{pB_{p-1,q+1}(1,1)}{q+1}+\frac{qB_{p+1,q-1}(1,1)}{p+1} ) \end{array} \tag*{(7)} \end{equation} For $B_{p,q}(1,1)$ there is the following recurrence relations \begin{equation} \begin{array}{c} B_{p,q}(1,1)=(q-1)!p!( \sum\limits_{k=0}^{p}\sum\limits_{j=0}^{q-1}C_{p+q-k-j-1}^{p-k}\frac{ (-1)^{p+q-k-j}B_{k,j}(1,1)}{k!j!} \\ -\sum\limits_{k=0}^{p-1}\sum\limits_{j=0}^{q-1}C_{p+q-k-j-1}^{p-k}\frac{(-1)^{p+q-k-j}\zeta (p+q-k-j)B_{k,j}(1,1)}{k!j!} )\end{array} . \tag*{(8)} \end{equation} By (8) we can get that \begin{equation} \begin{array}{c} B_{0,0}(1,1)=1,B_{0,1}(1,1)=-1,B_{0,2}(1,1)=2,B_{0,3}(1,1)=-6, \\ B_{1,1}(1,1)=2-\frac{\pi ^{2}}{6},B_{1,2}(1,1)=-6+\frac{\pi ^{2}}{3}+2\zeta (3), \\ B_{1,3}(1,1)=24-\pi ^{2}-\frac{\pi ^{4}}{15}-6\zeta (3),B_{2,2}(1,1)=24- \frac{4\pi ^{2}}{3}-\frac{\pi ^{4}}{90}-8\zeta (3), \\ B_{2,3}(1,1)=-120+6\pi ^{2}+\frac{\pi ^{4}}{6}+36\zeta (3)-2\pi ^{2}\zeta (3)+24\zeta (5) \\ B_{3,3}(1,1)=720-36\pi ^{2}-\pi ^{4}-\frac{23\pi ^{6}}{420}-216\zeta (3)+12\pi ^{2}\zeta (3)+36\zeta ^{2}(3)-144\zeta (5) \end{array} \tag*{(9)} \end{equation} By (7),(9) and $B_{p,q}(1,1)=B_{q,p}(1,1)$ we have \begin{equation} \begin{array}{c} I(1,0)=-\frac{\pi ^{2}}{24},I(1,1)=\frac{\zeta (3)}{4},I(1,2)=-\frac{\pi ^{4} }{576},I(1,3)=-\frac{\pi ^{2}\zeta (3)}{32}+\frac{9\zeta (5)}{16} \\ I(2,2)=-\frac{\pi ^{2}\zeta (3)}{24}+\frac{\zeta (4)}{2},I(2,3)=-\frac{\pi ^{6}}{2304}+\frac{9\zeta ^{2}(3)}{32}, \\ I(3,3)=-\frac{\pi ^{4}\zeta (3)}{128}-\frac{3\pi ^{2}\zeta (5)}{16}+\frac{ 45\zeta (7)}{16}. \end{array} \tag*{(10)} \end{equation}

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Welcome to MSE! Impressive answer, but could you please move your identifying information into your profile instead? If you like, add a comment like "If you'd like to contact me, see my profile." Thanks! – gnometorule Feb 18 '13 at 3:21
@Huizeng Qin: Welcome to Math Stack Exchange! I moved your contact information from your answer to your user profile: – Eric Naslund Feb 18 '13 at 4:08

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