# Evaluate $\int_{0}^{\large\frac{\pi}{4}} \ln {(\sin x)}\cdot\ln {(\cos x)} \left(\frac{\ln{(\sin x)}}{\cot x}+\frac{\ln {(\cos x)}}{\tan x}\right)dx$

How do I find the value of this integral?

$$I=\int_{0}^{\Large\frac{\pi}{4}} \ln {(\sin x)}\cdot\ln {(\cos x)} \left(\dfrac{\ln{(\sin x)}}{\cot x}+\dfrac{\ln {(\cos x)}}{\tan x}\right)dx$$

I tried substituting $t=\ln {(\sin x)}\cdot \ln {(\cos x)}$ and $t=\dfrac{\ln {(\sin x)}}{\ln {(\cos x)}}$, but it isn't working.

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Did you try to substitute $t=\sin x$? It may be a better option since $\frac{\ln(\sin x)}{\cot x} = \frac{\ln (\sin x)\sin x}{\cos x}$... – 5xum Jul 24 '14 at 10:39
No i didn't, but i will now. – Pkwssis Jul 24 '14 at 10:39
You can use the techniques I used to solve this problem – Mhenni Benghorbal Jul 24 '14 at 10:40
When $\sin x$ and $\cos x$ show up in the integral, it is often useful to make the following substitution: $$\tan{x \over 2}=t$$ With this sub the terms will be $$\sin x = {2t\over 1+t^2}\\ \cos x={1-t^2 \over 1+t^2}\\ dx = {2dt \over 1+t^2}$$ Try to make this and play around (you also need to pay attention to the integration borders, as they change with making the substitution). – Dmitry Kazakov Jul 24 '14 at 10:43
@DmitryKazakov I can't get to anything useful using that substitution...it keeps getting longer.. how did you do it? – Pkwssis Jul 24 '14 at 10:53

We will prove that $$I=-\frac{\pi^4}{2880}.$$ Indeed, let $$J=\int_0^{\pi/2}\log^2(\sin x)\log(\cos x)\tan x \,dx$$ It is easy to see that \eqalign{J&=\int_0^{\pi/4}\log^2(\sin x)\log(\cos x)\tan x \,dx+ \int_{\pi/4}^{\pi/2}\log^2(\sin x)\log(\cos x)\tan x \,dx\cr &=\int_0^{\pi/4}\log^2(\sin x)\log(\cos x)\tan x \,dx+ \int_{0}^{\pi/4}\log^2(\cos x)\log(\sin x)\cot x \,dx\cr &=I } Now, to calculate $J$ we make the substitution $t\leftarrow\sin^2x$: $$J=\frac{1}{16}\int_0^1\frac{\log(1-u)}{1-u}\log^2(u)\,du$$ But $$\frac{\log(1-u)}{1-u}=-\left(\sum_{n=0}^\infty u^n\right)\left(\sum_{n=1}^\infty \frac{u^n}{n}\right) =-\sum_{n=1}^\infty H_nu^n$$ where $H_n=\sum_{k=1}^n1/k$. Hence $$J=-\frac{1}{16}\sum_{n=1}^\infty H_n\int_0^1u^n\log^2(u)du =-\frac{1}{8}\sum_{n=1}^\infty\frac{ H_n}{(n+1)^3}$$ Finally, since $H_{n}=H_{n+1}-\frac{1}{n+1}$, we get $$J=\frac{1}{8}\zeta(4)-\frac{1}{8}\sum_{n=1}^\infty\frac{H_n}{n^3}\tag{1}$$

The sum $\sum_{n=1}^\infty\frac{H_n}{n^3}$ is known, it can be evaluated as follows, first we have $$H_n=\sum_{k=1}^\infty\left(\frac{1}{k}-\frac{1}{k+n}\right)= \sum_{k=1}^\infty \frac{n}{k(k+n)}$$ Thus $$\sum_{n=1}^\infty\frac{H_n}{n^3}=\sum_{k,n\geq1}\frac{1}{n^2k(n+k)} =\sum_{k,n\geq1}\frac{1}{k^2n(n+k)}$$ Taking the half sum we find $$\sum_{n=1}^\infty\frac{H_n}{n^3}=\frac{1}{2}\sum_{k,n\geq1}\frac{1}{kn(k+n)}\left(\frac{1}{k}+\frac{1}{n}\right)= \frac{1}{2}\sum_{k,n\geq1}\frac{1}{k^2n^2}=\frac{1}{2}\zeta^2(2)$$

Replacing in $(1)$ we obtain $$J=\frac{1}{8}\zeta(4)-\frac{1}{16}\zeta^2(2)=-\frac{\pi^4}{2880}.$$ as announced.$\qquad\square$

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Wow..thanks Professor Kouba!! What a great answer! – Pkwssis Jul 24 '14 at 11:38
Oh dear. Now I understand why Wolfram alpha couldn't find the solution. They should start scanning this site to improve their answers. – Matt B. Jul 24 '14 at 12:36
@Pkwssis: To evaluate $\sum_{n=1}^\infty\frac{H_n}{n^3}$ you can use the integral representation which I introduced for Euler sums $B(p,q)$. Your case is $B(1,3)$. Then you need to evaluate the corresponding integral which is not hard. – Mhenni Benghorbal Jul 24 '14 at 14:31
I use the $\Gamma$ function to evaluate this integral. Comparing to your solution, it is a little long. – xpaul Aug 21 '14 at 19:47
There is a simple derivation of $\displaystyle{\large\sum_{n = 1}^{\infty}{H_{n} \over n^{3}} = {\pi^{4} \over 72}}$. – Felix Marin Aug 21 '14 at 19:56

Here I use the Beta function to evaluate this integral and the calculation is a little bit tedious. As Omran Kouba did, \begin{eqnarray} I&=&\int_{0}^{\Large\frac{\pi}{2}} \ln^2 {(\sin x)}\cdot\ln {(\cos x)}\tan xdx. \end{eqnarray} Let $$J(a,b)=\int_0^{\frac{\pi}{2}}\sin^ax\cos^bx\tan xdx.$$ Then clearly $$\frac{\partial^3 J(0,0)}{\partial a^2\partial b}=I.$$ Now, by the change of variables $\sin x\to t$ and then $t^2\to u$, it is easy to see \begin{eqnarray} J(a,b)=\frac{\Gamma(1+\frac{a}{2})\Gamma(\frac{b}{2})}{2\Gamma(1+\frac{a+b}{2})}. \end{eqnarray} Noting $\Gamma'(x)=\Gamma(x)\psi_0(x), \Gamma''(x)=\Gamma(x)\psi_0^2(x)+\Gamma(x)\psi_1(a)$, we have \begin{eqnarray} \frac{\partial^2 J(a,b)}{\partial a^2}&=&\frac{\Gamma(1+\frac{a}{2})\Gamma(\frac{b}{2})}{8\Gamma(1+\frac{a+b}{2})}[\psi_0^2(1+\frac{a}{2})-2\psi_0(1+\frac{a}{2})\psi_0(1+\frac{a+b}{2})\\ &&+\psi_0^2(1+\frac{a+b}{2})+\psi_1(1+\frac{a}{2})-\psi_1(1+\frac{a+b}{2})]. \end{eqnarray} Using $$\Gamma(1)=1, \psi_0(1)=\gamma, \psi_1(1)=\frac{\pi^2}{6},$$ we obtain $$\frac{\partial^2 J(0,b)}{\partial a^2}=\frac{\Gamma(\frac{b}{2})}{48\Gamma(1+\frac{b}{2})}(6\gamma^2+\pi^2+12\gamma\psi_0(1+\frac{b}{2})+6\psi_0^2(1+\frac{b}{2})-6\psi_1(1+\frac{b}{2})))$$ and hence \begin{eqnarray} \frac{\partial^3 J(0,b)}{\partial a^2\partial b}&=&\frac{\Gamma(\frac{b}{2})}{96\Gamma(1+\frac{b}{2})}[-6\psi_0^3(1+\frac{b}{2})+6\psi_0^2(1+\frac{b}{2})(-2\gamma+\psi_0(1+\frac{b}{2}))\\ &&+\psi_0(\frac{b}{2})(6\gamma^2+\pi^2-6\psi_1(1+\frac{b}{2}))+12\gamma\psi_1(1+\frac{b}{2})+\gamma\psi_0(1+\frac{b}{2})(-6\gamma^2-\pi^2\\ &&+12\gamma\psi_0(\frac{b}{2})+18\gamma\psi_1(1+\frac{b}{2}))-6\gamma\psi_2(1+\frac{b}{2})). \end{eqnarray} Noting, for $z\in(0,1)$, $$\Gamma(1)=1, \Gamma(1-z)\Gamma(z)=\frac{\pi}{\sin(\pi z)}$$ we have $$\Gamma(\frac{b}{2})\approx\frac{\pi}{\sin(\frac{b\pi}{2})}, \psi_0(1+\frac{b}{2})=-\gamma+\frac{\pi^2b}{12}+O(b^2).$$ From this, we have \begin{eqnarray} \psi_0(\frac{b}{2})&=&-\frac{2}{b}-\gamma+\frac{\pi^2b}{12}+O(b^2)\\ \psi_0(1+\frac{b}{2})&=&-\gamma+\frac{\pi^2b}{12}+O(b^2)\\ \psi_1(1+\frac{b}{2})&=&\frac{\pi^2}{6}+\frac{b}{2}\psi_2(1)+\frac{\pi^4b^2}{120}+O(b^3)\\ \psi_2(1+\frac{b}{2})&=&\psi_2(1)+\frac{\pi^4b}{30}+O(b^2), \end{eqnarray} and hence \begin{eqnarray} \frac{\partial^3 J(0,b)}{\partial a^2\partial b}&\approx&\frac{b\pi^3}{11520\sin(\frac{b\pi}{2})}(-2\pi^2+b^2\pi^4+60b\psi_2(1)). \end{eqnarray} So $$\frac{\partial^3 J(0,0)}{\partial a^2\partial b}=\lim_{b\to0}\frac{\partial^3 J(0,b)}{\partial a^2\partial b}=\lim_{b\to0}\frac{b\pi^3}{11520\sin(\frac{b\pi}{2})}(-2\pi^2+b^2\pi^4+60b\psi_2(1))=-\frac{\pi^4}{2880}.$$

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