Find the value of $I=\displaystyle\int_0^{\pi/2}x^2\ln(\sin x)\ln(\cos x)\ \mathrm dx$

We have the information that $J=\displaystyle\int_0^{\pi/2}x\ln(\sin x)\ln(\cos x)\ \mathrm dx=\dfrac{(\pi\ln{2})^2}{8}-\dfrac{\pi^4}{192}$

  • $\begingroup$ if you represent ln(cos(x)) as ln(d(sinx)dx)? $\endgroup$ – dato datuashvili Mar 14 '13 at 5:31
  • $\begingroup$ Please refrain from using \displaystyle in the title. $\endgroup$ – user1551 Mar 14 '13 at 5:31
  • $\begingroup$ this one is answer ,but no step is shown from wolframalpha!enter image description here $\endgroup$ – dato datuashvili Mar 14 '13 at 5:37
  • $\begingroup$ Use sympy: from sympy import * x=Symbol('x') integrate(x**2*log(cos(x))*log(sin(x)),(x,0,pi/2)) $\endgroup$ – Jichao Mar 14 '13 at 5:56
  • $\begingroup$ I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Mar 12 '18 at 16:51

Tools Needed $$ \begin{align} \frac1{k(j-k)^2}&=\frac1{j^2k}-\frac1{j^2(k-j)}+\frac1{j(k-j)^2}\tag{1}\\ \frac1{k(j+k)^2}&=\frac1{j^2k}-\frac1{j^2(k+j)}-\frac1{j(k+j)^2}\tag{2}\\ \log(\sin(x))&=-\log(2)-\sum_{k=1}^\infty\frac{\cos(2kx)}{k}\tag{3}\\ \log(\cos(x))&=-\log(2)-\sum_{k=1}^\infty(-1)^k\frac{\cos(2kx)}{k}\tag{4}\\ \cos(2jx)\cos(2kx)&=\frac12\Big[\cos(2(j-k)x)+\cos(2(j+k)x)\Big]\tag{5}\\ \end{align} $$ $$ \int_0^{\pi/2}x^2\cos(2kx)\,\mathrm{d}x=\left\{ \begin{array}{} (-1)^k\frac\pi{4k^2}&\text{if }k\ne0\\ \frac{\pi^3}{24}&\text{if }k=0 \end{array}\right.\tag{6}\\ $$

Tool Use $$ \begin{align} &\int_0^{\pi/2}x^2\log(\sin(x))\log(\cos(x))\,\mathrm{d}x\\[12pt] &=\int_0^{\pi/2}x^2\left(\log(2)+\sum_{k=1}^\infty\frac{\cos(2kx)}{k}\right)\left(\log(2)+\sum_{k=1}^\infty(-1)^k\frac{\cos(2kx)}{k}\right)\,\mathrm{d}x\\[12pt] &=\log(2)^2\int_0^{\pi/2}x^2\,\mathrm{d}x +\log(2)\sum_{k=1}^\infty\frac1k\int_0^{\pi/2}x^2\cos(4kx)\,\mathrm{d}x\\ &+\sum_{j=1}^\infty\sum_{k=1}^\infty\frac{(-1)^k}{2jk}\int_0^{\pi/2}x^2\Big[\cos(2(j-k)x)+\cos(2(j+k)x)\Big]\,\mathrm{d}x\\[12pt] &=\frac{\pi^3}{24}\log(2)^2+\log(2)\frac\pi{16}\zeta(3)\\ &+\frac\pi8\sum_{j=1}^\infty\sum_{k=1}^\infty\frac{(-1)^j}{jk}\left[\mathrm{iif}\left(j=k,\frac{\pi^2}{6},\frac1{(j-k)^2}\right)+\frac1{(j+k)^2}\right]\\[12pt] &=\frac{\pi^3}{24}\log(2)^2+\log(2)\frac\pi{16}\zeta(3)\\ &+\frac\pi8\sum_{j=1}^\infty\frac{(-1)^j}{j}\sum_{k=1}^{j-1}\frac1{k(j-k)^2} +\frac\pi8\sum_{j=1}^\infty\frac{(-1)^j}{j^2}\frac{\pi^2}{6} +\frac\pi8\sum_{j=1}^\infty\frac{(-1)^j}{j}\sum_{k=j+1}^\infty\frac1{k(j-k)^2}\\ &+\frac\pi8\sum_{j=1}^\infty\frac{(-1)^j}{j}\sum_{k=1}^\infty\frac1{k(j+k)^2}\\[12pt] &=\frac{\pi^3}{24}\log(2)^2+\log(2)\frac\pi{16}\zeta(3)\\ &+\frac\pi8\sum_{j=1}^\infty\frac{(-1)^j}{j}\left(\frac2{j^2}H_{j-1}+\frac1jH_{j-1}^{(2)}\right) -\frac{\pi^5}{576} +\frac\pi8\sum_{j=1}^\infty\frac{(-1)^j}{j}\left(-\frac1{j^2}H_j+\frac1j\frac{\pi^2}{6}\right)\\ &+\frac\pi8\sum_{j=1}^\infty\frac{(-1)^j}{j}\left(\frac1{j^2}H_j-\frac1j\frac{\pi^2}{6}+\frac1jH_j^{(2)}\right)\\[12pt] &=\frac{\pi^3}{24}\log(2)^2+\log(2)\frac\pi{16}\zeta(3)\\ &+\frac\pi8\sum_{j=1}^\infty\frac{(-1)^j}{j}\left(\frac2{j^2}H_j+\frac2jH_j^{(2)}-\frac3{j^3}\right) -\frac{\pi^5}{576}\\[12pt] &=\frac{\pi^3}{24}\log(2)^2+\log(2)\frac\pi{16}\zeta(3)+\frac{11\pi^5}{5760} +\frac\pi4\sum(-1)^j\left(\frac1{j^3}H_j+\frac1{j^2}H_j^{(2)}\right)\\[12pt] &=\frac{\pi^3}{24}\log(2)^2+\log(2)\frac\pi{16}\zeta(3)-\frac{\pi^5}{960} -\frac\pi{16}\sum_{j=1}^\infty\frac1{j^3}H_{2j}\tag{7} \end{align} $$ Numerically, $(7)$ matches the integral. I'm working on the last harmonic sum. Both numerical integration and $(7)$ yield $0.0778219793722938643380944$.

Mathematica Help

Thanks to Artes' answer on Mathematica, I have verified that these agree to 100 places.

  • $\begingroup$ beatifull, Thank you my frend $\endgroup$ – math110 Mar 15 '13 at 15:41
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    $\begingroup$ @MhenniBenghorbal: Thanks! Using those functions, $(7)$ is $$ \frac{\pi^3}{24}\log(2)^2+\log(2)\frac\pi{16}\zeta(3)+\frac{11\pi^5}{5760} -\frac\pi4(A(1,3)+A(2,2)) $$ $\endgroup$ – robjohn Mar 15 '13 at 22:05
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    $\begingroup$ (+1) I also tried this approach, but just gave up because of the required amount of calculation. I purely admire all your efforts toward the conclusion. $\endgroup$ – Sangchul Lee Mar 16 '13 at 4:06
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    $\begingroup$ @robjohn: The value of that last Euler Sum is $$-\frac{\pi ^4}{15}-\frac{1}{3} \pi^2 \log^2(2)+\frac{\log^4(2)}{3}+8 \text{Li}_4\left(\frac{1}{2}\right)+7 \log(2)\zeta(3)$$ For it's proof you can refer to this page. $\endgroup$ – Shobhit Bhatnagar Jan 11 '14 at 8:55
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    $\begingroup$ Can someone please post the full answer, then? I mean, (7) with the sigma replaced by Integrals-and-Series's answer. $\endgroup$ – Akiva Weinberger Aug 24 '14 at 2:30

I'm still struggling with this integral, but I guess the following result may have a chance to be helpful:

\begin{align*} \int_{0}^{\frac{\pi}{2}} x^2 \log^2 \cos x \, dx &= \frac{11 \pi^5}{1440} + \frac{\pi^3}{24} \log^2 2 + \frac{\pi}{2}\zeta(3) \log 2 \tag{1} \\ &\approx 4.2671523609840988652 \cdots. \end{align*}

To prove this, let us consider the following identity

$$ \int_{0}^{\frac{\pi}{2}} \cos^{z}x \cos wx \, dx = \frac{\pi}{2^{z+1}} \binom{z}{\frac{z+w}{2}}.$$

You can find the proof of this identity at here. Thus it follows that

$$ \int_{0}^{\frac{\pi}{2}} x^2 \log^2 \cos x \, dx = - \left. \frac{\partial^4}{\partial z^2 \partial w^2} \frac{\pi}{2^{z+1}} \binom{z}{\frac{z+w}{2}} \right|_{(z, w) = (0, 0)}. $$

Performing a bunch of calculations, we obtain $(1)$. Similar idea shows that

$$ \int_{0}^{\frac{\pi}{2}} \log^2 \cos x \, dx = \left. \frac{\partial^2}{\partial z^2} \frac{\pi}{2^{z+1}} \binom{z}{\frac{z+w}{2}} \right|_{(z, w) = (0, 0)} = \frac{\pi^3}{24} + \frac{\pi}{2}\log 2. \tag{2} $$

Indeed, starting from the identity

$$ \log^2 \left( \frac{\sin 2x}{2} \right) = \log^2 \cos x + \log^2 \sin x + 2\log \cos x \log \sin x, $$

I obtained

\begin{align*}I &= -\frac{7}{8}\int_{0}^{\frac{\pi}{2}} x^2 \log^2 \cos x \, dx + \frac{\pi}{2} \int_{0}^{\frac{\pi}{2}} x \log^2 \cos x \, dx - \frac{3\pi^2}{32} \int_{0}^{\frac{\pi}{2}} \log^2 \cos x \, dx \\ &\quad -\frac{\log 2}{8}\int_{0}^{\pi} x^2 \log \sin x \, dx + \frac{\pi^3}{48} \log^2 2 \\ &\approx 0.077821979372293864338\cdots. \end{align*}

From the identity

$$ \log \sin x = -\log 2 - \sum_{n=1}^{\infty} \frac{\cos 2nx}{n}, $$

we obtain

$$\int_{0}^{\pi} x^2 \log \sin x \, dx = -\frac{\pi}{2} \zeta (3) - \frac{\pi^3}{3} \log 2. \tag{3}$$

Putting $(1)$, $(2)$ and $(3)$ together, I was able to derive

\begin{align*}I &= -\frac{61 \pi^5}{5760} - \frac{3\pi}{8} \zeta (3) \log 2 -\frac{\pi^3}{48} \log^2 2 + \frac{\pi}{2} \int_{0}^{\frac{\pi}{2}} x \log^2 \cos x \, dx. \end{align*}

I'm not sure if this formula will be helpful, since the last remaining integral seems to defy my techniques.

  • $\begingroup$ I meant to upvote this earlier, but got distracted. I had forgotten about the cosine expansion for $\log(\cos(x))$ and $\log(\sin(x))$ until you used one here. $\endgroup$ – robjohn Mar 16 '13 at 4:15
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    $\begingroup$ For the last integral you can use the following technique since we have $$\int_0^{\Large\frac\pi2}\cos^{n-1}x\cos ax\ dx=\frac{\pi}{2^n n\ \operatorname{B}\left(\frac{n+a+1}{2},\frac{n-a+1}{2}\right)}$$ $\endgroup$ – Tunk-Fey Sep 18 '14 at 13:02

Related problem: (I), (II). The contribution of this post is to evaluate the integral

$$ I = \int_0^{\pi/2}\ln(\sin x)\ln(\cos x)dx $$

symbolically. Now to find $I$, we use the first change of variables $ t = \sin(x) $ which results in

$$ I = \frac{1}{2}\int_{0}^{1}\frac{\ln(t)\ln(1-t^2)}{\sqrt{1-t^2}}dt. $$

Following it by the change of variables $u=t^2$ gives

$$ I = \frac{1}{8}\int_{0}^{1}\frac{ \ln(u) \ln(1-u) }{ \sqrt{u} \sqrt{1-u} } du .$$

To evaluate the last integral, we consider the integral

$$ F = \frac{1}{8}\int_{0}^{1}u^{a-\frac{1}{2}} (1-u)^{b-\frac{1}{2}} du = \beta(a+1/2,b+1/2) ,$$

where $\beta(u,v)$ is the beta function.

$$ \implies I = D_{b}\,D_{a} \beta(a+1/2,b+1/2)|_{a=0,b=0}= \frac{\pi}{48} \, \left( 24\, \left( \ln \left( 2 \right)\right)^{2} -{\pi }^{2} \right),$$

where $D_a=\frac{\partial }{\partial a}$ and $D_b=\frac{\partial }{\partial b}$.

  • $\begingroup$ Too trivial edit: $dt$ is missing at the end of the second formula. $\endgroup$ – Mark Hurd Mar 19 '13 at 1:51
  • $\begingroup$ @MarkHurd: Thank you. I really appreciate it. $\endgroup$ – Mhenni Benghorbal Mar 31 '13 at 12:30
  • $\begingroup$ @MhenniBenghorbal Could you please explain how you used the Beta Function in the last step? I couldn't understand. $\endgroup$ – User1234 Jun 4 '15 at 11:40

This is yet another partial answer, and a verification of some other claims.

Using $(4)$ and $(8)$ from this answer, we get $$ \int_0^{\pi/2}\log(\sin(x))\log(\cos(x))\,\mathrm{d}x=\frac\pi2\log(2)^2-\frac{\pi^3}{48}\tag{1} $$ Here is a way to extend kalpeshmpopat's suggestion about substituting $x\mapsto\frac\pi2-x$. Note that $g(x)=f(\sin(x))f(\cos(x))$ is even as a function of $x-\frac\pi4$; that is, $g(\frac\pi2-x)=g(x)$. Thus, if we multiply by an odd function of $x-\frac\pi4$, the integral over $[0,\frac\pi2]$ will be $0$.

Therefore, $$ \int_0^{\pi/2}\left(\frac\pi4-x\right)\log(\sin(x))\log(\cos(x))\,\mathrm{d}x=0\tag{2} $$ Using $(1)$ and $(2)$, we get $$ \begin{align} \int_0^{\pi/2}x\log(\sin(x))\log(\cos(x))\,\mathrm{d}x &=\frac\pi4\int_0^{\pi/2}\log(\sin(x))\log(\cos(x))\,\mathrm{d}x\\ &=\frac\pi4\left(\frac\pi2\log(2)^2-\frac{\pi^3}{48}\right)\\ &=\frac{\pi^2}{8}\log(2)^2-\frac{\pi^4}{192}\tag{3} \end{align} $$ We also have $$ \int_0^{\pi/2}\left(\frac\pi4-x\right)^3\log(\sin(x))\log(\cos(x))\,\mathrm{d}x=0\tag{4} $$ Which, along with $(1)$ and $(3)$, implies that $$ \begin{align} \int_0^{\pi/2}x^3\log(\sin(x))\log(\cos(x))\,\mathrm{d}x &=\frac{3\pi}{4}\int_0^{\pi/2}x^2\log(\sin(x))\log(\cos(x))\,\mathrm{d}x\\ &-\frac{3\pi^2}{16}\int_0^{\pi/2}x\log(\sin(x))\log(\cos(x))\,\mathrm{d}x\\ &+\frac{\pi^3}{64}\int_0^{\pi/2}\log(\sin(x))\log(\cos(x))\,\mathrm{d}x\\ &=\frac{3\pi}{4}\int_0^{\pi/2}x^2\log(\sin(x))\log(\cos(x))\,\mathrm{d}x\\ &-\frac{\pi^4}{64}\log(2)^2+\frac{\pi^6}{1536}\tag{5} \end{align} $$ Equation $(5)$ supports math110's claim that if we know $I_2$, we know $I_3$.

  • $\begingroup$ beatifull!,Thank you! Thank you $\endgroup$ – math110 Mar 15 '13 at 15:41

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{I \equiv \int_{0}^{\pi/2}x^{2}\ln\pars{\sin\pars{x}}\ln\pars{\cos\pars{x}} \,\dd x:\ {\Large ?}}$ Yet another partial idea...

First: Some reductions \begin{align} I&=\int_{0}^{\pi/2}x^{2}\, {\bracks{\ln\pars{\sin\pars{x}} + \ln\pars{\cos\pars{x}}}^{2} -\ln^{2}\pars{\sin\pars{x}} -\ln^{2}\pars{\cos\pars{x}} \over 2}\,\dd x \\[3mm]&=\half\int_{0}^{\pi/2}x^{2}\ln^{2}\pars{\sin\pars{x}\cos\pars{x}}\,\dd x -\half\int_{0}^{\pi/2}x^{2}\ln^{2}\pars{\sin\pars{x}}\,\dd x -\half\int_{0}^{\pi/2}x^{2}\ln^{2}\pars{\cos\pars{x}}\,\dd x \\[3mm]&=\half\int_{0}^{\pi/2}x^{2}\bracks{% \ln^{2}\pars{\sin\pars{2x}} - 2\ln\pars{2}\ln\pars{\sin\pars{2x}} + \ln^{2}\pars{2}} \,\dd x\\[3mm]& -\half\int_{0}^{\pi/2}x^{2}\ln^{2}\pars{\sin\pars{x}}\,\dd x -\half\int_{0}^{\pi/2}x^{2}\ln^{2}\pars{\cos\pars{x}} \,\dd x \\[3mm]&={\pi \over 4}\,\ln^{2}\pars{2} + {1 \over 16}\int_{0}^{\pi}x^{2}\ln^{2}\pars{\sin\pars{x}}\,\dd x - {1 \over 8}\ln\pars{2}\int_{0}^{\pi}x^{2}\ln\pars{\sin\pars{x}}\,\dd x \\[3mm]&-\half\int_{0}^{\pi/2}x^{2}\ln^{2}\pars{\sin\pars{x}}\,\dd x -\half\int_{0}^{\pi/2}x^{2}\ln^{2}\pars{\cos\pars{x}}\,\dd x \qquad\qquad\qquad\qquad\qquad\qquad\pars{1} \end{align}

Also, with $n = 1,2$: \begin{align} &\int_{0}^{\pi}x^{2}\ln^{n}\pars{\sin\pars{x}}\,\dd x= \int_{-\pi/2}^{\pi/2}\pars{x + {\pi \over 2}}^{2}\ln^{n}\pars{\cos\pars{x}}\,\dd x \\[3mm]&=\int_{0}^{\pi/2} \bracks{\pars{x + {\pi \over 2}}^{2} + \pars{-x + {\pi \over 2}}^{2}} \ln^{n}\pars{\cos\pars{x}}\,\dd x =\int_{0}^{\pi/2} \pars{2x^{2} + {\pi^{2} \over 2}}\ln^{n}\pars{\cos\pars{x}}\,\dd x \\[3mm]&=2\int_{0}^{\pi/2}x^{2}\ln^{n}\pars{\cos\pars{x}}\,\dd x + {\pi^{2} \over 2}\int_{0}^{\pi/2}\ln^{n}\pars{\cos\pars{x}}\,\dd x \end{align}

With this result, $\pars{1}$ is reduced to: \begin{align} I&={\pi \over 4}\,\ln^{2}\pars{2} -{3 \over 8}\int_{0}^{\pi/2}x^{2}\ln^{2}\pars{\cos\pars{x}}\,\dd x + {\pi^{2} \over 32}\ \overbrace{\int_{0}^{\pi/2}\ln^{2}\pars{\cos\pars{x}}\,\dd x} ^{\ds{{\pi^{3} \over 24} + {\pi \over 2}\,\ln^{2}\pars{2}}} \\[3mm]&-{1 \over 4}\,\ln\pars{2}\ \overbrace{\int_{0}^{\pi/2}x^{2}\ln\pars{\cos\pars{x}}\,\dd x} ^{\ds{-\,{\pi^{3} \over 24}\,\ln\pars{2} - {1 \over 4}\,\zeta\pars{3}}} -{\pi^{2} \over 16}\,\ln\pars{2}\ \overbrace{\int_{0}^{\pi/2}\ln\pars{\cos\pars{x}}\,\dd x} ^{\ds{-\,{\pi \over 2}\,\ln\pars{2}}} -\half\int_{0}^{\pi/2}x^{2}\ln^{2}\pars{\sin\pars{x}}\,\dd x \end{align}

$I$ is reduced to: $$ I = C -\half\int_{0}^{\pi/2}x^{2}\ln^{2}\pars{\sin\pars{x}}\,\dd x -{3 \over 8}\int_{0}^{\pi/2}x^{2}\ln^{2}\pars{\cos\pars{x}}\,\dd x $$ where $C$ is given by: \begin{align} C &= {\pi \over 4}\,\ln^{2}\pars{2} + {\pi^{5} \over 728} + {\pi^{3} \over 64}\,\ln^{2}\pars{2} + {\pi^{3} \over 96}\,\ln^{2}\pars{2} + {1 \over 16}\,\ln\pars{2}\zeta\pars{3} + {\pi^{3} \over 32}\,\ln^{2}\pars{2} \\[3mm]&= {\pi \over 4}\,\bracks{\ln\pars{2} + {1 \over 4}\,\zeta\pars{3}}\ln\pars{2} + {\pi^{5} \over 728} + {1 \over 16}\,\ln\pars{2}\zeta\pars{3} + {11\pi^{3} \over 192}\,\ln^{2}\pars{2} \end{align} $\tt @sos440$ user had already derived results which are related to the remaining integrals.


Hint: Replace x by π/2- x

then simplify so we will get one term same as I

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    $\begingroup$ When one writes Hint, one is asserting that the step which they indicate does lead to a proof. Yours does not. $\endgroup$ – Did Mar 14 '13 at 6:59
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    $\begingroup$ Let $$ I_k = \int_{0}^{\frac{\pi}{2}} x^k \log \sin x \log \cos x, \, dx $$ for notation convenience. If we follow your hint, the resulting equation, after some cancellations, does not involve $I_2$. Rather, it can be used to find the relationship between $I_0$ and $I_1$. $\endgroup$ – Sangchul Lee Mar 14 '13 at 7:00
  • $\begingroup$ This is clearly the crucial step. $\endgroup$ – Mark McClure Mar 14 '13 at 7:06
  • $\begingroup$ hello,sos440,$ I_{2}$ you can find value? Thank you ,I find if $I_{2}$ can evaluate,that we easy to find $I_{3}$ $\endgroup$ – math110 Mar 14 '13 at 7:36
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    $\begingroup$ 6 upvotes for a misleading answer. Not bad. $\endgroup$ – Did Feb 15 '14 at 21:57

This is not quite a complete answer but goes a good way towards showing that the idea of @kalpeshmpopat is not so far off the mark - if we want to answer the question that was orginally asked.

First, numerical investigation indicates that the correct integral is

$$I=\displaystyle\int_0^{\pi/2}x\ln(\sin x)\ln(\cos x)dx=\dfrac{(\pi\ln{2})^2}{8}-\dfrac{\pi^4}{192}.$$

Now, as @kalpeshmpopat points out, a simple substitution, together with the facts that $\cos(\frac{\pi}{2}-x)=\sin(x)$ and vice-versa, shows that

$$\displaystyle\int_0^{\pi/2}x\ln(\sin x)\ln(\cos x)dx=\int_0^{\pi/2}(\frac{\pi}{2}-x)\ln(\sin x)\ln(\cos x)dx.$$

Thus, if we add these two together we get

$$\displaystyle\int_0^{\pi/2} \frac{\pi}{2} \ln(\sin x)\ln(\cos x)dx=2I.$$

All that remains to show is that

$$\displaystyle\int_0^{\pi/2} \frac{\pi}{2} \ln(\sin x)\ln(\cos x)dx = \frac{1}{96} \pi ^2 \left(6 \log ^2(4)-\pi^2\right),$$

which Mathematica can do. It's getting late, but my guess on this last integral would be to expand $\ln(\cos(x))$ into a power series (which is easy, since we know $\ln(1+y)$) and try to integrate $x^n \ln(\sin(x))$.

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    $\begingroup$ Maybe you have realized this, but OP asks for an integral involving $x^2$, not the one you have worked on. OP also states that he may assume the value of a similar integral involving $x$, in case that helps in finding the answer. The question was always about the $x^2$-integral, but at first OP used the same name $I$ for both integrals, which was confusing. $\endgroup$ – alex.jordan Mar 14 '13 at 20:45
  • $\begingroup$ @alex.jordan Clearly, the question currently asks for an integral involving $x^2$. However, it was edited by someone other than the OP, so I don't think it's necessarily clear what was asked. It wasn't confusing - it was unclear. $\endgroup$ – Mark McClure Mar 14 '13 at 20:50
  • $\begingroup$ Hi Mark, do you know that you can see the complete edit history of a problem clicking the link displaying its last editor? $\endgroup$ – alex.jordan Mar 14 '13 at 20:58
  • $\begingroup$ @alex.jordan Yes. Clearly, the OP is not a native speaker. He easily could have been asking for a verification of an integral whose value was known. Don't get me wrong, I think that Raghib's edit is a perfectly reasonable interpretation. Even so, the more interesting question, it seems to me, is why is the $p=0$ case true? It seems pretty clear given the answers that the substitution $x\rightarrow\pi/2-x$ yields a way to express $I(p)$ in terms of $I(q)$ using all integers $q<p$. $\endgroup$ – Mark McClure Mar 14 '13 at 21:05
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    $\begingroup$ That substitution yields a way to express $I(p)$ in terms of $I(q)$ over $q<p$ when p is odd. Which is why it is not (immediately) helpful for $p=2$. $\endgroup$ – alex.jordan Mar 14 '13 at 21:11

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