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

Quanto
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math110
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9 Answers9

58

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.

robjohn
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  • @MhenniBenghorbal: to be honest, the last equation in $(7)$ requires a page and a half of computation. I can append those if requested, but I didn't want to unnecessarily clutter the answer. – robjohn Mar 15 '13 at 21:38
  • @robjohn: Check this. It may help you to evaluate $(7)$. – Mhenni Benghorbal Mar 15 '13 at 21:51
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    @MhenniBenghorbal: Thanks! Using those functions, $(7)$ is $$ \frac{\pi^3}{24}\log(2)^2+\log(2)\frac\pi{16}\zeta(3)+\frac{11\pi^5}{5760} -\frac\pi4(A(1,3)+A(2,2)) $$ – robjohn Mar 15 '13 at 22:05
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    (+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. – Sangchul Lee Mar 16 '13 at 04:06
  • @robjohn I believe it might be proved symbolically in Mathematica. I'll check if it can be done. By the way I'm Artes not Aries. – Artes Mar 16 '13 at 12:35
  • @robjohn: No problem. (+1) for stamina and endurance. – Mhenni Benghorbal Mar 16 '13 at 13:22
  • @Artes: Ah, my eyesight is not what it used to be. My apologies. – robjohn Mar 16 '13 at 14:27
  • @Artes: I know I have seen symbolic proofs on W|A, but I've never seen them in Mathematica. If it can be done, I would be greatly interested. – robjohn Mar 16 '13 at 14:39
  • Thank you robjohn! very nice. – math110 Mar 19 '13 at 04:09
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    @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. – Shobhit Bhatnagar Jan 11 '14 at 08:55
  • @IntegralsandSeries: Thanks. I had forgotten about this answer. I have done more work in Euler Series and I should revisit that sum. – robjohn Jan 11 '14 at 12:54
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    Can someone please post the full answer, then? I mean, (7) with the sigma replaced by Integrals-and-Series's answer. – Akiva Weinberger Aug 24 '14 at 02:30
  • @AkivaWeinberger For the record : $I=\dfrac {1} {16}\left(\dfrac {\pi^5} {20} +\pi^3\log^22- \dfrac13 \pi\log^42-6\pi \zeta(3)\log2-8\pi \text{ Li}_4(\frac {1} {2})\right)$ $= \dfrac {1} {28} \left(\dfrac {7\pi^5} {80} +\dfrac {71} {12} \pi\log^32\text{ Li}_1(\frac {1} {2})+9 \pi\log^22\text{ Li}_2(\frac {1} {2}) -12\pi\log2\text{ Li}_3(\frac {1} {2}) -14 \pi \text{ Li}_4(\frac {1} {2}) \right)$. – Wolfgang Nov 10 '19 at 11:30
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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.

Sangchul Lee
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  • 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. – robjohn Mar 16 '13 at 04:15
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    For the last integral you can use the following technique since we have $$\int_0^{\Large\frac\pi2}\cos^{n-1}x\cos ax\ dx=\frac{\pi}{2^n n\ \operatorname{B}\left(\frac{n+a+1}{2},\frac{n-a+1}{2}\right)}$$ – Tunk-Fey Sep 18 '14 at 13:02
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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 $I_3$.

robjohn
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10

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 $D_b=\frac{\partial }{\partial b}$.

6

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

Felix Marin
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  • The first term of $C$ should be $\frac{\pi^3}{48}\ln^2(2)$. This little mistake came from integrating $\int_0^{\pi/2} \ln^2(2)dx$ instead of $\int_0^{\pi/2} \ln^2(2)x^2 dx$ – Ali Shadhar Jun 13 '22 at 23:00
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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 $x^n \ln(\sin(x))$.

Mark McClure
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    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. – 2'5 9'2 Mar 14 '13 at 20:45
  • @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. – Mark McClure Mar 14 '13 at 20:50
  • Hi Mark, do you know that you can see the complete edit history of a problem clicking the link displaying its last editor? – 2'5 9'2 Mar 14 '13 at 20:58
  • @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 $q<p$. – Mark McClure Mar 14 '13 at 21:05
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    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 $p=2$. – 2'5 9'2 Mar 14 '13 at 21:11
1

Hint: Replace x by π/2- x

then simplify so we will get one term same as I

kalpeshmpopat
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1

I am going to continue @Felix answer above where he found

$$I = C-\frac12\int_0^{\pi/2}x^2\ln^2(\sin x)dx-\frac38\underbrace{\int_0^{\pi/2}x^2\ln^2(\cos x)dx}_{x\to \pi/2-x}$$

$$=C-\frac12\int_0^{\pi/2}x^2\ln^2(\sin x)dx-\frac38\int_0^{\pi/2}(\pi/2-x)^2\ln^2(\sin x)dx$$

$$=C-\int_0^{\pi/2}\left(\frac{7}{8}x^2-\frac{3\pi}{8}x+\frac{3\pi^2}{32}\right)\ln^2(\sin x)dx$$

Using

$$\ln^2(2\sin x)=\left(\frac{\pi}{2}-x\right)^2+2\sum_{n=1}^\infty\frac{H_{n-1}}{n}\cos(2nx),\quad 0<x<\pi\tag{1}$$

we have

$$\int_0^{\pi/2}\left(\frac{7}{8}x^2-\frac{3\pi}{8}x+\frac{3\pi^2}{32}\right)\ln^2(2\sin x)dx$$ $$=\int_0^{\pi/2}\left(\frac{\pi}{2}-x\right)^2\left(\frac{7}{8}x^2-\frac{3\pi}{8}x+\frac{3\pi^2}{32}\right)dx$$

$$+2\sum_{n=1}^\infty\frac{H_{n-1}}{n}\int_0^{\pi/2} \left(\frac{7}{8}x^2-\frac{3\pi}{8}x+\frac{3\pi^2}{32}\right)\cos(2nx)dx$$

$$=\frac{11\pi^5}{3840}+2\sum_{n=1}^\infty\frac{H_{n-1}}{n}\left(\frac{3\pi}{32 n^2}+\frac{(-1)^n \pi}{8n^2}\right)$$

$$=\frac{11\pi^5}{3840}+\frac{3\pi}{16}\sum_{n=1}^\infty\frac{H_{n-1}}{n^3}+\frac{\pi}{4}\sum_{n=1}^\infty\frac{(-1)^n H_{n-1}}{n^3}$$

$$=\frac{\pi}{2}\text{Li}_4(1/2)-\frac{7\pi^5}{2304}-\frac{\pi^3}{48}\ln^2(2)+\frac{7\pi}{16}\ln(2)\zeta(3)+\frac{\pi}{48}\ln^4(2).$$

What is remaining is to write $\ln^2(2\sin x)=\ln^2(2)+2\ln(2)\ln(\sin x)+\ln^2(\sin x)$ where the middle integral can be done using the Fourier series of $\ln(\sin x)$:

$$\int_0^{\pi/2}\left(\frac{7}{8}x^2-\frac{3\pi}{8}x+\frac{3\pi^2}{32}\right)\ln(\sin x)dx=-\frac{7\pi^3}{192}\ln(2).$$


Proof of (1): \begin{gather*} \sum_{n=1}^\infty(-1)^n\frac{H_{n-1}}{n}\cos(2nx)=\mathfrak{R}\sum_{n=1}^\infty(-1)^n\frac{H_{n-1}}{n}e^{2inx}\\ =\mathfrak{R}\sum_{n=1}^\infty\frac{H_{n-1}}{n}(-e^{2ix})^n\\ \{\text{replace $x$ by $-e^{2ix}$ in the generating function of $\sum_{n=1}^\infty H_{n-1} x^n/n$}\}\\ =\frac12\mathfrak{R}\ln^2(1+e^{2ix})\\ =\frac12\mathfrak{R}\left[\ln(2\cos x)+ix\right]^2\\ =\frac12\mathfrak{R}\left[\ln^2(2\cos x)+2ix\ln(2\cos x)-x^2\right]\\ =\frac12\ln^2(2\cos x)-\frac{x^2}{2}, \end{gather*}

replace $x$ by $\frac{\pi}{2}-x$ using $(\cos(2n(\frac{\pi}{2}-x)))=(-1)^n\cos(2nx)$ for integer $n$, we have

$$\ln^2(2\sin x)=\left(\frac{\pi}{2}-x\right)^2+2\sum_{n=1}^\infty\frac{H_{n-1}}{n}\cos(2nx),\quad 0<x<\pi$$

Ali Shadhar
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1

Presented below is a complete solution that evaluates the integral to the following close-form \begin{align} &\int_0^{\pi/2}x^2\ln(\sin x)\ln(\cos x)\ dx\\ =& -\frac\pi2 \text{Li}_4(\frac12)-\frac{3\pi}8\ln2\ \zeta(3) +\frac{\pi^5}{320}+\frac{\pi^3}{16}\ln^22-\frac\pi{48}\ln^42\tag1 \end{align} To derive (1), evaluate the integrals below sequentially, with $\int_0^{\pi/2}\ln^2(2\sin x)dx=\frac{\pi^3}{24}$ and $\int_0^{\pi/2}x^2 \ln^2(2\cos x) dx=\frac{11\pi^5}{1440}$ \begin{align} &\int_0^{\pi/2}x^2 \overset{x\to\frac\pi2-x}{\ln^2(2\sin x)}\ dx =\pi\int_0^{\pi/2}x \ln^2(2\sin x)\ dx-\frac{\pi^5}{360}\\ \\ &\int_0^{\pi/2}x^2 \ln^2(2\sin 2x)\overset{2x\to x}{dx} \\ =& \ \frac18\int_0^{\pi/2}(\pi^2-2\pi x+2x^2 ) \ln^2(2\sin x)\ dx=\frac{13\pi^5}{2880}\\ \\ &\int_0^{\pi/2}x^2\ln(2\sin x)\ln(2\cos x)\ dx\\ =& \ \frac12 \int_0^{\pi/2}[x^2 \ln^2(2\sin 2x)-x^2 \ln^2(2\sin x)-x^2 \ln^2(2\cos x)]\ dx\\ =& - \frac{\pi^5}{5760}-\frac\pi2 \int_0^{\pi/2}x \ln^2(2\sin x)\ dx\ \end{align}

Writing out both sides of the last integral and utilizing below to yield the close-form (1) \begin{align} \int_0^{\pi/2}x\ln(2\sin x)dx=\frac7{16}\zeta(3),\>\>\> \int_0^{\pi/2}x^2\ln(2\sin x)dx=\frac{3\pi}{16}\zeta(3)\\ \end{align} $$\int_0^{\pi/2}x\ln^2(\sin x)dx = \text{Li}_4(\frac12) -\frac{19\pi^4}{2880}+\frac{\pi^2}{12}\ln^22+\frac1{24}\ln^42 $$

Quanto
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