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

Solution 1:

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.

Solution 2:

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.

Solution 3:

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 $I_3$.

Solution 4:

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

Solution 5:

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