Compute $\int_{0}^{1}\frac{\ln(1+x)\ln(1+x^2)}{x}\,dx=\frac{\pi}{2}G-\frac{33}{32}\zeta(3)$

Question

I saw the following result and I am trying to prove it. $G$ is Catalan´s constant.

$$\boxed{\int_{0}^{1}\frac{\ln(1+x)\ln(1+x^2)}{x}\,dx=\frac{\pi}{2}G-\frac{33}{32}\zeta(3)}$$

I could not figure out any substitution or other trick to make it simple, so I proceeded expanding $\ln(1+x^2)$. But I ended up with a not so friendly expression. Any hint is welcome.

$$\int_{0}^{1}\frac{\ln(1+x)\ln(1+x^2)}{x}\,dx=\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}\int_{0}^{1}x^{2k-1}\ln(1+x)\,dx$$

Integrating by parts

\begin{align*} &=\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}\Biggl\{\frac{(x^{2k}-1)}{2k}\ln(1+x)\Bigg|_{0}^{1}-\frac{1}{2k}\int_{0}^{1}\frac{x^{2k}-1}{1+x}\,dx \Biggr\}\\ &=\frac{1}{2}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k^2}\int_{0}^{1}\frac{1-x^{2k}}{1+x}\,dx\\ &=\frac{1}{2}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k^2}\sum_{n=0}^{\infty}(-1)^n\int_{0}^{1}(1-x^{2k})x^n\,dx \end{align*}

Maybe, this last integral can be tackled in a different way.

Answer 1

\begin{align}J&=\int_{0}^{1}\frac{\ln(1+x)\ln(1+x^2)}{x}\,dx\\ &\overset{\text{IBP}}=-\underbrace{\int_0^1 \frac{\ln x\ln(1+x^2)}{1+x}dx}_{=A}-\underbrace{\int_0^1 \frac{2x\ln x\ln(1+x)}{1+x^2}dx}_{=B}\\ A&\overset{\text{IBP}}=\left[\left(\int_0^x \frac{\ln t}{1+t}dt\right)\ln(1+x^2)\right]_0^1 -\int_0^1 \frac{2x}{1+x^2}\left(\int_0^x \frac{\ln t}{1+t}dt\right)dx\\ &=-\frac{1}{12}\pi^2\ln 2-\int_0^1 \int_0^1\frac{2x^2\ln(tx)}{(1+x^2)(1+tx)}dtdx\\ &=-\frac{1}{12}\pi^2\ln 2+\int_0^1 \int_0^1\left(\frac{2\ln(tx)}{(1+t^2)(1+x^2)}-\frac{2tx\ln(tx)}{(1+t^2)(1+x^2)}-\frac{2\ln(tx)}{(1+x^2)(1+tx)}\right)dtdx\\ &=-\frac{1}{12}\pi^2\ln 2+4\int_0^1 \int_0^1 \frac{\ln t}{(1+t^2)(1+x^2)}dtdx-4\int_0^1 \int_0^1 \frac{tx\ln x}{(1+t^2)(1+x^2)}dtdx-\\&2\int_0^1\int_0^1 \frac{\ln(tx)}{(1+x^2)(1+tx)}dtdx\\ &\overset{\text{Fubini}}=-\frac{1}{12}\pi^2\ln 2-\pi\text{G}+\frac{1}{24}\pi^2\ln 2-2\int_0^1\int_0^1 \frac{\ln(tx)}{(1+x^2)(1+tx)}dtdx\\ &=-\frac{1}{24}\pi^2\ln 2-\pi\text{G}-2\int_0^1\int_0^1 \frac{\ln(tx)}{(1+x^2)(1+tx)}dtdx\\ &\overset{u(t)=tx}=-\frac{1}{24}\pi^2\ln 2-\pi\text{G}-2\int_0^1 \frac{1}{x(1+x^2)}\left(\int_0^x \frac{\ln u}{1+u}du\right)dx\\ &\overset{\text{IBP}}=-\frac{1}{24}\pi^2\ln 2-\pi\text{G}+\left[\big(\ln(1+x^2)-2\ln x\big)\left(\int_0^x \frac{\ln u}{1+u}du\right)\right]_0^1-\\&\int_0^1 \big(\ln(1+x^2)-2\ln x\big)\frac{\ln x}{1+x}dx\\ &=-\frac{1}{8}\pi^2\ln 2-\pi\text{G}+3\zeta(3)-A\\ A&=\boxed{-\dfrac{1}{16}\pi^2\ln 2-\dfrac{1}{2}\pi\text{G}+\dfrac{3}{2}\zeta(3)}\\ B^\prime&=\int_0^1 \frac{2x\ln x\ln(1-x)}{1+x^2}dx\\ B+B^\prime&\overset{y=x^2}=\frac{1}{2}\int_0^1 \frac{\ln y\ln(1-y)}{1+y}dy\\ B^\prime-B&\overset{y=\frac{1-x}{1+x}}=\int_0^1 \frac{2(1-y)\ln\left(\frac{1-y}{1+y}\right)\ln y}{(1+y)(1+y^2)}dy\\ &=2\int_0^1 \frac{\ln(1-y)\ln y}{1+x}dy-2\int_0^1 \frac{\ln(1+y)\ln y}{1+x}dy-(B^\prime-B)\\ &=\int_0^1 \frac{\ln(1-y)\ln y}{1+x}dy-\int_0^1 \frac{\ln(1+y)\ln y}{1+x}dy\\ B&=\frac{1}{2}\left(B+B^\prime\right)-\frac{1}{2}\left(B^\prime-B\right)\\ &=\frac{1}{2}\underbrace{\int_0^1 \frac{\ln(1+y)\ln y}{1+x}dy}_{=B_1}-\frac{1}{4}\underbrace{\int_0^1 \frac{\ln(1-y)\ln y}{1+x}dy}_{=B_2}\\ \end{align} \begin{align}B_1&=\int_0^1\frac{\ln(1+x)\ln x}{1+x}dx\\&=\frac{1}{2}\left(\int_0^1\frac{\ln^2(1+x)}{1+x}dx+\int_0^1\frac{\ln^2 x}{1+x}dx-\underbrace{\int_0^1 \frac{\ln^2\left(\frac{x}{1+x}\right)}{1+x}dx}_{y=\frac{x}{1+x}}\right)\\ &=\frac{1}{6}\ln^3 2+\frac{1}{2}\int_0^1\frac{\ln^2 x}{1+x}dx-\frac{1}{2}\int_0^{\frac{1}{2}}\frac{\ln^2 y}{1-y}dy\\ \end{align} On the other hand, \begin{align} B_1&\overset{y=\frac{1}{1+x}}=-\int_{\frac{1}{2}}^1\frac{\ln\left(\frac{1-y}{y}\right)\ln y}{y}dy\\ &=\frac{1}{3}\ln^3 2-\int_{\frac{1}{2}}^1 \frac{\ln\left(1-y\right)\ln y}{y}dy\\ &\overset{\text{IBP}}=\frac{1}{3}\ln^3 2-\frac{1}{2}\Big[\ln^2 y\ln(1-y)\Big]_0^1 +\frac{1}{2}\int_{\frac{1}{2}}^1 \frac{\ln^2 y}{1-y}dy\\ &=-\frac{1}{6}\ln^3 2-\frac{1}{2}\int_{\frac{1}{2}}^1 \frac{\ln^2 y}{1-y}dy\\ &=-\frac{1}{6}\ln^3 2-\frac{1}{2}\int_0^1 \frac{\ln^2 y}{1-y}dy+\frac{1}{2}\int_0^{\frac{1}{2}} \frac{\ln^2 y}{1-y}dy \end{align} Therefore, \begin{align}2B_1&=-\frac{1}{2}\int_0^1 \frac{2x\ln^2 x}{1-x^2}dx\\ &\overset{y=x^2}=-\frac{1}{8}\int_0^1 \frac{\ln^2 y}{1-y}dy\\ B_1&=\boxed{-\dfrac{1}{16}\int_0^1 \dfrac{\ln^2 y}{1-y}dy}\\ \end{align} \begin{align} B_2&\overset{\text{IBP}}=\left[\left(\int_0^x \frac{\ln t }{1+t}dt -\int_0^1 \frac{\ln x }{1+x}dx\right)\ln(1-x)\right]_0^1\\&+\int_0^1 \frac{1}{1-x}\left(\int_0^x \frac{\ln t }{1+t}dt-\int_0^1 \frac{\ln t }{1+t}dt\right)dx\\ &=\int_0^1 \int_0^1 \frac{1}{1-x}\left(\frac{x\ln(tx)}{1+tx}-\frac{\ln t}{1+t}\right)dtdx\\ &=\int_0^1 \int_0^1\left(\frac{\ln(tx)}{(1+t)(1-x)}-\frac{\ln(tx)}{(1+t)(1+tx)}-\frac{\ln t}{(1+t)(1-x)}\right)dtdx\\ &=-\frac{1}{6}\pi^2\ln 2-\int_0^1 \frac{1}{t(1+t)} \left(\int_0^t\frac{\ln u}{1+u}du\right)dt\\ &\overset{\text{IBP}}=-\frac{1}{6}\pi^2\ln 2+\left[\big(\ln(1+t)-\ln t \big)\left(\int_0^t\frac{\ln u}{1+u}du\right)\right]_0^1+\int_0^1\frac{\ln^2 t}{1+t}dt-\\&\int_0^1\frac{\ln(1+t)\ln t}{1+t}dt\\ &=\frac{3}{2}\zeta(3)-\frac{1}{4}\pi^2\ln 2-B_1\\ &=\frac{13}{8}\zeta(3)-\frac{1}{4}\pi^2\ln 2\\ B&=\boxed{\frac{1}{16}\pi^2\ln 2-\frac{15}{32}\zeta(3)}\\ \end{align} \begin{align}\boxed{J=\dfrac{1}{2}\pi\text{G}-\dfrac{33}{32}\zeta(3)}\end{align}

NB: I assume that, \begin{align}\int_0^1\frac{\ln t}{1+t}dt&=-\frac{1}{12}\pi^2, \int_0^1\frac{\ln t}{1-t}dt=-\frac{1}{6}\pi^2, \int_0^1\frac{\ln^2 t}{1+t}dt=\frac{3}{2}\zeta(3), \int_0^1\frac{\ln^2 t}{1-t}dt=2\zeta(3)\\ \end{align}

Answer 2

Starting from your last equation $$S=\frac{1}{2}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k^2}\sum_{n=0}^{\infty}(-1)^n\int_{0}^{1}(1-x^{2k})\,x^n\,dx$$ $$\int_{0}^{1}(1-x^{2k})\,x^n\,dx=\frac{2 k}{(n+1) (2 k+n+1)}$$ $$\sum_{n=0}^{\infty}(-1)^n\int_{0}^{1}(1-x^{2k})\,x^n\,dx=\log (2)-\Phi (-1,1,2 k+1)$$ $$S=\frac{1}{24} \pi ^2 \log (2)+\frac 12 \sum_{k=1}^{\infty}(-1)^{k}\,\frac{\Phi (-1,1,2 k+1)}{k^2}$$ and now, I am stuck !

Knowing the result gives a nice identity for the last summation $$\sum_{k=1}^{\infty}(-1)^{k}\,\frac{\Phi (-1,1,2 k+1)}{k^2}=\pi C-\frac{33 }{16}\zeta (3)-\frac{\pi ^2}{12} \log (2)$$

Answer 3

Note $\ln(1+x)= \frac12\ln(1-x^2)- \frac12\ln\frac{1-x}{1+x}$. Then

\begin{align} &\int_{0}^{1}\frac{\ln(1+x)\ln(1+x^2)}{x}\,dx=\frac12(J_1-J_2)\tag1 \end{align} \begin{align} J_1= &\int_0^1 \frac{\ln(1-x^2)\ln(1+x^2)}{x}\,dx =\frac14\int_0^1 {\frac{\overset{1-x^4\to x}{\ln^2(1-x^4)}- \ln^2\overset{\to x}{\frac{1-x^2}{1+x^2}}}x}\\ = &-\frac5{32}\int_0^1 \frac{\ln^2 x}{1-x}dx =-\frac5{16}\zeta(3)\tag2 \end{align}

\begin{align} J_2= & \int_0^1 \frac{\ln\frac{1-x}{1+x}\ln(1+x^2)}{x}\,dx = \int_0^1 \ln(1+x^2)\>d\left( \int_0^x \frac{\ln\frac{1-t}{1+t} }t\,dt\right)\\ = &\>\ln2 \int_0^1 \frac1t \ln \overset{\to t}{\frac{1-t}{1+t}}dt -\int_0^1 \frac{2x}{1+x^2} \overset{t=xy\> then \> ibp}{\left( \int_0^x \frac1t \ln\frac{1-t}{1+t}\,dt\right)}dx\\ =&\>2\ln2 \int_0^1 \frac{\ln t}{1-t^2}dt-4\int_0^1 \int_0^1 \frac{x^2\ln y}{(1+x^2)(1-x^2y^2)}dx\>dy\\ =& -\frac{\pi^2}4\ln2 -\pi G+\frac72\zeta(3) -2\int_0^1 \frac{y\ln y\ln\frac{1-y}{1+y}}{1+y^2}dy\tag3 \end{align} where the last equality is arrived with the decomposition $$\frac{x^2}{(1+x^2)(1-x^2y^2)} =\frac1{1+y^2}\left(-\frac1{1+x^2} +\frac1{2(1+xy)} +\frac1{2(1-xy)} \right) $$ and $\int_0^1 \frac{\ln t}{1-t^2}dt= -\frac{\pi^2}8$, $ \int_0^1 \frac{\ln y}{1+y^2}dy =-G$, $\int_0^1 \frac{\ln^2y}{1-y^2}dy=\frac74\zeta(3) $. Evaluate the remaining integral in (3) as follows \begin{align} & \int_0^1 \frac{y\ln y\ln\overset{\to y} {\frac{1-y}{1+y}}}{1+y^2}dy\\ = &\frac12 \int_0^1 \frac{\ln y\ln(1-y)}{1+y}dy - \frac12 \int_0^1 \frac{\ln y\ln(1+y)}{1+y}dy\\ \overset{ibp} =&\frac12 \int_0^1 \frac{\ln y\ln(1+y)}{1-y}dy -J_1 + \frac14 \int_0^1 \frac{\ln^2(1+y)}{y}dy\\ =&\frac12 \int_0^1 \ln(1+y)\>d\left( \int_0^y \frac{\ln t}{1-t}dt \right)_{t=yx}+ \frac5{16}\zeta(3)+ \frac1{16}\zeta(3)\\ =& \frac12 \ln2\int_0^1 \frac{\ln t}{1-t}dt -\frac12 \int_0^1\int_0^1 \frac{y\ln y+\overset{x\leftrightarrow y}{y \ln x}}{(1+y)(1-yx)} dxdy + \frac3{8}\zeta(3)\\ =& -\frac{\pi^2}{12}\ln2 -\frac12 \int_0^1\int_0^1 \frac{\ln y}{1-yx} -\frac{\ln y}{(1+y)(1+x)} dxdy + \frac3{8}\zeta(3)\\ = & -\frac{\pi^2}{12}\ln2 +\frac12 \zeta(3)-\frac{\pi^2}{24}\ln2+ \frac3{8}\zeta(3) = -\frac{\pi^2}8 \ln2 +\frac78\zeta(3)\tag4 \end{align} Substitute (4) into (3), and then (2) and (3) into (1) to obtain

$$\int_{0}^{1}\frac{\ln(1+x)\ln(1+x^2)}{x}\,dx= \frac{\pi}{2}G-\frac{33}{32}\zeta(3)$$ As a by-product

$$\int_{0}^{1}\frac{\ln(1-x)\ln(1+x^2)}{x}\,dx= \frac12(J_1+J_2)=-\frac{\pi}{2}G+\frac{23}{32}\zeta(3)$$

Answer 4

My solution is a little extensive, but something different here. Let's start by noting that $\displaystyle 2\Re\left\{\ln(1-ix)\right\}=\ln\left(1+x^2\right)$, so we can write: \begin{equation*} \int_0^1\frac{\ln(1+x)\ln\left(1+x^2\right)}{x}\mathrm dx=\Re\left\{\int_0^1\frac{2\ln(1+x)\ln(1-ix)}{x}\mathrm dx\right\},\tag{1} \end{equation*} using $2ab=a^2+b^2-(a-b)^2$ with $a=\ln(1+x)$ and $b=\ln(1-ix)$, the inner integral brackets in (1) is: \begin{align*} \int_0^1\frac{2\ln(1+x)\ln(1-ix)}{x}\mathrm dx&=\int_0^1\left\{\ln^2(1-ix)+\ln^2(1+x)-\ln^2\left(\frac{1+x}{1-ix}\right)\right\}\frac{\mathrm dx}{x}\\ &\hspace{-3cm}=\int_0^1\frac{\ln^2(1-ix)}{x}\mathrm dx+\int_0^1\frac{\ln^2(1+x)}{x}\mathrm dx-\int_0^1\ln^2\left(\frac{1+x}{1-ix}\right)\frac{\mathrm dx}{x}.\tag{2} \end{align*} Let's evaluate $\displaystyle \int_0^1\frac{\ln^2(1-ix)}{x}\mathrm dx$, so let $1-ix\mapsto x$: \begin{align*} \int_0^1\frac{\ln^2(1-ix)}{x}\mathrm dx&=\int_{1-i}^1\frac{\ln^2(x)}{1-x}\mathrm dx\\ &=\sum_{n=1}^\infty\int_{1-i}^1 x^{n-1}\ln^2(x)\mathrm dx\\ &=\sum_{n=1}^\infty\frac{\mathrm d^2}{\mathrm dn^2}\left[\frac1n-\frac{(1-i)^n}{n}\right]\\ &\hspace{-2cm}=2\zeta(3)-\sum_{n=1}^\infty\left[\ln^2(1-i)\frac{(1-i)^n}{n}-2\ln(1-i)\frac{(1-i)^n}{n^2}+2\frac{(1-i)^n}{n^3}\right]\\ &=2\zeta(3)+\ln^2(1-i)\ln(i)+2\ln(1-i)\text{Li}_2(1-i)-2\text{Li}_3(1-i),\tag{3} \end{align*} where we have the values: \begin{align*} \begin{cases} \ln^2(1-i)\ln(i)&=\displaystyle\frac{3}{4}\zeta(2)\ln(2)-\left(\frac{\pi^3}{32}-\frac\pi 8\ln^2(2)\right)i\\ \\ 2\ln(1-i)\text{Li}_2(1-i)&=\displaystyle-\frac{\pi G}{2}-\frac{3}{8}\zeta(2)\ln(2)-2\left(\frac12 G\ln(2)+\frac{\pi^3}{64}+\frac\pi 8\ln^2(2)\right)i\\ \\ 2\Re\left\{\zeta(3)-\text{Li}_3(1-i)\right\}&=\displaystyle\frac{29}{32}\zeta(3)-\frac38\zeta(2)\ln(2) \end{cases} \end{align*} The above values can be derived using the Trilog and Dilog identities: \begin{align*} \therefore\quad\text{Li}_3(z)+\text{Li}_3(1-z)+\text{Li}_3\left(1-\frac1z\right)&=\zeta(3)+\frac{\ln^3(z)}{6}+\zeta(2)\ln(z)-\frac{\ln^2(z)\ln(1-z)}{2}\\ \text{Li}_2(z)+\text{Li}_2(1-z)&=\zeta(2)-\ln(z)\ln(1-z) \end{align*} and the $\displaystyle\ln(i)=\frac{i\pi}{2}$, $\displaystyle\ln(1-i)=\frac12\ln(2)-\frac{i\pi}{4}$ values. As we are dealing with the real part of the values above then $\Re\left\{\text{Li}_3(1-i)+\text{Li}_3(\overline{1-i})\right\}=2\Re\left\{\text{Li}_3(1-i)\right\}$, leaving the third equation of the system justified. Therefore: \begin{equation*} \therefore\ \Re\left\{\int_0^1\frac{\ln^2(1-ix)}{x}\mathrm dx\right\}=-\frac{\pi G}2+\frac{29}{32}\zeta(3).\tag{4} \end{equation*}

Now, for $\displaystyle \int_0^1\frac{\ln^2(1+x)}{x}\mathrm dx$ we have straightforward the result: \begin{equation*} \therefore\quad\int_0^1\frac{\ln^2(1+x)}{x}\mathrm dx=\frac14\zeta(3),\tag{5} \end{equation*} in which it can be easily calculated.

For $\displaystyle \int_0^1\ln^2\left(\frac{1+x}{1-ix}\right)\frac{\mathrm dx}{x}$ in (2), take the replacement $\displaystyle \frac{1+x}{1-ix}\mapsto x$ yielding: \begin{align*} \int_0^1\ln^2\left(\frac{1+x}{1-ix}\right)\frac{\mathrm dx}{x}&=(1+i)\int_{1+i}^1\frac{(1+ix)\ln^2(x)}{(1-x)(1+ix)^2}\mathrm dx\\ &=(1+i)\int_{1+i}^1\frac{\ln^2(x)}{1+i}\left(\frac{1}{1-x}+\frac{i}{1+ix}\right)\mathrm dx\\ &=\int_{1+i}^1\frac{\ln^2(x)}{1-x}\mathrm dx+i\int_{1+i}^1\frac{\ln^2(x)}{1+ix}\mathrm dx\\ &=\sum_{n=1}^\infty \frac{\mathrm d^2}{\mathrm dn^2}\int_{1+i}^1 x^{n-1}\mathrm dx-\sum_{n=1}^\infty(-i)^n\frac{\mathrm d^2}{\mathrm dn^2}\int_{1+i}^1 x^{n-1}\mathrm dx,\tag{6} \end{align*} where, the terms in common in both sums above, are: \begin{equation*} \therefore\quad\frac{\mathrm d^2}{\mathrm dn^2}\int_{1+i}^1 x^{n-1}\mathrm dx=-\frac{\ln^2\left(1+i\right)(1+i)^n}{ n}-\frac{2(1+i)^n}{n^3}+\frac{2}{n^3}+\frac{2\ln\left(1+i\right)(1+i)^n}{n^2}. \end{equation*} Applying the summation in (6), we have: \begin{align*} &=\ln^2(1+i)\ln(-i)-2\text{Li}_3\left(1+i\right)+2\ln(1+i)\text{Li}_2(1+i)+2\zeta(3)\\ &\quad-\ln^2(1+i)\ln\left(i\right)+2\text{Li}_3\left(1-i\right)-2\ln(1+i)\text{Li}_2\left(1-i\right)-2\text{Li}_3(-i)\\ &=\ln^2(1+i)\left\{\ln(-i)-\ln(i)\right\}+2\ln(1+i)\left\{\text{Li}_2(1+i)-\text{Li}_2(1-i)\right\}\\ &\quad+2\left\{\text{Li}_3(1-i)-\text{Li}_3(1+i)\right\}+2\zeta(3)-2\text{Li}_3(-i) \end{align*}

Assuming the following values: \begin{align*} \begin{cases} \ln^2(1+i)\left\{\ln(-i)-\ln(i)\right\}&=\displaystyle \frac32\zeta(2)\ln(2)+\frac1{16}\left(\pi^3-4\pi\ln^2(2)\right)i\\ \\ \displaystyle 2\ln(1+i)\left\{\text{Li}_2(1+i)-\text{Li}_2(1-i)\right\}&=\displaystyle-\pi G-\frac32\zeta(2)\ln(2)+\left(2G\ln(2)+\frac\pi 2\ln^2(2)\right)i\\ \\ \displaystyle 2\Re\left\{\text{Li}_3(1-i)-\text{Li}_3(1+i)\right\}&=2\Re\left\{\text{Li}_3(1-i)-\text{Li}_3(\overline{1-i})\right\}=0\\ \\ \displaystyle 2\left\{\zeta(3)-\text{Li}_3(-i)\right\}&=\displaystyle \frac{35}{16}\zeta(3)-\frac{\pi^3}{16}i. \end{cases} \end{align*}

We have finally: \begin{equation*} \therefore\ \Re\left\{\int_0^1\ln^2\left(\frac{1+x}{1-ix}\right)\frac{\mathrm dx}{x}\right\}=-\pi G+\frac{35}{16}\zeta(3).\tag{7} \end{equation*}

Collecting (7), (5) and (4) so taking (2) and consequently (1), the results follows naturally: \begin{align*} \therefore\quad\int_0^1\frac{\ln(1+x)\ln(1+x^2)}{x}\mathrm dx=\frac{\pi G}{2}-\frac{33}{32}\zeta(3).\qquad\blacksquare \end{align*}