What's the value of $\int_0^1\frac{1}{2y} \ln(y) \ln^2(1-y) \, dy$?

Question

I came across this integral while doing a different problem:

$$ \int_0^1\frac{1}{2y} \ln (y)\ln^2(1-y) \, dy$$

I think we can evaluate this integral by differentiating the common integral representation of the beta function, but it seems to get a bit messy.

Answer 1

We start by introducing the integral $$ I(a,b)=\frac{1}{2}\int_0^1y^{a-1}(1-y)^b\,dy=\frac{1}{2}B(a,1+b), $$ where $B$ denotes the beta function. Note that this integral is singular at $a=0$ and $b=-1$. Since $\partial_a y^a=y^a\ln y$ we are led to calculate $$ \partial_{a,b,b}I(a,b)=\frac{1}{2}\int_0^1 y^{a-1}(1-y)^b\ln y\bigl(\ln(1-y)\bigr)^2\,dy $$ as $a$ and $b$ tend to $0$. We will below inser the "non-dangerous" point $b=0$. In other words, we want to calculate $$ \partial_{a,b,b}B(a,1+b)\mid_{a\to 0^+,b\to 0}. $$ When differentiating the beta function, polygammas appear. Indeed, $$ \begin{aligned} \partial_bB(a,1+b)&=B(a,1+b)\bigl(\psi_0(1+b)-\psi_0(1+a+b)\bigr)\\ \partial_{b,b}B(a,1+b)&=B(a,1+b)\Bigl(\bigl(\psi_0(1+b)-\psi_0(1+a+b)\bigr)^2 +\psi_1(1+b)-\psi_1(1+a+b)\Bigr). \end{aligned} $$ Next, we can actually insert $b=0$ before we differentiate with respect to $a$ and take the limit $a\to 0$. We should differentiate the function (here we have used the facts that $\psi_0(1)=-\gamma$ (Euler's constant) and that $\psi_1(1)=\pi^2/6$) $$ f(a)=B(a,1)\Bigl(\bigl(\gamma+\psi_0(1+a)\bigr)^2+\frac{\pi^2}{6}-\psi_1(1+a)\Bigr) $$ and calculate $\lim_{a\to 0^+}f'(a)$. We get that $$ \begin{aligned} f'(a)&=B(a,1)\bigl(\psi_0(a)-\psi_0(1+a)\bigr)\Bigl(\bigl(\gamma+\psi_0(1+a)\bigr)^2+\frac{\pi^2}{6}-\psi_1(1+a)\Bigr)\\ &\quad+B(a,1)\Bigl(2\bigl(\gamma+\psi_0(1+a)\bigr)\psi_1(1+a)-\psi_2(1+a)\Bigr) \end{aligned} $$ Next, we use the (non-obvious) expansions around $a=0$ $$ \begin{aligned} B(a,1)&=\frac{1}{a}+O(1)\\ \psi_0(a)&=-\frac{1}{a}-\gamma+O(a)\\ \psi_0(1+a)&=-\gamma+\frac{\pi^2}{6}a+O(a^2)\\ \psi_1(1+a)&=\frac{\pi^2}{6}+\psi_2(1)a+\frac{\pi^4}{30}a^2+O(a^3)\\ \psi_2(1+a)&=\psi_2(1)+\frac{\pi^4}{15}a+O(a^2). \end{aligned} $$ to find that, as $a\to0^+$, $$ \begin{aligned} f'(a)&\approx -\frac{1}{a^2}\Bigl(\bigl(\frac{\pi^2}{6}a\bigr)^2-\psi_2(1)a-\frac{\pi^4}{30}a^2\Bigr)+\frac{1}{a}\Bigl(2\frac{\pi^2}{6}a\frac{\pi^2}{6}-\psi_2(1)-\frac{\pi^4}{15}a\Bigr)+O(a)\\ &=-\frac{\pi^4}{180}+O(a) \end{aligned} $$ as $a\to 0^+$. We conclude that $$ \partial_{a,b,b}B(a,1+b)\mid_{a\to 0^+,b\to 0}=-\frac{\pi^4}{180}. $$ Finally, dividing by $2$ (remember, we had a one-half in front of the beta function in the beginning), we get that $$ \int_0^1\frac{1}{2y}\ln y(\ln(1-y))^2\,dy=-\frac{1}{360}\pi^4. $$

Answer 2

There is a closed form antiderivative that can be found using repeated integration by parts: $$\begin{align}\int\frac{\ln y\cdot\ln^2(1-y)}{2y}dy=\frac{\ln^4(1-y)}8+\frac{\ln y\cdot\ln ^3(1-y)}6+\left(\frac{\pi^2}{12}-\frac{\ln ^2y}2\right)\cdot\ln ^2(1-y)\\ +\left[\left(\frac{\pi^2}6+\operatorname{Li}_2\left(\tfrac y{y-1}\right)\right)\cdot\ln y-\operatorname{Li}_3(1-y)-\operatorname{Li}_3\left(\tfrac y{y-1}\right)\right]\cdot\ln(1-y)\\ +\left[\vphantom{\Large|}\zeta(3)-\operatorname{Li}_3(1-y)\right]\cdot\ln y+\operatorname{Li}_4(1-y)-\operatorname{Li}_4(y)-\operatorname{Li}_4\left(\tfrac y{y-1}\right)\color{gray}{+C}\end{align}$$

Answer 3

Another one:

Write $\log^2(1-y)=\sum_{n,m=1}^{\infty}\frac{y^{m+n}}{mn}$

we obtain (exchanging summation and integration)

$$ 2I=\sum_{n,m=1}^{\infty}\frac{1}{mn}\int_0^1\log(y)y^{m+n-1}=\\ -\underbrace{\sum_{n,m=1}^{\infty}\frac{1}{mn}\frac{1}{(m+n)^2}}_{S} $$

the double sum can be tackeled by writing (i shamelessly benefit from this awesome answer)

$$ S=\sum_{n,m=1}^{\infty}\frac{1}{mn}\frac{1}{(m+n)^2}=\frac{1}{2}\sum_{n,m=1}^{\infty}\frac{1}{m^2n^2}-\frac{1}{2}\sum_{n,m=1}^{\infty}\frac{1}{(m+n)^2n^2}-\frac{1}{2}\sum_{n,m=1}^{\infty}\frac{1}{(m+n)^2m^2} $$

shifting arguments in the last two sums gives

$$ 2S=\sum_{n,m=1}^{\infty}\frac{1}{mn}\frac{1}{(m+n)^2}-\sum_{m=1,n<m}^{\infty}\frac{1}{n^2m^2}-\sum_{m=1,n>m}^{\infty}\frac{1}{n^2m^2} $$

which yields

$$ 2S=\sum_{n=m=1}^{\infty}\frac{1}{m^2n^2}=\zeta(4)=\frac{\pi^4}{90} $$

and therefore

$$ I=-S/2=-\frac{\pi^4}{360} $$

Answer 4

This question, for some reason, popped up in the Top Questions tab, and I thought I'd share another way to solve this integral using Harmonic Numbers. I hope you guys don't mind!

First, we use integration by parts on $u=\log^2(1-x)$ to get$$\begin{align*}\int\limits_0^1dx\,\frac {\log x\log^2(1-x)}{x} & =\frac 12\log^2x\log^2(1-x)\,\Biggr\rvert_0^1+\int\limits_0^1dx\,\frac {\log^2x\log(1-x)}{1-x}\\ & =\int\limits_0^1dx\,\frac {\log^2x\log(1-x)}{1-x}\end{align*}$$Now use the fact that

$$H(x)=-\frac {\log(1-x)}{1-x}=\sum\limits_{n\geq1}H_nx^n$$

And substitute to get$$\begin{align*}\int\limits_0^1dx\,\frac {\log x\log^2(1-x)}{x} & =-\sum\limits_{n\geq1}H_n\int\limits_0^1dx\,x^n\log^2x\\ & =-\lim\limits_{\mu\to0}\sum\limits_{n\geq1}H_n\frac {\partial^2}{\partial\mu^2}\frac 1{n+\mu+1}\\ & =-2\sum\limits_{n\geq1}\frac {H_n}{(n+1)^3}\end{align*}$$Split the sum up and use a well-known formula due to Euler

$$\sum\limits_{n\geq1}\frac {H_n}{n^m}=\frac 12(m+2)\zeta(m+1)-\frac 12\sum\limits_{n=1}^{m-2}\zeta(m-n)\zeta(n+1)$$

Therefore$$I=2\zeta(4)-5\zeta(4)+\zeta^2(2)=-\frac {\pi^4}{180}$$Our desired integral is half of that, so take half and the answer is$$\int\limits_0^1dx\,\frac {\log x\log^2(1-x)}{2x}\color{blue}{=-\frac {\pi^4}{360}}$$

Answer 5

I offer up yet another approach, this one relying on the Maclaurin series expansion for $\ln^2 (1 - x)$. It is similar to that used by @Frank W.

As was shown here $$\ln^2 (1 - x) = 2 \sum_{n = 2}^\infty \frac{H_{n - 1} x^n}{n}, \qquad |x| < 1.$$ Here $H_n$ denotes the Harmonic number. So for the integral we may write \begin{align} \int_0^1 \frac{1}{2x} \ln x \ln^2 (1 - x) \, dx &= \sum_{n = 2}^\infty \frac{H_{n - 1}}{n} \int_0^1 x^{n - 1} \ln x \, dx = -\sum_{n = 2}^\infty \frac{H_{n - 1}}{n^3}, \end{align} after integrating by parts. From properties for the Harmonic number, since $$H_n = H_{n - 1} + \frac{1}{n},$$ the infinite sum can be rewritten as \begin{align} \int_0^1 \frac{1}{2x} \ln x \ln^2 (1 - x) \, dx &= \sum_{n = 2}^\infty \frac{1}{n^4} - \sum_{n = 2}^\infty \frac{H_n}{n^3} = \sum_{n = 1}^\infty \frac{1}{n^4} - \sum_{n = 1}^\infty \frac{H_n}{n^3}. \end{align} Values for each of these sums are well known. For the first $$\sum_{n = 1}^\infty \frac{1}{n^4} = \zeta (4) = \frac{\pi^4}{90}.$$ For the second sum $$\sum_{n = 1}^\infty \frac{H_n}{n^3} = \frac{\pi^4}{72}.$$ (for various proofs of this result, see here) Thus $$\int_0^1 \frac{1}{2x} \ln x \ln^2 (1 - x) \, dx = \frac{\pi^4}{90} - \frac{\pi^4}{72} = -\frac{\pi^4}{360}.$$

Answer 6

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{{1 \over 2}\int_{0}^{1}{\ln\pars{y}\ln^{2}\pars{1 - y} \over y}\,\dd y = -\,{\pi^{4} \over 360}:\ {\large ?}}$.

\begin{align} &\bbox[10px,#ffd]{\ds{{1 \over 2}\int_{0}^{1}{\ln\pars{y}\ln^{2}\pars{1 - y} \over y}\,\dd y}} = \left.{1 \over 2}\,{\partial^{3} \over \partial\nu^{2}\partial\mu}\int_{0}^{1}{y^{\mu}\bracks{\pars{1 - y}^{\nu} - 1} \over y}\,\dd y\,\right\vert_{\ {\large\mu\ =\ 0^{+}} \atop {\large\nu\ =\ 0}} \\[5mm] = &\ {1 \over 2}\,{\partial^{3} \over \partial\nu^{2}\partial\mu} \bracks{{\Gamma\pars{\mu}\Gamma\pars{\nu + 1} \over \Gamma\pars{\mu + \nu + 1}} - {1 \over \mu}}_{\ {\large\mu\ =\ 0^{+}} \atop {\large\nu\ =\ 0}}\quad \pars{~\Gamma:\ Gamma\ Function~} \\[5mm] = &\ {1 \over 2}\,{\partial^{3} \over \partial\nu^{2}\partial\mu} \bracks{{\pi \over \Gamma\pars{1 - \mu}\sin\pars{\pi\mu}}\,{\Gamma\pars{\nu + 1} \over \Gamma\pars{\mu + \nu + 1}} - {1 \over \mu}}_{\ {\large\mu\ =\ 0^{+}} \atop {\large\nu\ =\ 0}} \\[5mm] = &\ {1 \over 2}\,{\partial^{3} \over \partial\nu^{2}\partial\mu} \bracks{{1 \over \mu}\,{\Gamma\pars{\nu + 1} \over \Gamma\pars{1 - \mu}\Gamma\pars{\mu + \nu + 1}} + {\pi^{2} \over 6}\,\mu} _{\ {\large\mu\ =\ 0^{+}} \atop {\large\nu\ =\ 0}} \\[5mm] = &\ {1 \over 2}\,{\partial^{3} \over \partial\nu^{2}\partial\mu} \bracks{\left.{1 \over 2}\,\partiald[2]{}{x}{\Gamma\pars{\nu + 1} \over \Gamma\pars{1 - x}\Gamma\pars{x + \nu + 1}}\,\right\vert_{\ x\ =\ 0^{+}}\mu + {\pi^{2} \over 6}\,\mu}_{\ {\large\mu\ =\ 0^{+}} \atop {\large\nu\ =\ 0}} \\[5mm] = &\ {1 \over 2}\,{\partial^{2} \over \partial\nu^{2}} \bracks{\left.{1 \over 2}\,\partiald[2]{}{x}{\Gamma\pars{\nu + 1} \over \Gamma\pars{1 - x} \Gamma\pars{x + \nu + 1}}\,\right\vert_{\ x\ =\ 0^{+}} + {\pi^{2} \over 6}} _{\ {\large\mu\ =\ 0^{+}} \atop {\large\nu\ =\ 0}} = {1 \over 4}\,{\partial^{4} \over \partial\nu^{2}\partial\mu^{2}} {\nu \choose \mu + \nu}_{\ {\large\mu\ =\ 0^{+}} \atop {\large\nu\ =\ 0}} \\[5mm] = &\ {1 \over 4}\,\partiald[2]{}{\nu} \bracks{-\,{\pi^{2} \over 6} + H^{2}_{\nu} - \Psi\, '\pars{1 + \nu}} _{\ \nu\ =\ 0}\quad \pars{~H_{z}:\ Harmonic\ Number~} \\[5mm] = &\ {1 \over 4}\bracks{2\Psi\, '^{2}\pars{1} + 2H_{0}\,\Psi\,''\pars{1}- \Psi\, '''\pars{1}} \qquad\qquad\qquad\qquad \left\{\begin{array}{lcr} \ds{\Psi\, '\pars{1}} & \ds{=} & \ds{\pi^{2} \over 6} \\ \ds{\Psi\, '''\pars{1}} & \ds{=} & \ds{\pi^{4} \over 15} \\ \ds{H_{0}} & \ds{=} & \ds{0} \end{array}\right. \\[5mm] = &\ {1 \over 4}\bracks{2\pars{\pi^{2} \over 6}^{2} + 0 - {\pi^{4} \over 15}} = \bbx{-\,{\pi^{4} \over 360}} \end{align}

Answer 7

Here is a solution without using Beta function:

Lets start with subbing $1-y=x$, we get $$\int_0^1\frac{\ln y\ln^2(1-y)}{y}dy=\int_0^1\frac{\ln^2x\ln(1-x)}{1-x}dx$$

We have the identity

$$\int_0^1\frac{x^{n}\ln^m(x)\ln(1-x)}{1-x}\ dx=\frac12\frac{\partial^m}{\partial n^m}\left(H_n^2+H_n^{(2)}\right)$$

Set $m=2$ then let $n$ approach $0$ we get

$$\int_0^1 \frac{\ln^2x\ln(1-x)}{1-x}dx=\frac12\frac{\partial^2}{\partial n^2}\left(H_n^2+H_n^{(2)}\right)_{n\to 0}\\=\frac12\left(4H_nH_n^{(3)}+2\left(H_n^{(2)}\right)^2+6H_n^{(4)}-4\zeta(2)H_n^{(2)}-4\zeta(3)H_n-\zeta(4)\right)_{n\to 0}\\=-\frac12\zeta(4)$$

Answer 8

Noting that the digamma function $\displaystyle f(x)=-\operatorname{Li_2}(x)=\int_0^x \frac{\ln (1-u)}{u} d u=-\sum_{n=1}^{\infty} \frac{x^n}{n^2}\tag*{} $ $\displaystyle \begin{aligned}I & =\int_0^1 \ln x \ln (1-x) d(f(x) )\\& =[f(x) \ln x \ln (1-x)]_0^1-\int_0^1 f(x)\left[\frac{\ln (1-x)}{x}-\frac{\ln x}{1-x}\right]dx \\& =-\int_0^1 f(x) f^{\prime}(x) d x+\int_0^1 \frac{f(x) \ln x}{1-x} d x \\& =-\int_0^1 f(x) d(f(x))-\int_0^1 \sum_{n=1}^{\infty} \frac{x^n}{n^2} \cdot \frac{\ln x}{1-x} d x \\& =\left[-\frac{f^2(x)}{2}\right]_0^1-\sum_{n=1}^{\infty} \frac{1}{n^2} \int_0^1 \frac{x^n \ln x}{1-x} d x\end{aligned}\tag*{} $ By the definition of $f(x)$ and integration by parts, we get $\displaystyle \begin{aligned}I & =-\frac{1}{2} \zeta^2(2)-\sum_{n=1}^{\infty} \frac{1}{n^2} \cdot \sum_{k=0}^{\infty} \int_0^1 x^{n+k} \ln x d x \\& =-\frac{1}{2} \zeta^2(2)+\sum_{n=1}^{\infty} \frac{1}{n^2}\left[\sum_{k=1}^{\infty} \frac{1}{(n+k+1)^2}\right]\end{aligned}\tag*{} $ Considering the square of the zeta function, we have $\displaystyle \begin{aligned}\zeta^2(2)= & \left(\sum_{n=1}^{\infty} \frac{1}{n^2}\right)^2=\left(\sum_{j=1}^{\infty} \frac{1}{j^2}\right)\left(\sum_{k=1}^{\infty} \frac{1}{k^2}\right) \\= & \sum_{n=1}^{\infty} \frac{1}{n^4}+2 \cdot \frac{1}{1^2}\left(\frac{1}{2^2}+\frac{1}{3^2}+\cdots\right) +2 \cdot \frac{1}{2^2}\left(\frac{1}{3^2}+\frac{1}{4^2}+\cdots\right) +2 \cdot \frac{1}{3^2}\left(\frac{1}{4^2}+\frac{1}{5^2}+\cdots\right) +\cdots \\= & \zeta(4)+2 \sum_{n=1}^{\infty} \frac{1}{n^2} \sum_{k=0}^{\infty} \frac{1}{(n+k+1)^2}\end{aligned}\tag*{} $ Plugging back yields $\displaystyle \begin{aligned}I & =-\frac{1}{2} \zeta^2(2)+\frac{1}{2}\left( \zeta^2(2)-\zeta(4)\right) =-\frac{\zeta(4)}{2}=-\frac{\pi^4}{180} \end{aligned}\tag*{} $