# The other ways to calculate $\int_0^1\frac{\ln(1-x^2)}{x}dx$

Prove that $$\int_0^1\frac{\ln(1-x^2)}{x}dx=-\frac{\pi^2}{12}$$ without using series expansion.

An easy way to calculate the above integral is using series expansion. Here is an example \begin{align} \int_0^1\frac{\ln(1-x^2)}{x}dx&=-\int_0^1\frac{1}{x}\sum_{n=0}^\infty\frac{x^{2n}}{n} dx\\ &=-\sum_{n=0}^\infty\frac{1}{n}\int_0^1x^{2n-1}dx\\ &=-\frac{1}{2}\sum_{n=0}^\infty\frac{1}{n^2}\\ &=-\frac{\pi^2}{12} \end{align} I am wondering, are there other ways to calculate the integral without using series expansion of its integrand? Any method is welcome. Thank you. (>‿◠)✌

After substituting $y=x^2$, we obtain $$\int_0^1\frac{\ln(1-x^2)}{x}\ dx=\frac12\int_0^1\frac{\ln(1-y)}{y}\ dy$$ Using the fact that $$\frac{\ln(1-x)}{x}=-\int_0^1\frac{1}{1-xy}\ dy$$ then $$\frac12\int_0^1\frac{\ln(1-x)}{x}\ dx=-\frac12\int_{x=0}^1\int_{y=0}^1\frac{1}{1-xy}\ dy\ dx.$$ Using transformation variable by setting $(u,v)=\left(\frac{x+y}{2},\frac{x-y}{2}\right)$ so that $(x,y)=(u-v,u+v)$ and its Jacobian is equal to $2$. Therefore $$-\frac12\int_{x=0}^1\int_{y=0}^1\frac{1}{1-xy}\ dy\ dx=-\iint_A\frac{du\ dv}{1-u^2+v^2},$$ where $A$ is the square with vertices $(0,0),\left(\frac{1}{2},-\frac{1}{2}\right), (1,0),$ and $\left(\frac{1}{2},\frac{1}{2}\right)$. Exploiting the symmetry of the square, we obtain \begin{align} \iint_A\frac{du\ dv}{1-u^2+v^2}=\ &2\int_{u=0}^{\Large\frac12}\int_{v=0}^u\frac{dv\ du}{1-u^2+v^2}+2\int_{u=\Large\frac12}^1\int_{v=0}^{1-u}\frac{dv\ du}{1-u^2+v^2}\\ =\ &2\int_{u=0}^{\Large\frac12}\frac{1}{\sqrt{1-u^2}}\arctan\left(\frac{u}{\sqrt{1-u^2}}\right)\ du\\ &+2\int_{u=\Large\frac12}^1\frac{1}{\sqrt{1-u^2}}\arctan\left(\frac{1-u}{\sqrt{1-u^2}}\right)\ du. \end{align} Since $\arctan\left(\frac{u}{\sqrt{1-u^2}}\right)=\arcsin u$, and if $\theta=\arctan\left(\frac{1-u}{\sqrt{1-u^2}}\right)$ then $\tan^2\theta=\frac{1-u}{1+u}$ and $\sec^2\theta=\frac{2}{1+u}$. It follows that $u=2\cos^2\theta-1=\cos2\theta$ and $\theta=\frac12\arccos u=\frac\pi4-\frac12\arcsin u$. Thus \begin{align} \iint_A\frac{du\ dv}{1-u^2+v^2} &=2\int_{u=0}^{\Large\frac12}\frac{\arcsin u}{\sqrt{1-u^2}}\ du+2\int_{u=\Large\frac12}^1\frac{1}{\sqrt{1-u^2}}\left(\frac\pi4-\frac12\arcsin u\right)\ du\\ &=\bigg[(\arcsin u)^2\bigg]_{u=0}^{\Large\frac12}+\left[\frac\pi2\arcsin u-\frac12(\arcsin u)^2\right]_{u=\Large\frac12}^1\\ &=\frac{\pi^2}{36}+\frac{\pi^2}{4}-\frac{\pi^2}{8}-\frac{\pi^2}{12}+\frac{\pi^2}{72}\\ &=\frac{\pi^2}{12} \end{align} and the result follows.

• Why does the half disappear in this equation $$-\frac12\int_{x=0}^1\int_{y=0}^1\frac{1}{1-xy}\ dy\ dx=-\iint_A\frac{du\ dv}{1-u^2+v^2}?$$ – Anastasiya-Romanova 秀 Aug 10 '14 at 4:57
• @V-Moy Because of the Jacobian. In simple words, when we apply transformation variable technique, let's say from $(x,y)$ to $(u,v)$ then you must use Jacobian in this case, therefore $$\int f(x,y)\ dx\ dy=\int f(u,v)\ |J|\ du\ dv,$$ where $J$ is the Jacobian. Perhaps you are familiar with the transformation variable technique when solving Gaussian integral, we transform the variable from Cartesian to polar coordinate and you will see the factor $r$ appears. – Tunk-Fey Aug 10 '14 at 5:34
• The answer that you posted makes me learn multivariable calculus more comprehensive first. I need time to understand your answer – Anastasiya-Romanova 秀 Aug 11 '14 at 7:30
• @Tunk-Fey (+1) Brilliant! What made you think of using that 2-variable substitution? Also, I think you have a typo in your very first equation; the RHS should end in $dy$, not $dx$. – David H Aug 12 '14 at 8:31
• @Anastasiya-Romanova: In other words, $$du \wedge dv = \frac{dx+dy}{2}\wedge\frac{dx-dy}{2} = \frac{2\, dy \wedge dx}{4}$$ so $$dy \wedge dx = 2 \, du \wedge dv$$ when we make the substitution. – Daniel McLaury Nov 10 '14 at 5:22

Substitute $u=x^2$. Then,

\begin{align} \int_{0}^{1}\frac{\ln{(1-x^2)}}{x}\mathrm{d}x &=\int_{0}^{1}\frac{\ln{(1-u)}}{\sqrt{u}}\cdot\frac{\mathrm{d}u}{2\sqrt{u}}\\ &=\frac12\int_{0}^{1}\frac{\ln{(1-u)}}{u}\mathrm{d}u\\ &=-\frac12\operatorname{Li}_2{(u)}\bigg{|}_{0}^{1}\\ &=-\frac12\operatorname{Li}_2{(1)}\\ &=-\frac{\pi^2}{12}. \end{align}

I fully anticipate there will be a not small number of people who call this cheating, but it's certainly cheating with style!

Edit: Here's a slightly more satisfying result. Substitute $u=x^2$ first like before, and next substitute $u=1-e^{-w}$. Then,

\begin{align} \int_{0}^{1}\frac{\ln{(1-x^2)}}{x}\mathrm{d}x &=\int_{0}^{1}\frac{\ln{(1-u)}}{\sqrt{u}}\cdot\frac{\mathrm{d}u}{2\sqrt{u}}\\ &=\frac12\int_{0}^{1}\frac{\ln{(1-u)}}{u}\mathrm{d}u\\ &=\frac12\int_{0}^{\infty}\frac{-w}{1-e^{-w}}(e^{-w})\mathrm{d}w\\ &=-\frac12\int_{0}^{\infty}\frac{w}{e^w-1}\mathrm{d}w\\ &=-\frac12\Gamma{(2)}\zeta{(2)}\\ &=-\frac12\zeta{(2)}. \end{align}

So now the question becomes do you accept that $\zeta{(2)}=\frac{\pi^2}{6}$. This still leaves a bit to be desired, but a lot more has written about $\zeta{(2)}$ than $\operatorname{Li}_2{(1)}$.

• I was thinking of this but it begs the question "how are these values of $Li_2(x)$ determined?". If it's with the series expansion then it doesn't $really$ change much :) – Jam Aug 8 '14 at 9:20
• Thanks for your answer but I am really sorry, this answer doesn't satisfy me since, in my opinion, I am considering dilog is a series, but I always upvote to those who answer my question. I also notice that with changing variable we will get derivative of beta function, but it's still a dilog function. – Anastasiya-Romanova 秀 Aug 8 '14 at 9:24
• @V-Moy I agree with you, and it doesn't satisfy me either. It's a nice tease though. – David H Aug 8 '14 at 9:36
• Hey... that's cheating! – Bennett Gardiner Aug 12 '14 at 23:36

Using the dilogarithm $\mathrm{Li}_2\;$ and the particular values for $0,1,-1\;$you get: $$\int_0^1\frac{\ln(1-x^2)}{x}dx= \int_0^1\frac{\ln(1-x)(1+x)}{x}dx= \int_0^1\frac{\ln(1+x)}{x}dx + \int_0^1\frac{\ln(1-x)}{x}dx= -\mathrm{Li}_2(-x)\Big{|}_0^1 - \mathrm{Li}_2(x)\Big{|}_0^1 =\frac{\pi^2}{12}-\frac{\pi^2}{6} = -\frac{\pi^2}{12}$$

• Heh, I thought some people might cry foul for invoking the value of $\operatorname{Li}_2{(1)}$, but you have invoke $\operatorname{Li}_2{(-1)}$ on top of that. You, sir, are cheating twice as much as me! :) – David H Aug 8 '14 at 9:24
• Thanks for your answer but I am really sorry, this answer doesn't satisfy me since, in my opinion, I am considering dilog is a series, but I always upvote to those who answer my question. I also notice that with changing variable we will get derivative of beta function, but it's still a dilog function. – Anastasiya-Romanova 秀 Aug 8 '14 at 9:25

Here is a route without power series expansion. Observing that, by the change of variable $x=\sin u$, we have $$\int_0^1\frac{\ln(1-x^2)}{x}\mathrm{d}x=\int_0^{\pi/2}\ln(\cos^2 u)\frac{\cos u}{\sin u}\mathrm{d}u$$ You may use the Fourier series expansion $$\log(\cos u)=\sum_{k=1}^\infty(-1)^{k}\frac{1-\cos(2kx)}{k}=\sum_{k=1}^\infty(-1)^{k}\frac{\sin^2(kx)}{k}$$ to obtain $$\int_0^1\frac{\ln(1-x^2)}{x}\mathrm{d}x=2\sum_{k=1}^\infty\! \frac{(-1)^{k}}{k}\!\!\int_0^{\pi/2}\frac{\sin^2(kx)}{\sin u}\cos u \:\mathrm{d}u=-\sum_{k=1}^\infty\frac{\frac11+ ...+\frac1{2k+1}}{2k(2k-1)} =-\frac{\pi^2}{12}.$$

• The OP states: "prove without using series expansion". – Tunk-Fey Aug 8 '14 at 17:56
• @Tunk-Fey Sorry, I thought 'prove without using power series expansion'... – Olivier Oloa Aug 8 '14 at 18:06
• Thanks for your answer Mr. Olivier Oloa but as Mr. Tunk-Fey said, the proof must not use series expansion. +1 anyway. – Anastasiya-Romanova 秀 Aug 10 '14 at 5:00

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\color{#66f}{\large\int_{0}^{1}{\ln\pars{1 - x^{2}} \over x}\,\dd x} =\half\int_{0}^{1}{\ln\pars{1 - x} \over x}\,\dd x =\half\int_{0}^{1}{\ln\pars{x} \over 1 - x}\,\dd x \\[3mm]&=-\,\half\lim_{\mu \to 0}\partiald{}{\mu}\int_{0}^{1} {1 - x^{\mu} \over 1 - x}\,\dd x =-\,\half\lim_{\mu \to 0}\partiald{\color{#f00}{\Psi\pars{\mu + 1}}}{\mu} =-\,\half\,\Psi'\pars{1} \\[3mm]&=-\,\half\,\color{#f0f}{\zeta\pars{2}} =-\,\half\,\color{maroon}{\pi^{2} \over 6}=\color{#66f}{\large -\,{\pi^{2} \over 12}} \end{align}

• Thanks Mr. Felix Marin, but as I said in the previous comment. I consider dilog is a series. – Anastasiya-Romanova 秀 Aug 10 '14 at 4:59
• @V-Moy There is a list that you can check. – Felix Marin Aug 10 '14 at 5:14
• Thanks for the link. Anyway, I have upvoted your answer yesterday – Anastasiya-Romanova 秀 Aug 11 '14 at 7:31
• @V-Moy ${\rm Li}_{2}\pars{z}$ is a series when $\left\vert\,z\,\right\vert \leq 1$. For other values ( besides a branch-cut ) it's found by means of an analytical continuation. However, I didn't use the DiLog: I used the DiGamma function instead. Thanks. – Felix Marin Aug 30 '14 at 6:25