Calculate $\int_0^1\frac{\ln(1-x^2)}{x}dx$ without series expansion.

Question

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

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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 whether there are other ways to calculate the integral without using series expansion of its integrand?

Any method is welcome. Thank you. (>‿◠)✌

Answer 1

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.

Answer 2

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!

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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 $\operatorname{Li}_2{(1)}$.

Answer 3

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

Answer 4

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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}

See $\underline{\color{#f00}{6.3.22}}$, $\underline{\color{#f0f}{6.4.2}}$ and $\underline{\color{maroon}{23.2.24}}$.

Answer 5

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 $$ \int_0^1\frac{\ln(1-x^2)}{x}\mathrm{d}x=2\sum_{k=1}^\infty\! \frac{(-1)^{k}}{k}\!\!\int_0^{\pi/2}\frac{\sin^2(kx)}{\sin u}\cos u \:\mathrm{d}u=-\sum_{k=1}^\infty\frac{\frac11+ ...+\frac1{2k+1}}{2k(2k-1)} =-\frac{\pi^2}{12}. $$

Answer 6

The integral $I$ can be evaluated by the parametrised integral $$ \begin{aligned} I(a)&=\int_0^1 \frac{\ln \left(1-a x^2\right)}{x} d x \\ I^{\prime}(a)&=-\int_0^1 \frac{x}{1-a x^2} d x \\ &=\left.\frac{1}{2 a} \ln \left(1-a x^2\right)\right|_0 ^1 \\ &=\frac{1}{2 a} \ln (1-a) \\ I&=\int_0^1 \frac{1}{2 a} \ln (1-a) d x \\ &=\frac{1}{2}[-\operatorname{Li_2}(a)]_0^1 \\ &=-\frac{1}{2} \operatorname{Li_2}(1) \\ &=-\frac{\pi^2}{12} \end{aligned} $$