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As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem) $$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}.$$ However, Euler was Euler and he gave other proofs.

I believe many of you know some nice proofs of this, can you please share it with us?

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    $\begingroup$ @J.M. Thanks. But Euler could very well be a good tag I believe. $\endgroup$
    – AD.
    Oct 30 '10 at 10:16
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    $\begingroup$ Robin Chapman has a collection of proofs on his homepage: empslocal.ex.ac.uk/people/staff/rjchapma/etc/zeta2.pdf $\endgroup$ Oct 30 '10 at 10:32
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    $\begingroup$ makes no sense to have an Euler tag... maybe Eulerian but that's pushing it. $\endgroup$
    – anon
    Oct 30 '10 at 10:46
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    $\begingroup$ Probably Robin should answer with a link to his note. I know I've pointed people to it when they ask precisely this, and they've always been more than satisfied! $\endgroup$ Oct 30 '10 at 14:09
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    $\begingroup$ What I like the most about this thread is that I know most of the proofs that I've seen posted up to this time, it makes me think that perhaps I was given adequate mathematical education after all :) $\endgroup$
    – Asaf Karagila
    Nov 1 '10 at 12:57

48 Answers 48

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Another proof i have (re?)discovered.

I want to prove that,

$\displaystyle J:=\int_0^1 \frac{\ln(1+x)}{x}dx=\frac{\pi^2}{12}$

Let $f$, be a function, such that, for $s\in[0;1]$,

$\displaystyle f(s)=\int_0^{\frac{\pi}{2}} \arctan\left(\frac{\sin t}{\cos t+s}\right)\,dt$

Observe that,

$\begin{align} f(0)&=\int_0^{\frac{\pi}{2}}\arctan\left(\frac{\sin t}{\cos t}\right)\,dt\\ &=\int_0^{\frac{\pi}{2}} t\,dt\\ &=\left[\frac{t^2}{2}\right]_0^{\frac{\pi}{2}}\\ &=\frac{\pi^2}{8} \end{align}$

For $t$ in $\left[0,\frac{\pi}{2}\right]$,

$\begin{align} \frac{\sin t}{\cos t+1}&=\frac{2\sin\left(\frac{t}{2}\right)\cos\left(\frac{t}{2}\right)}{\cos^2\left(\frac{t}{2}\right)-\sin^2\left(\frac{t}{2}\right)+1}\\ &=\frac{2\sin\left(\frac{t}{2}\right)\cos\left(\frac{t}{2}\right)}{2\cos^2\left(\frac{t}{2}\right)}\\ &=\tan\left(\frac{t}{2}\right) \end{align}$

Therefore,

$\begin{align} f(1)&=\int_0^{\frac{\pi}{2}}\arctan\left(\frac{\sin t}{\cos t+1}\right)\,dt\\ &=\int_0^{\frac{\pi}{2}}\arctan\left(\tan\left(\frac{t}{2}\right)\right)\,dt\\ &=\int_0^{\frac{\pi}{2}} \frac{t}{2}\,dt\\ &=\left[\frac{t^2}{4}\right]_0^{\frac{\pi}{2}}\\ &=\frac{\pi^2}{16} \end{align}$

For $s$ in $[0,1]$,

$\begin{align} f^\prime(s)&=-\int_0^{\frac{\pi}{2}}\frac{\sin t}{1+2s\cos t+s^2}\,dt\\ &=\left[\frac{\ln(1+2s\cos t+s^2)}{2s}\right]_0^{\frac{\pi}{2}}\\ &=\frac{1}{2}\frac{\ln\left(1+s^2\right)}{s}-\frac{\ln\left(1+s\right)}{s} \end{align}$

Therefore,

$\begin{align} f(1)-f(0)&=\int_0^1 f^\prime(s)ds\\ &=\frac{1}{2}\int_0^1\frac{\ln\left(1+s^2\right)}{s}\,ds-\int_0^1 \frac{\ln\left(1+s\right)}{s}\,ds\\ \end{align}$

In the first integral perform the change of variable $y=s^2$, therefore,

$\displaystyle f(1)-f(0)=-\frac{3}{4}J$

But,

$\begin{align} f(1)-f(0)&=\frac{\pi^2}{16}-\frac{\pi^2}{8}\\ &=-\frac{\pi^2}{16} \end{align}$

Therefore,

$\boxed{\displaystyle J=\frac{\pi^2}{12}}$

PS:

To obtain the value of $J$ knowing that $\displaystyle \zeta(2)=-\int_0^1 \frac{\ln(1-x)}{x}dx$

$\begin{align} \int_0^1 \frac{\ln(1+t)}{t}\,dt+\int_0^1 \frac{\ln(1-t)}{t}\,dt=\int_0^1 \frac{\ln(1-t^2)}{t}\,dt \end{align}$

Perform the change of variable $y=t^2$ in RHS integral,

$\begin{align} \int_0^1 \frac{\ln(1+t)}{t}\,dt+\int_0^1 \frac{\ln(1-t)}{t}\,dt=\frac{1}{2}\int_0^1 \frac{\ln(1-t)}{t}\,dt \end{align}$

Therefore,

$\begin{align} \int_0^1 \frac{\ln(1+t)}{t}\,dt=-\frac{1}{2}\int_0^1 \frac{\ln(1-t)}{t}\,dt \end{align}$

$\boxed{\displaystyle \int_0^1 \frac{\ln(1+t)}{t}\,dt=\frac{1}{2}\zeta(2)}$

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    $\begingroup$ I have never seen such soluton - well done ! It reminds me a similar trick (substitutions and taking the difference) as done in evaluation to obtain $\int_{0}^{\pi}\ln\sin x \;\mathrm{d}x=-\frac{\pi}{2}\ln 2$ $\endgroup$
    – Machinato
    Aug 25 '17 at 12:15
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This is, by no measure, the best nor the simplest approach, but I think the approach is pretty peculiar.

We estimate the number $N(x)$ of integer solutions to $a^2+b^2+c^2+d^2\leq x$ as $x\rightarrow\infty$. On one hand, this is the number of lattice points inside the the $4$-ball of radius $\sqrt{x}$, which has volume $\frac{1}{2}\pi^2x^2$, hence $N(x)=\frac{\pi^2}{2}x^2+O(x^{3/2})$.

On the other hand, let $r_4(n)$ be the number of solutions to $a^2+b^2+c^2+d^2=n$. Following the derivation in the book by Iwaniec-Kowalski, by Jacobi's four-square identity we can write $$N(x)=\sum_{n\leq x}r_4(n)=8\sum_{m\leq x}(2+(-1)^m)\sum_{dm\leq x,d\text{ odd}} d \\ =8\sum_{m\leq x}(2+(-1)^m)\left(\frac{x^2}{4m^2}+O\left(\frac{x}{m}\right)\right)\\ =2x^2\sum_{m\leq x}(2+(-1)^m)m^{-2}+O(x\log x)\\ =3x^2\zeta(2)+O(x\log x)$$ (I have copied the steps as they were in the book, it's a neat exercise to justify every transition). In particular, we have $$\zeta(2)=\lim\limits_{x\rightarrow\infty}\frac{N(x)}{3x^2}=\frac{\pi^2}{6}.$$

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  • $\begingroup$ (+1) I wonder if one can do the same by only exploiting the fact that the average value of $r_2(n)$ is $\pi$ by Gauss circle problem. $\endgroup$ Nov 9 '17 at 4:50
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by using Fourier series of $f(x)=1, x\in[0,1]$ $$1=\sum_{n=1}^\infty\frac{4}{(2n-1)\pi}\sin (2n-1)\pi x$$ integrate both sides when integration limits are $x=0 \rightarrow 1$ $$\int_{0}^{1}1.dx=\int_{0}^{1} \sum_{n=1}^\infty\frac{4}{(2n-1)\pi}\sin (2n-1)\pi x dx$$ $$1=\sum_{n=1}^\infty\frac{8}{(2n-1)^2\pi^2}$$ $$\sum_{n=1}^\infty\frac{1}{(2n-1)^2}=\frac{\pi^2}{8}$$ then we use the equality series $$\sum_{n=1}^\infty\frac{1}{n^2}=\sum_{n=1}^\infty\frac{1}{(2n-1)^2}+\sum_{n=1}^\infty\frac{1}{(2n)^2}$$ simplify it to get $$\sum_{n=1}^\infty\frac{1}{n^2}=\frac{4}{3}\sum_{n=1}^\infty\frac{1}{(2n-1)^2}$$ so, $$\sum_{n=1}^\infty\frac{1}{n^2}=\frac{4}{3}\frac{\pi^2}{8}=\frac{\pi^2}{6}$$

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The sum can be written as the integral: $$\int_0^{\infty} \frac{x}{e^x-1} dx $$ This integral can be evaluated using a rectangular contour from 0 to $\infty$ to $\infty + \pi i$ to $ 0$ .

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I propose a solution... Consider for $n\in\mathbb N^*$ : $$(1) : \int_0^\pi \left(\alpha t+\beta t^2\right)\cos(nt)\,\mathrm dt = \dfrac{1}{n^2} $$ Integrate by parts : $$ \int_0^\pi t\cos(nt)\,\mathrm dt = \underbrace{\left.\dfrac{t\sin(nt)}{n}\right\vert_0^\pi}_{=\,0} -\int_0^\pi \dfrac{\sin(nt)}{n}\,\mathrm dt = -\underbrace{\int_0^{n\pi}\dfrac{\sin x}{n^2}\,\mathrm dx}_{\mathrm{substitution\;by\;}x=nt} = \dfrac{\cos(n\pi)-1}{n^2}$$ and $$ \begin{split}\int_0^\pi t^2\cos(nt)\,\mathrm dt &= \underbrace{\left.\dfrac{t^2\sin(nt)}{n}\right\vert_0^\pi}_{=\,0} - \int_0^\pi \dfrac{2t\sin(nt)}{n}\,\mathrm dt = \left.\dfrac{2t\cos(nt)}{n^2}\right\vert_0^\pi - \int_0^\pi \dfrac{2\cos(nt)}{n^2}\,\mathrm dt \\&= \dfrac{2\pi\cos(n\pi)}{n^2} - \underbrace{\int_0^{n\pi}\dfrac{2\cos x}{n^3}\,\mathrm dx}_{\mathrm{substition\;by\;}x=nt} = \dfrac{2\pi\cos(n\pi)}{n^2}- \underbrace{\left.\dfrac{2\sin x}{n^3}\right\vert_0^{n\pi}}_{=\,0} \\&=\dfrac{2\pi\cos(n\pi)}{n^2} \end{split}$$ Thus $$ \int_0^\pi \left(\alpha t+\beta t^2\right)\cos(nt)\,\mathrm dt = \alpha \cdot \dfrac{\cos(n\pi)-1}{n^2} + \beta\cdot\dfrac{2\pi\cos(n\pi)}{n^2} $$ We deduce that $\alpha = -1$ and $\beta = 1/2\pi$ satisfies $(1)$.

Since for $x\in\mathbb R\backslash 2\pi\mathbb Z$ : $$ \sum_{k=1}^n \cos(kx) =-\dfrac{1}{2} + \dfrac{\sin(nx+x/2)}{2\sin(x/2)} $$ we have $$ \begin{split}\sum_{k=1}^n \dfrac{1}{k^2} &= \sum_{k=1}^n \int_0^\pi \left(\dfrac{t^2}{2\pi}-t\right)\cos(kt)\,\mathrm dt \\&= \int_0^\pi \left(\dfrac{t^2}{2\pi}-t\right)\sum_{k=1}^n \cos(kt)\,\mathrm dt\\ &= -\dfrac{1}{2}\int_0^\pi \left(\dfrac{t^2}{2\pi}-t\right)\mathrm dt + \int_0^\pi \left(\dfrac{t^2}{2\pi}-t\right)\cdot \dfrac{\sin(nt+t/2)}{2\sin(t/2)}\,\mathrm dt \end{split}$$ However $\sin(nt+t/2) = \sin(t/2)\cos(nt)+\sin(nt)\cos(t/2)$. Let $\phi$ and $\psi$ such that $$\phi(t) = \dfrac{t^2}{4\pi}-\dfrac{t}{2} \;\mathrm{and}\; \psi(t) = \left(\dfrac{t^2}{2\pi}-t\right)\cdot\dfrac{\cos(t/2)}{2\sin(t/2)}$$ so that $$\int_0^\pi \left(\dfrac{t^2}{2\pi}-t\right)\cdot \dfrac{\sin(nt+t/2)}{2\sin(t/2)} = \int_0^\pi \phi(t)\cos(nt)\,\mathrm dt + \int_0^\pi \psi(t)\sin(nt)\,\mathrm dt$$ $\phi$ is continuous on $[0,\pi]$. And $\psi$ can be extended at $t=0$. Indeed as $t\to 0$ $$ \psi(t) = \underbrace{\dfrac{t\cos(t/2)}{2\pi}}_{\to\, 0}\cdot\underbrace{\dfrac{t/2}{\sin(t/2)}}_{\to \ 1}- \underbrace{\cos(t/2)}_{\to \, 1}\cdot\underbrace{\dfrac{t/2}{\sin(t/2)}}_{\to\, 1} \xrightarrow[t\to 0]{} -1 $$ Therefore, $\psi$ is continuous (by extension) on $[0,\pi]$.

There remains to apply Lebesgue-Riemann Lemma, which tells us that : $$ \int_0^\pi \phi(t)\cos(nt)\,\mathrm dt \xrightarrow[n\to \infty]{} 0\;\;\mathrm{and}\;\;\int_0^\pi \psi(t)\sin(nt)\,\mathrm dt\xrightarrow[n\to \infty]{} 0$$ Consequently $$ \int_0^\pi \left(\dfrac{t^2}{2\pi}-t\right)\cdot \dfrac{\sin(nt+t/2)}{2\sin(t/2)} \xrightarrow[n\to \infty]{} 0 $$ and $$ \sum_{k=1}^n \dfrac{1}{k^2} \xrightarrow[n\to \infty]{} -\dfrac{1}{2}\int_0^\pi \left(\dfrac{t^2}{2\pi}-t\right)\mathrm dt$$ Now, we can evaluate this integral : $$-\dfrac{1}{2}\int_0^\pi \left(\dfrac{t^2}{2\pi}-t\right)\mathrm dt = -\dfrac{1}{2}\left[\dfrac{t^3}{6\pi}-\dfrac{t^2}{2}\right]_0^\pi = -\dfrac{1}{2}\left[\dfrac{\pi^2}{6}-\dfrac{\pi^2}{2}\right] = \dfrac{\pi^2}{6} $$ Then... the desired result : $$ \boxed{\sum_{k=1}^\infty \dfrac{1}{k^2} = \dfrac{\pi^2}{6}}$$

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$$ \begin{align} \log(2\cos(x)) &=\log\left(e^{ix}+e^{-ix}\right)\tag{1a}\\ &=ix+\log\left(1+e^{-2ix}\right)\tag{1b}\\ &=-ix+\log\left(1+e^{2ix}\right)\tag{1c}\\ &=\cos(2x)-\frac{\cos(4x)}2+\frac{\cos(6x)}3-\cdots\tag{1d} \end{align} $$ Explanation:
$\text{(1a)}$: $2\cos(x)=e^{ix}+e^{-ix}$
$\text{(1b)}$: factor out $e^{ix}$
$\text{(1c)}$: factor out $e^{-ix}$
$\text{(1d)}$: average $\text{(1b)}$ and $\text{(1c)}$ using the power series for $\log(1+x)$ $$ \begin{align} \sum_{k=1}^\infty\frac1{k^2} &=\frac1{2\pi}\int_0^{2\pi}\sum_{k=1}^\infty\frac{e^{ikx}}k\sum_{k=1}^\infty\frac{e^{-ikx}}k\,\mathrm{d}x\tag{2a}\\ &=\frac1{2\pi}\int_0^{2\pi}\left|\log(1-e^{ix})\right|^2\,\mathrm{d}x\tag{2b}\\ &=\frac1{2\pi}\int_{-\pi}^\pi\left|\log(1+e^{ix})\right|^2\,\mathrm{d}x\tag{2c}\\ &=\frac1{2\pi}\int_{-\pi}^\pi\left|\,\log\left(2\cos\left(\frac x2\right)\right)+\frac{ix}2\,\right|^{\,2}\,\mathrm{d}x\tag{2d}\\ &=\frac1{2\pi}\int_{-\pi}^\pi\left(\log\left(2\cos\left(\frac x2\right)\right)^2+\frac{x^2}4\right)\,\mathrm{d}x\tag{2e}\\ &=\frac{\pi^2}{12}+\frac1{2\pi}\int_{-\pi}^\pi\left(\cos(x)-\frac{\cos(2x)}2+\frac{\cos(3x)}3-\cdots\right)^2\,\mathrm{d}x\tag{2f}\\ &=\frac{\pi^2}{12}+\frac12\sum_{k=1}^\infty\frac1{k^2}\tag{2g}\\ &=\frac{\pi^2}6\tag{2h} \end{align} $$ Explanation:
$\text{(2a)}$: use the orthogonality of $e^{ijx}$ and $e^{ikx}$ when $j\ne k$
$\text{(2b)}$: use the power series for $\log(1+x)$
$\text{(2c)}$: substitute $x\mapsto x+\pi$
$\text{(2d)}$: $1+e^{ix}=2\cos(x/2)e^{ix/2}$
$\text{(2e)}$: $\left|\,x+iy\,\right|^2=x^2+y^2$
$\text{(2f)}$: apply $(1)$
$\text{(2g)}$: use the orthogonality of $\cos(jx)$ and $\cos(kx)$ for $j\ne k$
$\text{(2h)}$: subtract the original from twice $\text{(2g)}$

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    $\begingroup$ This is fantastic $\endgroup$ Jul 9 '19 at 4:28
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Let $X$ be an independent Laplace random variable with $X\sim L(0,1) = \frac12 \exp{(-|x|)}$, then its characteristic function :

$$\varphi_X(t)=\mathbb{E}[e^{itX}]=\frac{1}{1+t^2} \newcommand{\var}[1]{\mathrm{var}\left[#1\right]}$$

By symmetry $\mathbb{E}[X]=0$ we write (generally) :

$$\varphi_X(t)=\mathbb{E}[e^{itX}]=\mathbb{E}[1+itX-t^2X^2+\cdots\,]=1-\var{X}t^2+O(t^3)\tag{A}$$

since $\var{X}=\mathbb{E}[X^2]-\mathbb{E}[X]^2=\mathbb{E}[X^2]$, for our case : $$\frac{1}{1+t^2}=1-t^2+O(t^3) \rightarrow \var{X}=1$$ Now consider a set of such variables $X_n$, independent of each other, for a construction of a new random variable $Y$ as follows :

$$Y=\sum_{n=1}^{\infty}\frac{X_n}{n}$$

Then, taking variance of both sides :

$$\var{Y}=\var{\sum_{n=1}^{\infty}\frac{X_n}{n}}=\sum_{n=1}^{\infty}\var{\frac{X_n}{n}}=\sum_{n=1}^{\infty}\frac{\var{X_n}}{n^2}=\var{X}\zeta(2)=\zeta(2)\tag{B}$$

However, for characteristic function instead, using properities of characteristic function :

$$\varphi_Y(t)=\varphi_{\sum_{n=1}^{\infty}X_n/n}\left(t\right)=\prod_{n=1}^\infty \varphi_{X}\left(\frac{t}{n}\right) = \prod_{n=1}^\infty \frac{1}{1+\frac{t^2}{n^2}}=\frac{\pi t}{\sinh \pi t} = 1-\frac{\pi^2}{6}t^2+O(t^3)\tag{C}$$

Combining it with $(A)$ and $(B)$ we get : $$\zeta(2)=\var{Y}=\frac{\pi^2}{6}$$

NOTE : As long as the set $\{X_n\}_{n\in\mathbb{N}}$ consist of independent variables with idential pdf. the steps are the same upto $\var{X}=1$. So, there might be distributions for which the product in $C$ is easily evaluable.

NOTE2 : by mystake I posted just a fisrst sentence of this answer, so after deleting, this is the second copy

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  • $\begingroup$ This is equivalent to Euler's infinite product with sin(x)/x, no? Take the reciprocal and let $x=i\pi t$. Otherwise I am not sure how you conclude the product equals $\frac{\pi t}{\sinh{\pi t}}$ $\endgroup$
    – nimish
    Feb 26 at 6:11
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This is a similar proof as posted by Hans Lundmark, but I find it to be a little simpler. I ran across this approach in a Dover copy of The USSR Olympiad Problem Book. It is also based on the observation that $$\cot^2x<\frac{1}{x^2}<\csc^2x\,.$$

We first have the trig identity $$\sin(2n+1)\alpha=\sum_{k=0}^n(-1)^k\binom{2n+1}{2k+1}\cos^{2(n-k)}\alpha\sin^{2k+1}\alpha$$ which is arguably the hardest part of this proof. This directly manipulates into $$\sin(2n+1)\alpha=\sin^{2n+1}\alpha\sum_{k=0}^n(-1)^k\binom{2n+1}{2k+1}\cot^{2(n-k)}\alpha\,.$$ This formula reveals that the $n$ distinct quantities below $$\cot^2\frac{\pi}{2n+1},\quad\cot^2\frac{2\pi}{2n+1},\quad\ldots,\quad\cot^2\frac{n\pi}{2n+1}$$ are the roots of the polynomial $$\sum_{k=0}^n(-1)^k\binom{2n+1}{2k+1}x^{n-k}\,.$$ After scaling by the lead coefficient, Viete's Formulas then imply that $$\sum_{k=1}^n \cot^2\frac{k\pi}{2n+1}=\frac{n(2n-1)}{3}$$ By another elementary trig identity, we also get $$\sum_{k=1}^n \csc^2\frac{k\pi}{2n+1}=\frac{2n(n+1)}{3}$$ The inequality above then gives us $$\frac{n(2n-1)}{3}<\frac{(2n+1)^2}{\pi^2}\left(1+\frac{1}{2^2}+\cdots+\frac{1}{n^2}\right)<\frac{2n(n+1)}{3}$$ which gives us the desired conclusion after taking limits.

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  • $\begingroup$ I later found out that this is one of the three different proofs of the Basel Problem that was presented in "Proofs from the Book". $\endgroup$
    – user123641
    May 1 '18 at 15:54
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Here is an interesting solution that evaluates three sums, one of which is $\zeta(2).$

Let us start with the double integral

\begin{equation} \tag{1}\label{Double Integral} \int_{0}^{1} \int_{0}^{\sqrt{1-x^2}} \frac{1}{\sqrt{x^2+y^2} \ (1+x^2+y^2)} \ dy \ dx. \end{equation} A quick polar coordinates transformation $x=r \cos(\theta), y= r \sin(\theta)$ transforms \eqref{Double Integral} into $$\int_{0}^{\frac{\pi}{2}}\int_{0}^{1} \frac{1}{1+r^2} \ dr \ d \theta=\frac{\pi^2}{8}.$$

Hence, \eqref{Double Integral} is equal to $\frac{\pi^2}{8}.$

Now integrate \eqref{Double Integral} with respect to $y$ using the fact

$$\int \frac{1}{\sqrt{x^2+y^2} (1+x^2+y^2)} \ dy = \frac{\tanh^{-1} \left( \frac{y}{\sqrt{1+x^2}\sqrt{x^2+y^2}} \right)}{\sqrt{1+x^2}}$$ to see that \eqref{Double Integral} becomes

\begin{equation} \tag{2} \label{arctanh} \int_{0}^{1} \frac{\tanh^{-1} \left( \frac{\sqrt{1-x^2}}{\sqrt{1+x^2}} \right)}{\sqrt{1+x^2}} \ dx. \end{equation}

Next, observe that \eqref{arctanh} is equal to the double integral

\begin{equation} \tag{3} \label{double integral 2} \int_{0}^{1} \int_{0}^{\sqrt{1-x^2}} \frac{1}{1+x^2-y^2} \ dy \ dx, \end{equation} which can be confirmed by integrating the inner integral of \eqref{double integral 2} with respect to $y.$ Now here comes the interesting part. Use polar coordinates with $x=r\cos(\theta),y=r\sin(\theta)$ on \eqref{double integral 2} and then with $x=r\sin(\theta),y=r\cos(\theta)$ on \eqref{double integral 2} and average the two to see that \eqref{double integral 2} is the same as

$$\frac{1}{2}\int_{0}^{\frac{\pi}{2}} \int_{0}^{1} \frac{r}{1+r^2\cos(2\theta)} \ dr \ d \theta + \frac{1}{2}\int_{0}^{\frac{\pi}{2}} \int_{0}^{1} \frac{r}{1-r^2\cos(2\theta)} \ dr \ d \theta \,$$ which simplifies down to \begin{align} \int_{0}^{\frac{\pi}{2}} \frac{\ln(1+\cos(2\theta))}{4\cos(2\theta)}-\frac{\ln(1-\cos(2\theta))}{4\cos(2\theta)} \ d \theta & = \int_{0}^{\frac{\pi}{2}} \frac{\ln \left(\frac{1+\cos(2\theta)}{1-\cos(2\theta)} \right)}{4\cos(2\theta)} \ d \theta\\ & = \int_{0}^{\frac{\pi}{2}} -\frac{\ln(\tan^2(\theta))}{4\cos(2\theta)} \ d \theta \tag{4} \label{double angle} \\ & = \int_{0}^{\infty} \frac{\ln(u)}{2(u^2-1)} \ du \tag{5} \label{pi^2/4} \end{align} with \eqref{double angle} following from simplifying the logarithmic term with the double angle formulas $$\sin^2(\theta)=\frac{1-\cos(2\theta)}{2}, \quad \cos^2(\theta)=\frac{1+\cos(2\theta)}{2},$$ and \eqref{pi^2/4} following from the substitution $u=\tan(\theta).$

Splitting \eqref{pi^2/4} into $$ \int_{0}^{1} \frac{\ln(u)}{2(u^2-1)} \ du + \int_{1}^{\infty} \frac{\ln(u)}{2(u^2-1)} \ du,$$ a substitution $u=\frac{1}{t}$ on the second term shows \eqref{pi^2/4} is equal to $$ 2 \int_{0}^{1} \frac{\ln(u)}{2(u^2-1)} \ du = \int_{0}^{1} \frac{\ln(u)}{u^2-1} \ du.$$ Hence, we have \begin{align} \tag{6} \label{pi^2/8} \int_{0}^{1} \frac{\ln(u)}{u^2-1} \ du = \frac{\pi^2}{8} \end{align}

Now following the other users' answers, convert the integrand in the left hand side of \eqref{pi^2/8} into a geometric series, apply the Monotone Convergence Theorem to interchange sum and integral, to see we have $$\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} = \frac{\pi^2}{8},$$ and observing \begin{align} \zeta(2) & =\sum_{n=1}^{\infty} \frac{1}{(2n)^2}+ \sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} \\ & = \frac{1}{4} \zeta(2) + \frac{\pi^2}{8}, \end{align} we see $$\zeta(2)=\frac{\pi^2}{6}.$$

Those are the first two sums. We refer back to \eqref{arctanh}. Make the substitution $u=\sqrt{\frac{{1-x^2}}{{1+x^2}} }$ and simplify to see \eqref{arctanh} becomes to get: \begin{equation} \tag{7} \label{complicated sub} \int_{0}^{1} \frac{\sqrt{2} u\tanh^{-1}(u)}{\sqrt{1-u^2}(1+u^2)}\ du. \end{equation} Substituting $u=\tanh(\theta)$ transforms \eqref{complicated sub} into \begin{align} \sqrt{2}\int_{0}^{\infty}\frac{\theta(e^{2\theta}-1)}{e^{4\theta}+1}\ d\theta \end{align} and substituting $z=e^{\theta}$ shows that \eqref{arctanh} is the same as $$\sqrt{2}\int_{1}^{\infty}\frac{(z^2-1)\ln(z)}{z^4+1}\ dz,$$ and splitting the region of integration as with \eqref{pi^2/4} to get \eqref{pi^2/8}, we see \eqref{arctanh} is

\begin{equation} \tag{7} \label {crazy integral} \sqrt{2}\int_{0}^{1}\frac{(t^2-1)\ln(t)}{t^4+1}\ dt. \end{equation}

Expanding this integrand into a geometric series and integrating term by term, we see that \begin{align}\frac{\pi^2}{8} & =\sqrt{2}\left(\sum_{n=0}^{\infty}\frac{(-1)^n}{(4n+1)^2}-\frac{(-1)^n}{(4n+3)^2}\right) \\ & =\sqrt{2}\sum_{n=0}^{\infty}\frac{(-1)^n}{(4n+1)^2} +\sqrt{2} \sum_{n=-\infty}^{-1}\frac{(-1)^n}{(4n+1)^2} \\ & = \sqrt{2}\sum_{n=-\infty}^{\infty}\frac{(-1)^n}{(4n+1)^2}. \end{align} Thus, \begin{align} \sum_{n=-\infty}^{\infty}\frac{(-1)^n}{(4n+1)^2}=\frac{\pi^2}{8\sqrt{2}}, \end{align} which is the third sum.

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Following proof rely on this integral identity :

$$\int_{a}^{1}\frac{\arccos x}{\sqrt{x^2-a^2}}\mathrm{d}x=-\frac{\pi}{2}\ln a\qquad ;\,a\in(0,1]$$

We will prove it later on. Now, let's make a power series :

$$\zeta(2)=\sum_{n=1}^{\infty}\frac{1}{n^2}=\int_0^1\frac{1}{x}\sum_{n=1}^{\infty}\frac{x^n}{n}\,\mathrm{d}x=-\int_0^1\frac{\ln(1-x)}{x}\,\mathrm{d}x=-\int_0^1\frac{\ln x}{1-x}\,\mathrm{d}x$$

Inserting the formula above we get :

$$\zeta(2)=\frac{2}{\pi}\int_0^1\int_{x}^{1}\frac{\arccos y}{(1-x)\sqrt{y^2-x^2}}\,\mathrm{d}y\,\mathrm{d}x$$

Interchanging the order of integration :

$$\zeta(2)=\frac{2}{\pi}\int_0^1\int_{0}^{y}\frac{\arccos y}{(1-x)\sqrt{y^2-x^2}}\,\mathrm{d}x\,\mathrm{d}y\tag{A}$$

But, with help of substitution $x=y \cos{\theta}$ and universal $t=\tan\frac{\theta}{2}$ : $$\int_{0}^{y}\frac{\mathrm{d}x}{(1-x)\sqrt{y^2-x^2}}=\int_{0}^{\frac{\pi}{2}}\frac{\mathrm{d}\theta}{1-y \cos{\theta}}=\int_{0}^{1}\frac{\frac{2\mathrm{d}t}{1+t^2}}{1-y\frac{1-t^2}{1+t^2}}=\frac{\pi-\arccos{y}}{\sqrt{1-y^2}}$$

Plugging this to $(A)$ we get :

$$ \begin{align*}&\zeta(2)=\frac{2}{\pi}\int_{0}^{1}\frac{\pi\arccos{y}-\arccos^2 y}{\sqrt{1-y^2}}\,\mathrm{d}y=\frac{2}{\pi}\left(\frac{\pi}{2}\arccos^2 y- \frac{1}{3}\arccos^3 y\right)\bigg{|}_{1}^{0}= \\ \\ &\frac{2}{\pi}\left(\frac{\pi}{2}\left(\frac{\pi}{2}\right)^2-\frac{1}{3}\left(\frac{\pi}{2}\right)^3\right) = \frac{2}{\pi}\left(\frac{\pi}{2}\right)^3 \left(1-\frac{1}{3}\right) =\frac{\pi^2}{6} \end{align*}$$

ADDENDUM : Proof of the apriori integral :

$$\begin{align*}&\int_{a}^{1}\frac{\arccos x}{\sqrt{x^2-a^2}}\mathrm{d}x=\int_{a}^{1}\frac{\arccos\left(\frac{x}{y}\right)}{\sqrt{x^2-a^2}}\bigg{|}_{y=x}^{y=1}\mathrm{d}x=\int_{a}^{1}\int_{x}^{1}\frac{x}{y}\frac{\mathrm{d}y\,\mathrm{d}x}{\sqrt{x^2-a^2}\sqrt{y^2-x^2}} = \\ \\ & \int_{a}^{1}\int_{a}^{y}\frac{x}{y}\frac{\mathrm{d}x\,\mathrm{d}y}{\sqrt{x^2-a^2}\sqrt{y^2-x^2}} = \frac{\pi}{2}\int_{a}^{1}\frac{\mathrm{d}y}{y} = -\frac{\pi}{2}\ln a \end{align*}$$

Where the inner integral was computed via substitution $x^2=a^2\cos^2\theta+y^2\sin^2\theta$ it is clear, taking differential, that $2x\;\mathrm{d}x=2\left(y^2-a^2\right)\sin\theta\cos\theta\;\mathrm{d}\theta$, then :

$$(x^2-a^2)(y^2-x^2)=(a^2\cos^2\theta+y^2\sin^2\theta-a^2)(y^2-a^2\cos^2\theta-y^2\sin^2\theta)=(y^2\sin^2\theta-a^2\sin^2\theta)(y^2\cos^2\theta-a^2\cos^2\theta)=(y^2-a^2)^2\sin^2\theta\cos^2\theta$$

Or $$\sqrt{x^2-a^2}\sqrt{y^2-x^2}= \left(y^2-a^2\right)\sin\theta\cos\theta\ = x\,\mathrm{d}x$$

Therefore :

$$\int_{a}^{y}\frac{x\,\mathrm{d}x}{\sqrt{x^2-a^2}\sqrt{y^2-x^2}}=\int_{0}^{\frac{\pi}{2}}\frac{x\,\mathrm{d}x}{x\,\mathrm{d}x}=\frac{\pi}{2}$$

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Define $f$ on $[0;2\pi]$,

$\displaystyle f(a)=\int_0^1 \dfrac{\ln(x^2-2x\cos(a)+1)}{x}dx\tag 0$

Theorem:

For all $a\in [0;2\pi]$,

$f(a)=-\dfrac{1}{2}a^2+\pi a-\dfrac{\pi^2}{3}\tag 1$

For all $a\in [0;2\pi]$,

$\displaystyle f\left(\frac{a}{2}\right)+f\left(\pi-\frac{a}{2}\right)=\frac{f(a)}{2}\tag 2$

Proof:

$\begin{align} f\left(\frac{a}{2}\right)+f\left(\pi-\frac{a}{2}\right)&=\int_0^1 \frac{\ln\left(\left(x^2-2x\cos\left(\frac{a}{2}\right)+1\right)\left(x^2+2x\cos\left(\frac{a}{2}\right)+1\right) \right)}{x}dx\\ &=\int_0^1 \frac{\ln\left(x^4-2x^2\cos(a)+1\right)}{x}dx\\ \end{align}$

Perform the change of variable $y=x^2$ in the latter integral to obtain (2).

According to theorems about functions défined by integrals $f^{\prime\prime}$ exists and it is continuous.

Derive twice (2),

For all $a\in [0;2\pi]$,

$\displaystyle f^{\prime\prime}\left(\frac{a}{2}\right)+f^{\prime\prime}\left(\pi-\frac{a}{2}\right)=2f^{\prime\prime}(a)\tag 3$

$f^{\prime\prime}$ is continuous on $[0;2\pi]$ therefore this fonction has a maximum $M$ and a minimum $m$ that are obtainable.

Therefore it exists $a_0\in[0;2\pi]$ such that $f^{\prime\prime}(a_0)=M$.

Plug $a_0$ into (3),

$\displaystyle f^{\prime\prime}\left(\frac{a_0}{2}\right)+f^{\prime\prime}\left(\pi-\frac{a_0}{2}\right)=2f^{\prime\prime}(a_0)=2M$

But, $ f^{\prime\prime}\left(\frac{a_0}{2}\right)\leq M$ et $f^{\prime\prime}\left(\pi-\frac{a_0}{2}\right)\leq M$ according to the définition of $M$.

therefore $f^{\prime\prime}\left(\frac{a_0}{2}\right)=f^{\prime\prime}\left(\pi-\frac{a_0}{2}\right)=M$

By recurrence reasoning, for all $n\geq 1$, natural integer,

$f^{\prime\prime}\left(\frac{a_0}{2^n}\right)=M\tag 4$

$f^{\prime\prime}$ is continuous in $0$ therefore taking $n$ to infinity in (4) one obtains,

$M=f^{\prime\prime}(0)$.

Considering $m$ the minimum of $f^{\prime\prime}$ using the same way it can be proved that,

$m=f^{\prime\prime}(0)$

Since $m=M$, therefore $f^{\prime\prime}$ is a constant function.

therefore, there exist $\alpha,\beta,\gamma$ real such that,

For all $a\in[0;2\pi]$,

$f(a)=\alpha a^2+\beta a+\gamma\tag 5$

Plug (5) into (3), one obtains:

$\alpha\pi+\dfrac{\beta}{2}=0$ and $\alpha \pi^2 +\beta \pi+\dfrac{3}{2}\gamma=0$

On the other hand, for all $a\in [0;2\pi]$,

$\displaystyle f^\prime(a)= 2\sin a\int_0^1\dfrac{1}{x^2-2x\cos a+1}dx$

If $a=\dfrac{\pi}{2}$ one obtains,

$\begin{align} f^\prime\left(\dfrac{\pi}{2}\right)&=2\int_0^1 \dfrac{1}{x^2+1}dx\\ &=2\times \dfrac{\pi}{4}\\ &=\dfrac{\pi}{2} \end{align}$

Taking derivative of (5), one obtains for all $a\in [0;2\pi]$,

$f^\prime(a)=2\alpha a+\beta$

Therefore,

$\alpha \pi +\beta=\dfrac{\pi}{2}$

One have obtained a linear system of three equations in $\alpha,\beta,\gamma$.

To achieve the proof of the theorem solve it.

To get the value of $\zeta(2)$, apply the theorem with $a=0$, one obtains,

$\displaystyle \int_0^1 \dfrac{\ln(1-x)}{x}dx=-\dfrac{\pi^2}{6}$

And then, continue in usual way, expand the integrand...

From, Euler's integrals, H. Haruki and S. Haruki, The mathematical gazette, 1983.

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  • $\begingroup$ This one is very simple and straightforward to follow. However, I would like to see the argument for continuity for the second derivative $f''(x)$ to be more explained (or am I so blind to see it?). $\endgroup$
    – Machinato
    Aug 25 '17 at 12:20
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    $\begingroup$ It uses general theorems about derivation under the integral sign as Lebesgue's dominated convergence theorem (but weaker theorems do exist for the Riemann integral). $\endgroup$
    – FDP
    Aug 25 '17 at 17:21
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Using Fourier's expansion of $f(x)=x(1-x)$, we get $$a_{0}=\frac{1}{6}$$ $$a_{n}=\frac{1}{n^2\pi^2}$$ $$b_{n}=0$$ Therefore ,we have $$x(1-x)=\frac{1}{6}-\sum_{n=1}^{\infty}\frac{\cos2xn\pi}{(n\pi)^2}$$ putting $x=0$ we get $$\sum _{k=1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6}$$

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I really like this one. Consider $f(x)=x^2-\pi^2$. Compute it's Fourier expansion to obtain $$f(x)=\frac{2}{3}\pi^2-4\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\cos nx.$$ Now let $x=\pi$, then it quickly follows that $$4\zeta(2)=\frac{2}{3}\pi^2\implies \zeta(2)=\frac{\pi^2}{6}.$$

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    $\begingroup$ This is the same as the first part of this one. $\endgroup$
    – AD.
    Oct 24 '16 at 12:31
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Since $\int_0^1 \frac{dx}{1+x^2}=\frac{\pi}{4}$, we have

$$\frac{\pi^2}{16}=\int_0^1\int_0^1\frac{dydx}{(1+x^2)(1+y^2)}\overset{t=xy}{=}\int_0^1\int_0^x\frac{dtdx}{x(1+x^2)(1+t^2/x^2)}$$

$$=\frac12\int_0^1\int_t^1\frac{dxdt}{x(1+x^2)(1+t^2/x^2)}\overset{x^2\to x}{=}\frac12\int_0^1\left(\int_{t^2}^1\frac{dx}{(1+x)(x+t^2)}\right)dt$$

$$=-\frac12\int_0^1\frac{\ln\left(\frac{4t^2}{(1+t^2)^2}\right)}{1-t^2}dt\overset{t=\frac{1-x}{1+x}}{=}-\frac12\int_0^1\frac{\ln\left(\frac{1-x^2}{1+x^2}\right)}{x}dx$$

$$\overset{x^2\to x}{=}-\frac14\int_0^1\frac{\ln\left(\frac{1-x}{1+x}\right)}{x}dx=-\frac14\int_0^1\frac{\ln\left(\frac{(1-x)^2}{1-x^2}\right)}{x}dx$$

$$=-\frac12\int_0^1\frac{\ln(1-x)}{x}dx+\frac14\underbrace{\int_0^1\frac{\ln(1-x^2)}{x}dx}_{x^2\to x}$$

$$=-\frac38\int_0^1\frac{\ln(1-x)}{x}dx\Longrightarrow \int_0^1\frac{-\ln(1-x)}{x}dx=\frac{\pi^2}{6}$$


Remark:

This solution can be considered a proof that $\zeta(2)=\frac{\pi^2}{6}$ as we have $\int_0^1\frac{-\ln(1-x)}{x}dx=\text{Li}_2(x)|_0^1=\text{Li}_2(1)=\zeta(2)$

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I found this proof on YouTube but I did little changes:

Lets start with

\begin{align} I&=\int_0^{\pi/2}\ln(2\cos x)\ dx=\int_0^{\pi/2}\ln\left(e^{ix}(1+e^{-2ix})\right)\ dx\\ &=\int_0^{\pi/2}ix\ dx-\sum_{n=1}^\infty \frac{(-1)^n}{n}\int_0^{\pi/2}e^{-2inx}\ dx\\ &=\frac{\pi^2}{8}i-\sum_{n=1}^\infty\frac{(-1)^n}{n}\left(-\frac{(-1)^n-1}{2in}\right)\\ &=\frac{\pi^2}{8}i-\frac12i\left(\zeta(2)-\operatorname{Li}_2(-1)\right)\\ &=\frac{\pi^2}{8}i-\frac12i\left(\zeta(2)+\frac12\zeta(2)\right)\\ &=i\left(\frac{\pi^2}{8}-\frac34\zeta(2)\right) \end{align}

By comparing the imaginary parts, we have

$$0=\frac{\pi^2}{8}-\frac34\zeta(2)\Longrightarrow\zeta(2)=\frac{\pi^2}{6}$$

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Let $f(x)=\frac 12-x$ on the interval $[0, 1)$, and extend $f$ to be periodic on $ \mathbb{R} $.

By definition, \begin{align*} \hat f(0)=\int_0^1 f(x)dx=\int_0^1 \left(\frac 12-x\right)dx=0. \end{align*} And for $ \kappa\ne 0 $: \begin{align*} \hat f(\kappa)&=\int_{0}^{1}f(x)e^{-2\pi i\kappa x }dx=\int_0^1\left( \frac 12 -x \right)e^{-2\pi i\kappa x}dx=-\int_0^1xe^{-2\pi i \kappa x}dx\\ &=\frac{1}{2\pi i\kappa }\int_{0}^{1}xd(e^{-2\pi i\kappa x})=\left.\frac{1}{2\pi i\kappa}xe^{-2\pi i\kappa x}\right|_0^1+\frac{1}{2\pi i\kappa}\int_0^1 e^{-2\pi i\kappa x}dx\\ &=\frac{1}{2\pi i\kappa}. \end{align*}

By the Parseval identity \begin{align*} \int_{0}^{1}|f(x)|^2dx=\sum_{k=-\infty}^{\infty}|\hat{f}(k)|^2=|\hat{f}(0)|^2+2\sum_{k=1}^{\infty}|\hat{f}(k)|^2=2\sum_{k=1}^{\infty}\frac{1}{4\pi^2 k^2}. \end{align*} On the other hand, \begin{align*} \int_{0}^{1}|f(x)|^2dx&=\int_{0}^{1}\left( \frac{1}{2}-x \right)^2 dx=\frac 14-\frac 12+\frac 13=\frac 1{12}. \end{align*} Hence, we have $$ \sum_{k=1}^{\infty}\frac{1}{k^2}=\frac{\pi^2}{6}. $$


Remark: This is an exercise(Chapter 8.13 on page 254 ) in Folland's book.

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Here's mine. I'm answering late, I know that, but I am still answering it.
We'll use the expansion of $\tanh^{-1}$: $$\frac{1}{2}\log\frac{1+y}{1-y}=\sum_{n\geq0}\frac{y^{2n+1}}{2n+1},\quad|y|<1$$ We start with this inequality:
$$\int_{-1}^{1}\int_{-1}^{1}\frac{1}{1+2xy+y^2}dy\,dx=\int_{-1}^{1}\frac{1}{1+2xy+y^2}dx\,dy$$ The LHS of this equality gives: $$\int_{-1}^{1}\int_{-1}^{1}\frac{1}{1+2xy+y^2}dy\,dx=\int_{-1}^{1}\frac{\arctan \frac{x+y}{\sqrt{1-x^2}}}{\sqrt{1-x^2}}dx\Biggr|_{y=-1}^{y=1}\\ \quad\,\,\quad\quad\quad\quad\quad\quad\quad\quad\quad=\int_{-1}^{1}\frac{\pi}{2\sqrt{1-x^2}}dx=\frac{\pi^2}{2}$$ The RHS of the former equality yields: \begin{align} \int_{-1}^{1}\int_{-1}^{1}\frac{1}{1+2xy+y^2}dy\,dx&=\int_{-1}^{1}\frac{\log(1+2xy+y^2)}{2y}dy\Biggr|_{x=-1}^{x=1}\\ &=\int_{-1}^{1}\frac{\log\frac{1+y}{1-y}}{y}dy\\ &=2\int_{-1}^{1}\sum_{n\geq0}\frac{y^{2n}}{2n+1}dy\\ &=4\sum_{n\geq0}\frac{1}{(2n+1)^2} \end{align} Hence, $$\sum_{r\geq0}\frac{1}{(2n+1)^2}=\frac{\pi^2}{8}$$ Now $$\frac{3}{4}\zeta(2)=\zeta(2)-\frac{1}{4}\zeta(2)=\sum_{n\geq 1}\frac{1}{n^2}=\sum_{m\geq1}\frac{1}{(2m)^2}=\sum_{r\geq0}\frac{1}{(2r+1)^2}=\frac{\pi^2}{8}$$ Solving this we get $$\zeta(2)=\frac{\pi^2}{6}$$ as desired. Source:https://www.emis.de/journals/GM/vol16nr4/ivan/ivan.pdf
Here are more proofs.

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The following proof is by Khalaf Ruhemi ( he is not a MSE member)

By partial fraction decomposition, we have $$\frac{y}{(1+y^2)(y^2+x^2)}=\frac{1}{x^2-1}\left(\frac{y}{1+y^2}-\frac{y}{y^2+x^2}\right).$$ Integrate both sides from $y=0$ to $y=\infty$, \begin{gather*} \int_0^\infty\frac{y}{(1+y^2)(y^2+x^2)}\mathrm{d}y=\frac{1}{x^2-1}\int_0^\infty\left[\frac{y}{1+y^2}-\frac{y}{y^2+x^2}\right]\mathrm{d}y\\ =\frac{1}{x^2-1}\left[\frac12\ln(1+y^2)-\frac12\ln(y^2+x^2)\right]_0^\infty=\frac{1}{2(x^2-1)}\left[\ln\left(\frac{1+y^2}{y^2+x^2}\right)\right]_0^\infty\\ =\frac{1}{2(x^2-1)}\left[\ln(1)-\ln\left(\frac{1}{x^2}\right)\right]=\frac{1}{2(x^2-1)}\left[2\ln(x)\right]=\frac{\ln(x)}{x^2-1}. \end{gather*} Next, integrate both sides from $x=0$ to $x=\infty$ \begin{gather*} \int_0^\infty\frac{\ln(x)}{x^2-1}\mathrm{d}x=\int_0^\infty\int_0^\infty\frac{y}{(1+y^2)(y^2+x^2)}\mathrm{d}y\,\mathrm{d}x\\ \{\text{change the order of integration}\}\\ =\int_0^\infty\frac{1}{1+y^2}\left[\int_0^\infty\frac{y\,\mathrm{d}y}{y^2+x^2}\right]\mathrm{d}y\\ =\int_0^\infty\frac{1}{1+y^2}\left[\arctan\left(\frac{x}{y}\right)\right]_0^\infty dy=\int_0^\infty\frac{1}{1+y^2}\left[\frac{\pi}{2}-0\right] \mathrm{d}y\\ =\frac{\pi}{2}\int_0^\infty\frac{1}{1+y^2} dy=\frac{\pi}{2}\arctan(y)\bigg|_0^\infty=\frac{\pi}{2}\cdot\frac{\pi}{2}=\frac{\pi^2}{4}. \end{gather*} Thus, \begin{gather*} \frac{\pi^2}{4}=\int_0^\infty\frac{\ln(x)}{x^2-1}\mathrm{d}x=\left(\int_0^1+\int_1^\infty\right)\frac{\ln(x)}{x^2-1}\mathrm{d}x\\ =\int_0^1\frac{\ln(x)}{x^2-1}\mathrm{d}x+\underbrace{\int_1^\infty\frac{\ln(x)}{x^2-1}\mathrm{d}x}_{x\to1/x}\\ =2\int_0^1\frac{\ln(x)}{x^2-1}\mathrm{d}x=-\int_0^1\frac{\ln(x)}{1-x}\mathrm{d}x-\int_0^1\frac{\ln(x)}{1+x}\mathrm{d}x\\ \left\{\text{use $\frac{1}{1+x}=\frac{1}{1-x}-\frac{2x}{1-x^2}$ in the second integral}\right\}\\ =-2\int_0^1\frac{\ln(x)}{1-x}\mathrm{d}x+2\underbrace{\int_0^1\frac{x\ln(x)}{1-x^2}\mathrm{d}x}_{x^2\to x}\\ =-2\int_0^1\frac{\ln(x)}{1-x}\mathrm{d}x+\frac12\int_0^1\frac{\ln(x)}{1-x}\mathrm{d}x\\ =-\frac32\int_0^1\frac{\ln(x)}{1-x}\mathrm{d}x\overset{1-x\to y}{=}-\frac32\int_0^1\frac{\ln(1-y)}{y}\mathrm{d}y\\ \{\text{expand $\ln(1-y)$ in series}\}\\ =\frac32\sum_{n=1}^\infty \frac{1}{n}\int_0^1 y^{n-1}\mathrm{d}y=\frac32\sum_{n=1}^\infty\frac{1}{n^2}=\frac{3}{2}\zeta(2). \end{gather*} So we have $$\frac{\pi^2}{4}=\frac{3}{2}\zeta(2)\Longrightarrow \zeta(2)=\frac{\pi^2}{6}.$$

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  • $\begingroup$ This looks identical to Daniele Ritelli's solution to the Basel Problem. We would like to refer you to our joint AMS Publication for a generalization of this particular proof: ams.org/journals/qam/2018-76-03/S0033-569X-2018-01499-3 $\endgroup$ Mar 5 at 23:02
  • $\begingroup$ @Vivek Kaushik can you provide a link to his solution? $\endgroup$ Mar 6 at 18:13
  • $\begingroup$ Here is his paper: arxiv.org/abs/1208.5981. The joint publication listed earlier also references some other authors with similar solutions. $\endgroup$ Mar 6 at 18:59
  • $\begingroup$ @Vivek Kaushik thanks $\endgroup$ Mar 7 at 0:09
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