43
$\begingroup$

"The number $\frac 16 \pi^2$ turns up surprisingly often and frequently in unexpected places." - Julian Havil, Gamma: Exploring Euler's Constant.


It is well-known, especially in 'pop math,' that $$\zeta(2)=\frac1{1^2}+\frac1{2^2}+\frac1{3^2}+\cdots = \frac{\pi^2}{6}.$$ Euler's proof of which is nice. I would like to know where else this constant appears non-trivially. This is a bit broad, so here are the specifics of my question:

  1. We can fiddle with the zeta function at arbitrary even integer values to eek out a $\zeta(2)$. I would consider these 'appearances' of $\frac 16 \pi^2$ to be redundant and ask that they not be mentioned unless you have some wickedly compelling reason to include it.
  2. By 'non-trivially,' I mean that I do not want converging series, integrals, etc. where it is obvious that $c\pi$ or $c\pi^2$ with $c \in \mathbb{Q}$ can simply be 'factored out' in some way such that it looks like $c\pi^2$ was included after-the-fact so that said series, integral, etc. would equal $\frac 16 \pi^2$. For instance, $\sum \frac{\pi^2}{6\cdot2^n} = \frac 16 \pi^2$, but clearly the appearance of $\frac 16\pi^2$ here is contrived. (But, if you have an answer that seems very interesting but you're unsure if it fits the 'non-trivial' bill, keep in mind that nobody will actually stop you from posting it.)

I hope this is specific enough. This was my attempt at formally saying 'I want to see all the interesting ways we can make $\frac 16 \pi^2$.' With all that being said, I will give my favorite example as an answer below! :$)$


There used to be a chunk of text explaining why this question should be reopened here. It was reopened, so I removed it.

$\endgroup$
9
  • 3
    $\begingroup$ This video on youtube gives a novel way to evaluate this sum, essentially by high-school geometry involving triangles and circles. $\endgroup$ Jan 18, 2020 at 2:36
  • 4
    $\begingroup$ @Jam I think the 'natural appearances' of $\pi$ are significantly different from the natural appearances of $\pi^2$. You could square the solutions given to the question you link, though they'd hardly be compelling solutions to this question. For instance, $$\int_{-\infty}^\infty e^{-x^2}~dx = \sqrt \pi$$ is certainly a suitable answer to the question you've linked, but I wouldn't say $$\frac{\left( \int_{-\infty}^\infty e^{-x^2}~dx\right)^4}6 = \frac 16\pi^2$$ belongs here. Likewise, I don't think the sqrt of some of the answers here would be compelling responses to the question you linked. $\endgroup$ Jan 18, 2020 at 17:12
  • 3
    $\begingroup$ I believe that "natural appearances of $\pi$" and "natural appearances of $\pi^2\over6$" are two very different things. $\endgroup$
    – ViHdzP
    Jan 18, 2020 at 20:34
  • 1
    $\begingroup$ probability of primality is ${\pi^2\over 6}^{\pi(\sqrt{n})}$ but that' just contrived. $\endgroup$
    – user645636
    Jan 20, 2020 at 2:45
  • 1
    $\begingroup$ @RoddyMacPhee Correct me if I'm wrong but wouldn't it be $\left(\frac{6}{\pi^2}\right)^{\pi\sqrt n}$, otherwise it is increasing wrt. $n$? $\endgroup$
    – Jam
    Jan 20, 2020 at 20:34

25 Answers 25

33
$\begingroup$

Let $I(n)$ be the probability that two integers chosen randomly from $[1,n]$ are coprime. Then, $$\lim_{n \to \infty} I(n)=\frac{6}{\pi^2}.$$ So, you could say the odds that two randomly-chosen positive integers are coprime is $1$ in $\frac{\pi^2}6$.

$\endgroup$
5
  • 4
    $\begingroup$ But does the proof of this use: $\sum 1/n^2 =\pi^2/6$? Or is it proved independently of that. $\endgroup$
    – GEdgar
    Jan 18, 2020 at 1:37
  • $\begingroup$ @GEdgar I am unsure. When I say I want to avoid $\sum \frac 1{n^2}$, I mean that I do not want it to be obvious that it appears in the derivation of a solution. In this case, if the proof does use $\zeta(2)$, it certainly isn't obvious and is therefore interesting enough to include! I will append a source when I find one. $\endgroup$ Jan 18, 2020 at 1:42
  • 5
    $\begingroup$ The probability is $\prod_{p\in\Bbb P}(1-p^{-2})=1/\zeta(2)$, so the proof does use the Basel problem's solution, but it also uses the factorization of $\zeta$ over primes. So there are two "non-obvious" parts to the proof, Basel being the second: the first is that you can change that product into (one over) a sum on $\Bbb N$. $\endgroup$
    – J.G.
    Jan 18, 2020 at 10:47
  • 2
    $\begingroup$ So, since $\zeta(2) = \pi^2/6$ is used in the proof of this, it is not an independent case where $\pi^2/6$ arises. I would conjecture this is also true of most (or all?) of the other answers here. $\endgroup$
    – GEdgar
    Jan 18, 2020 at 12:55
  • $\begingroup$ This is clearly zeta(2) so no upvote from me. Sorry. No downvote either. $\endgroup$
    – mick
    Jan 31, 2020 at 17:09
27
$\begingroup$

Define a continuous analog of the binomial coefficient as

$$\binom{x}{y}=\frac{\Gamma(x+1)}{\Gamma(y+1)\Gamma(x-y+1)}.$$

While exploring integrals of the form

$$\int_{-\infty}^\infty\prod_{n=1}^m\binom{x_n}{t}\,\mathrm dt$$

I was surprised the first time I saw

$$\int_{-\infty}^\infty\binom{1}{t}^3\,\mathrm dt=\frac{3}{2}+\frac{6}{\pi^2}$$

show up.

$\endgroup$
2
  • $\begingroup$ Interesting answer. Do you perhaps know of any articles or other references that delve into these types of integrals in detail? $\endgroup$
    – Max Muller
    Jun 17 at 11:20
  • $\begingroup$ The only places I've seen similar material are in Ramanujan's Notebook's, part 1, pages 302-304; part 2, pages 225-227. However, the material is quite brief and I believe it would be best to find something more in depth. $\endgroup$
    – dxdydz
    Jun 18 at 2:35
19
$\begingroup$

Unexpected at first glance is $$2\sum_{m\ge1}\sum_{n\ge1}\frac{(-1)^n}{n^3}\sin(\tfrac{n}{m^{2k}})=\frac{1}{6}\zeta(6k)-\frac{\pi^2}{6}\zeta(2k).$$ A generalization may be found here.

Perhaps more unexpected is $$\sqrt3 \int_0^\infty \frac{\arctan x}{x^2+x+1} \, dx=\frac{\pi^2}{6},$$ which is proven here.

Even nicer is $$\frac1{12}\int_0^{2\pi}\frac{x\,dx}{\phi-\cos^2 x}=\frac{\pi^2}6,$$ which can be seen here. Here $\phi=\frac{1+\sqrt5}2$ is the golden ratio.

A pleasing logarithmic integral is $$\frac83\int_1^{1+\sqrt2}\frac{\ln x}{x^2-1}dx=\frac{\pi^2}6-\frac23\ln^2(1+\sqrt2),$$ proven here.

Another nice trigonometric integral: $$2\int_0^{\pi/2}\cot^{-1}\sqrt{1+\csc x}\, dx=\frac{\pi^2}{6},$$ from here.


Edit: as was stated in the comments of this answer, it's the $\pi^2$ that counts, though un-scaled integrals evaluating to $\pi^2/6$ are best. With this in mind, I present a nice $\zeta$-quotient integral involving $\pi^2$: $$\int_0^\infty \left(\frac{\tanh(x)}{x^3}-\frac{1}{x^2\cosh^2(x)}\right)\, dx=\frac{7\zeta(3)}{\pi^2}=\frac{7\zeta(3)}{6\zeta(2)},$$ shown here.


I just derived another identity: $$\int_0^\infty\frac{(x^2+1)\arctan x}{x^4+\tfrac14x^2+1}dx=\frac{\pi^2}{6}.$$ Since I just found this identity I present the proof. In the link I provided after the second identity it is shown that $$f(a)=\int_0^\infty \frac{\arctan x}{x^2+2ax+1}dx=\frac{\pi}{4\sqrt{1-a^2}}\left(\frac\pi2-\phi(a)\right)\qquad |a|<1$$ where $\phi(a)=\arctan\frac{a}{\sqrt{1-a^2}}$. First off, notice that $\phi(-a)=-\phi(a)$. Thus $$j(a)=\frac12(f(a)+f(-a))=\int_0^\infty\frac{(x^2+1)\arctan x}{x^4+2(1-2a^2)x^2+1}dx=\frac{\pi^2}{8\sqrt{1-a^2}}.$$ Hence $$j(\sqrt{7}/4)=\int_0^\infty\frac{(x^2+1)\arctan x}{x^4+\tfrac14x^2+1}dx=\frac{\pi^2}{6}.$$


Expect more nice examples as I gather the best ones.

$\endgroup$
6
  • 4
    $\begingroup$ These are nice examples, but in all of them you are cheating by adding a constant in front of the integral/sum in such a way the final constant multiplying pi^2 in the result is 1/6. $\endgroup$ Jan 18, 2020 at 8:04
  • 2
    $\begingroup$ OP here - I think these answers are all perfectly valid, as the most characteristic part of $\frac 16 \pi^2$ (which would be the $\pi^2$ term) arises naturally. This may kind of blur the focus of my question, but I’d much rather have these integrals than not :) Though if clathratus digs up an integral that evaluates to our desired constant with no scaling needed, that would be all the more interesting! $\endgroup$ Jan 18, 2020 at 10:24
  • 1
    $\begingroup$ Similar to your second integral, $$\frac{4}{3} \int_0^\infty \frac{atan{x}}{1 + x^2} dx = \frac{16}{3} \int_0^1 \frac{atan{x}}{1 + x^2} dx=\frac{\pi^2}{6} $$ $\endgroup$ Jan 19, 2020 at 8:11
  • 2
    $\begingroup$ What a brilliant compilation!(+1) Here's another one of the logatirhmic form: $$\int_0^\phi\frac{\ln x}{x^2-1}dx=\frac{\pi^2}{6}-\frac{3}{4}\ln^2\phi$$ where $\phi=\frac{1+\sqrt{5}}{2}$ $\endgroup$
    – user632577
    Jan 26, 2020 at 2:43
  • $\begingroup$ @EdwardH. wow! that one is spectacular! source? $\endgroup$
    – clathratus
    Jan 26, 2020 at 2:46
19
$\begingroup$

Problem 11953 from AMM (January 2017) asked for the evaluation of the following double integral whose value turns out to be equal to $\frac{\pi^2}{6}$. $$\int_0^\infty \!\!\!\int_0^\infty \frac{\sin x \sin y \sin (x + y)}{xy(x + y)} \, dx \, dy = \frac{\pi^2}{6}.$$

Problem 2074 from Mathematics Magazine (June 2019) asked for the following evaluation of a limit of a sum whose value turns out to be equal to $\frac{\pi^2}{6}$. $$\lim_{n \to \infty} \sum_{k = 1}^n \frac{(-1)^{k + 1}}{k} \binom{n}{k} H_k = \frac{\pi^2}{6}.$$ Here $H_n = \sum_{k = 1}^n \frac{1}{k}$ denotes the $n$th Harmonic number.

And here are a few sums: $$\sum_{n = 1}^\infty \frac{H_n}{n2^{n - 1}} = \frac{\pi^2}{6}.$$

$$\sum_{n = 1}^\infty \frac{H_n}{n (n + 1)} = \frac{\pi^2}{6}.$$

$$\frac{3}{2} \sum_{n = 0}^\infty \left (\frac{1}{(6n + 1)^2} + \frac{1}{(6n + 5)^2} \right ) = \frac{\pi^2}{6}.$$ and $$\sum_{n = 1}^\infty \frac{3}{n^2 \binom{2n}{n}} = \frac{\pi^2}{6}.$$

And a few more sums, this time involving the variant harmonic number term $\Lambda_n$ were $$\Lambda_n = 1 + \frac{1}{3} + \cdots + \frac{1}{2n - 1} = \sum_{k = 1}^n \frac{1}{2k - 1} = H_{2n} - \frac{1}{2} H_n.$$

$$\sum_{n = 1}^\infty \frac{\Lambda_n}{n(2n - 1)} = \frac{\pi^2}{6},$$ $$\sum_{n = 1}^\infty (-1)^{n+ 1} \left (\frac{2n + 1}{n(n+ 1)} \right )^2 \Lambda_n = \frac{\pi^2}{6},$$ and $$2\sum_{n = 1}^\infty \frac{(-1)^{n + 1} \Lambda_n}{3^{n - 1} n} = \frac{\pi^2}{6}.$$

Some function values: $$\zeta (2) = \operatorname{Li}_2 (1) = \frac{\pi^2}{6},$$ where $\zeta$ denotes the Riemann zeta function while $\operatorname{Li}_2 (x)$ is the dilogarithm. $$6 \operatorname{Li}_2 \left (\frac{1}{2} \right ) - 6 \operatorname{Li}_2 \left (\frac{1}{4} \right ) - 2 \operatorname{Li}_2 \left (\frac{1}{8} \right ) + \operatorname{Li}_2 \left (\frac{1}{64} \right ) = \frac{\pi^2}{6}.$$

And some strange integrals: $$\int_0^1 (x^{-x})^{{{(x^{-x})}^{(x^{-x})}}^\cdots} \, dx = \frac{\pi^2}{6},$$ and $$\int_0^\infty \frac{dx}{\operatorname{Ai}^2 (x) + \operatorname{Bi}^2(x)} = \frac{\pi^2}{6},$$ where $\operatorname{Ai}(x)$ and $\operatorname{Bi}(x)$ denote the Airy functions of the first and second kinds, respectively.

$\endgroup$
13
$\begingroup$

Here is the simplest,

$$\int_0^1 \frac{\ln x}{x-1}\ \mathrm{d}x=\frac{\pi^2}6$$

$\endgroup$
14
  • 4
    $\begingroup$ That is simply $\text{Li}_2(0)=\sum_{n=1}^\infty \frac1{n^2}=\frac{\pi^2}{6}$ which is not different from $\zeta(2)$. So, this contributes nothing to the OP. $\endgroup$
    – Mark Viola
    Jan 18, 2020 at 4:48
  • 10
    $\begingroup$ Not sure agree. The integral could be obtained with elementary mehods. To write it as an expansion, or in advanced functions, is fine, but unnecessary. Besides, any $\frac{\pi^2}6$-example could be equated to $Li_2(0)$. $\endgroup$
    – Quanto
    Jan 18, 2020 at 4:58
  • 3
    $\begingroup$ Is there a proof of this not using knowledge of the value of $\zeta(2)$ ? $\endgroup$
    – GEdgar
    Jan 18, 2020 at 13:21
  • 2
    $\begingroup$ @GEdgar Do you mind if I answer you the question instead? All we need is to connect $\displaystyle \int_0^1 \frac{\log(x)}{1-x}\textrm{d}x$ to $\displaystyle \int_0^1 \frac{\log(x)}{1-x^2}\textrm{d}x$, which is trivial to do, and then follow the procedure from the penultimate row of page $56$, going backwards, from (Almost) Impossible Integrals, Sums, and Series. Done. So, no need to use the knowledge of the value of $\zeta(2)$. $\endgroup$ Jan 18, 2020 at 15:08
  • 2
    $\begingroup$ I like that you changed this from 'this may be the simplest' to 'here is the simplest.' After seeing the rest of the responses this question has produced, I would make that change too. $\endgroup$ Jan 20, 2020 at 1:38
12
$\begingroup$

In terms of the two real branches of the Lambert W function

$\require{begingroup} \begingroup$ $\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}$

\begin{align} \int_0^1 \frac{\Wp(-\tfrac t{\mathrm e})\,(\Wp(-\tfrac t{\mathrm e})-\Wm(-\tfrac t{\mathrm e}))} {t\,(1+\Wp(-\tfrac t{\mathrm e}))\,(1+\Wm(-\tfrac t{\mathrm e}))}\, dt &=\frac{\pi^2}6 \tag{1}\label{1} . \end{align}


Edit

And another one, with different integrand curve:

\begin{align} \int_0^1 \frac{\Wp(-\tfrac t\e)+t\,(1+\Wm(-\tfrac t\e))}{t\,(1+\Wm(-\tfrac t\e))} \, dt &=\frac{\pi^2}6 \tag{2}\label{2} . \end{align}

The intersection point of the integrands in \eqref{1} and \eqref{2} can be found exactly at $t=\tfrac1\Omega-1\approx 0.763222834$, where $\Omega$ is Omega constant, $\Omega \e^{\Omega }=1,\ \Omega=\W(1)\approx 0.56714329$ (thanks, @omegadot).

Also, one more:

\begin{align} \int_0^1 \ln\left(\frac{-\Wm(-t\,\exp(-t))}t\right) \, dt &= \int_0^1 -t-\Wm(-t\,\exp(-t)) \, dt =\frac{\pi^2}6 \tag{3}\label{3} . \end{align}

$\endgroup$

$\endgroup$
6
  • $\begingroup$ That is a very nice example. Do you mind if I ask, where does it come from? $\endgroup$
    – omegadot
    Jan 18, 2020 at 9:25
  • 4
    $\begingroup$ @omegadot: Well, it's just $\int_0^1 \frac{\ln x}{x-1}$ in disguise, using parametric representation of the real branches of the Lambert W function. $\endgroup$
    – g.kov
    Jan 18, 2020 at 12:13
  • $\begingroup$ How we do love the Omega constant: $\operatorname{W}_0(1) = \Omega$. $\endgroup$
    – omegadot
    Jan 26, 2020 at 5:06
  • $\begingroup$ @omegadot: "What’s in a name?.." $\endgroup$
    – g.kov
    Jan 26, 2020 at 6:04
  • 1
    $\begingroup$ @omegadot: Even better, $\mathrm{e}$ is just a shortcut for $\Omega^{-1/\Omega}$. $\endgroup$
    – g.kov
    Jan 26, 2020 at 11:00
9
$\begingroup$

Related, but certainly not in an immediately obvious way, to $\zeta(2)$ is the density of the squarefree numbers.

Call a natural number squarefree if no square larger than $1$ divides it (e.g. 12 is not squarefree because 4 divides it, but 30 is squarefree). Let $S$ be the set of squarefree numbers. Then $$ \lim_{n\to\infty}\frac{\#([1..n]\cap S)}{n} = \frac{6}{\pi^2}. $$

See here: http://mathworld.wolfram.com/Squarefree.html

$\endgroup$
8
$\begingroup$

I'll give you three cute examples from the book, (Almost) Impossible Integrals, Sums, and Series.

A particular case of the generalization from Section $1.11$, page $7$ $$i)\ 1- \int_0^1 \left(2 x+ 2^2 x^{2^2-1}+2^3 x^{2^3-1}+2^4 x^{2^4-1}+\cdots\right) \frac{\log(x)}{1+x} \textrm{d}x=\frac{\pi^2}{6}.$$

A particular case of the generalization from Section $1.38$, page $25$ $$ii) \ \frac{1}{2}\int_0^{ \infty} \int_0^{\infty}\frac{x -y}{e^x-e^y} \textrm{d}x \textrm{d}y=\int_0^{ \infty} \int_0^y\frac{x -y}{e^x-e^y} \textrm{d}x \textrm{d}y=\frac{\pi^2}{6}.$$ The first example from Section $1.17$, page $10$ $$\frac{6}{7\zeta(3)}\int _0^1 \int _0^1 \frac{\displaystyle \log \left(\frac{1}{x}\right)-\log \left( \frac{1}{y} \right)}{\displaystyle \log \left(\log \left(\frac{1}{x}\right)\right)-\log \left(\log \left(\frac{1}{y}\right)\right)} \textrm{d}x \textrm{d}y =\frac{6}{\pi^2}.$$

Another curious sum of (crazy) integrals leading to the same value which was proposed by the author of the mentioned book is

$$\frac{\pi^2}{6}=\frac{4}{3}\int_0^{\pi/2} \log \left(\frac{\left(x^2\sin^2(x)+ \pi ^2/4 \cos ^2(x)\right)^{x/2}}{x^x}\right)\sec ^2(x) \textrm{d}x$$ $$-\frac{2}{3} \int_0^1 \frac{\log \left(\left(x^2+\left(1-x^2\right) \cos (\pi x)+1\right)/2\right)}{x-x^3} \textrm{d}x,$$

but also the amazing $\zeta(2)\zeta(3)$ product in the harmonic series (with zeta tail) form

$$\frac{1}{2\zeta(3)}\sum_{n=1}^{\infty} \frac{H_n^2}{n}\left(\zeta(2)-1-\frac{1}{2^2}-\cdots-\frac{1}{n^2}\right)=\frac{\pi^2}{6},$$

or

$$\frac{\pi^2}{6}=4\sum_{n=1}^{\infty}\biggr(2n\biggr(1-\frac{1}{2^{2n+1}}\biggr)\zeta(2n+1)-2\log(2)\biggr(1-\frac{1}{2^{2n}}\biggr)\zeta(2n)$$ $$-\frac{1}{2^{2n}}\sum_{k=1}^{n-1}(1-2^{k+1})\zeta(k+1)(1-2^{2n-k})\zeta(2n-k)\biggr).$$

$\endgroup$
7
$\begingroup$

Here is a crazy-looking integral, which I believe I originally saw on the (now abandoned) integrals and series forum:

$$\int_{0}^{1}\frac{1}{\sqrt{1-x^2}}\arctan\left(\frac{88\sqrt{21}}{36x^2+215}\right)dx=\frac{\pi^2}{6}$$

$\endgroup$
1
  • 3
    $\begingroup$ Yes, that is from I&S. I took a whack at it in the comments there, but the closest I could get at the time was the similar looking integral $$\int_0^1\text{arctan}\left(\frac{88\sqrt{21}}{36x^2+215}\right)\,\mathrm dx=\text{arctan}\left(\frac{88\sqrt{21}}{251}\right)+\frac{\alpha+\bar{\alpha}}{2742}$$ where $\alpha=w_1w_2\text{arctan}\left(\frac{6}{w_1}\right)$, $w_1=\sqrt{215+i88\sqrt{21}}$, and $w_2=i(\bar{w_1})^2.$ Sad the forum has been inactive for so long, it was a fun place to post. $\endgroup$
    – dxdydz
    Jan 19, 2020 at 1:21
7
$\begingroup$

A somewhat surprising occurence, which can be seen immediatelly via the Euler product, appears in the study of visible points of lattice.

Given a lattice $\Gamma \subset \mathbb R^d$, meaning $\Gamma=\mathbb Z v_1 \oplus ... \oplus\mathbb Z v_d$ for some $\mathbb R$ basis $v_1,.., v_d$ of $\mathbb R^d$, the visible points of $\Gamma$ are defined as $$V:= \{ z=n_1v_1+...+n_dv_d : n_1,.., n_d \in \mathbb Z , \mbox{ gcd } (n_1,.., n_d)=1 \}$$

The, we have the following result, (see Prop.~6 in Diffraction from visible lattice points and k-th power free integers)

Proposition The natural density of $V$ is $$ \mbox{dens}(V)=\frac{1}{ \det(A) \zeta(d) } $$ where $A$ is the matrix with columns $v_1,v_2,...,v_d$. Here, natural density means the density calculated with respect to the sequence $A_n=[-n,n]^d$, note that this set can have a different density with respect to other sequences.

In particular, the visible sets of $\mathbb Z^2$, given by $$V=\{ (n,m) \in \mathbb Z^2 : \mbox{gcd}(m,n) =1 \}$$ have natural density $\frac{1}{\zeta(2)}$.


The so called "cut and project" formalism establises a connection between the above example and some sets in compact groups, which appeared in my research area recently.

Consider the group $$\mathbb K:= \prod_{p \in P} \left( \mathbb Z^2 / p \mathbb Z^2 \right)$$ where $p$ denotes the set of all primes. $\mathbb K$ is a compact Abelian group, and hence has a probability Haar measure $\theta_{\mathbb K}$.

Now, $\phi(m,n) := \left( (m,n)+p \mathbb Z^2 \right)_{p \in P}$ defines an embedding of $\mathbb Z^2$ into $\mathbb K$.

Define the set $$W:= \prod_{p \in P} \left( \bigl(\mathbb Z^2 / p \mathbb Z^2\bigr) \backslash \{ (0,0) + p\mathbb Z^2 \} \right)$$

Then, the visible points of $\mathbb Z^2$ are exactly $$V= \phi^{-1}(W)$$

The set $W$, which is used in the study of diffraction of $V$, has the following properties:

  • $W$ is closed and hence compact.
  • $W$ has empty interior (hence is fractal shape).
  • $\theta_{\mathbb K}(W) = \frac{1}{\zeta(2)}$

The last property is where I was going to, and it is intuitively not that hard to see once you identify $\theta_{\mathbb K}(W)$ as the product of the counting measures on $\mathbb Z^2 / p \mathbb Z^2$: this immediatelly gives $$\theta_{\mathbb K}(W) = \prod_{p \in P}\frac{p^2-1}{p^2}$$

P.S. There are similar appearences of $\zeta(n)$ in the study of $k$th power free integers, that is all the integers $n \in \mathbb Z$ which are not divisible by the $k$th power of any prime, for a fixed $k$.

$\endgroup$
7
$\begingroup$

Consider the following picture:enter image description here, centered at the origin of $\mathbf{R}^{2}.$ It is a concentric arrangement of circles $\color{red}{\text{(- this should be discs ?)}}$; each circle has radius $1/n.$ We can think of it as an infinite bulls-eye. The sum of the areas shaded in red is equal to $\frac{1}{2}\pi\zeta(2).$ In particular $$\sum_{k=1}^\infty \int_0^1 \frac{\sin (\pi (2k - 1)/ r)}{2k - 1} r \, dr = \frac{\pi}{8}\left(1-\zeta(2)\right), $$ Surprisingly if you take this arrangement and rotate it about the $x-$axis then you have a similar arrangement with circles being replaced by $3-$balls each with radius $1/n.$ In this case the sum of volumes shaded in red is equal to $\pi\zeta(3).$


Update: It dawned on me that I can in fact extend the notion "volume shaded in red" to higher dimensions.

Let $K_{i}$ be the $n-$ ball at the center of the origin of Euclidean $n-$space, $\mathbf{E}^{n},$ with radius $\frac{1}{i}$ and whose volume I denote by $\mu\left(K_{i}\right).$ Consider $$ \sum\limits_{i=1}^{\infty}(-1)^{i+1}\mu\left(K_{i}\right). $$ A closed form for this quantity is known whenever $n$ is an even number: $$ (-1)^{1+\frac{n}{2}}{\left(2^{n-1}-1 \right)B_{n}\above 1.5pt \Gamma(1+\frac{n}{2})\Gamma(1+n)}\pi^{\frac{3}{2}n}. $$ Inspections shows the numerator of the rational part is the sequence A036280(n/2). You can check in the case that $n=2$ the quantity computes to $\frac{1}{2}\pi\zeta(2).$

$\endgroup$
6
$\begingroup$

Two simple trigonometric integrals are $$\frac{4}{3}\int_0^1 \frac{\arctan{x}}{1+x^2}dx =\frac{\pi^2}{6}$$ and

$$\frac{4}{3}\int_0^1 \frac{\arcsin{x}}{\sqrt{1-x^2}}dx = \frac{\pi^2}{6}$$


Using inverse hyperbolic functions:

$$\frac{10}{3} \int_0^1\frac{\operatorname{arcsinh}\left({\frac{x}{2}}\right)}{x}dx=\frac{\pi^2}{6}$$

$$\frac{4}{3} \int_0^1 \frac{\operatorname{arctanh}{x}}{x} dx = \frac{\pi^2}{6}$$


From series $$ \sum_{k=0}^\infty \frac{1}{((k+1)(k+2))^2} = \frac{\pi^2}{3}-3 $$

and

$$ \sum_{k=0}^\infty \frac{k}{((k+1)(k+2))^2} = 5- \frac{\pi^2}{2} $$

$\frac{\pi^2}{6}$ arises directly when cancelling out the integer terms:

$$ \sum_{k=0}^\infty \frac{5+3k}{((k+1)(k+2))^2} =\frac{\pi^2}{6}$$

Similarly, $$\frac{8}{3}\sum_{k=0}^\infty \frac{4k+5}{(2k+1)^2(2k+3)^2} = \frac{\pi^2}{6}$$


More series and integrals are available at http://oeis.org/A013661

$\endgroup$
5
$\begingroup$

This paper gives a polynomial-time approximation algorithm for the Minimum Equivalent Digraph (MEG) problem, with approximation ratio $\pi^2/6$.

The problem is, given a directed graph, to find a min-size subset $S$ of the edges that preserves all reachability relations between pairs of vertices. (That is, for every pair $u, v$ of vertices, if there is a path from $u$ to $v$ in the original graph, then there is such a path that uses only edges in $S$.) The problem is NP-hard. This was the first poly-time algorithm with approximation ratio less than 2.

$\endgroup$
5
$\begingroup$

Integral representations are given by

$$2\int_0^1 x \left \lfloor{\frac1x}\right \rfloor \ dx=2\int_1^\infty\frac{1}{t}\lfloor{t\rfloor}\frac{dt}{t^2}=2\sum_{n=1}^{\infty}\int_n^{\infty}t^{-3}\,dt=\sum_{n=1}^{\infty}n^{-2}=\frac{\pi^2}{6}$$

and

$$\frac{4}{3}\int_0^1\int_0^1\frac{1}{1-x^2y^2}\,dxdy=\frac{\pi^2}{6}$$

Then, for every $-1 < \alpha \le 1$,

$$\int_0^\infty\frac{\log(1+\alpha x)}{x(1+x)}\,dx= \log(\alpha)\log(1-\alpha)+\text{Li}_2(\alpha)$$

and when $\alpha=1$, this integral becomes

$$\int_0^\infty\frac{\log(1+ x)}{x(1+x)}\,dx= \frac{\pi^2}{6}$$

$\endgroup$
5
$\begingroup$

This I found neat

$$\int_0^\pi \frac{\log(\frac{\cos x}{2}+1)}{\cos x} dx=\frac{\pi^2}{6}$$

$\endgroup$
5
$\begingroup$

Here is one, $$-\sum_{n=0}^{\infty}\left[\zeta(2n)-\zeta(2n+2)-\zeta(2n+3)+\zeta(2n+4)\right]=\frac{\pi^2}{6}$$

$\endgroup$
3
  • $\begingroup$ I found it, by experimental mathematic approach. $\endgroup$
    – Sibawayh
    May 8, 2021 at 21:08
  • $\begingroup$ Are you sure the sum starts from $n=0$? I thought the zeta function is undefined at $n=0$. $\endgroup$ Jun 11, 2021 at 10:26
  • 1
    $\begingroup$ @A-Level Student - $\zeta(0) = -\frac{1}{2}$. $\endgroup$
    – omegadot
    Jan 7 at 4:55
5
$\begingroup$

This is one of those amazing series for $1/\pi^2$. You can find them in this paper by G. Almkvist and J. Guillera.

$$\left(\frac{2}{5}\right)^{3}\sum_{n=0}^{\infty}\frac{(6n)!}{n!^{6}10^{6n}}(532n^{2}+126n+9)=\frac{6}{\pi^{2}}$$

$\endgroup$
4
$\begingroup$

How about

$$\int_0^1 dx \, \log{x} \, \log{(1-x)} = 2 - \frac{\pi^2}{6} $$

$\endgroup$
1
  • 3
    $\begingroup$ Combining with $$\int_0^1 x(1-x)\log{x}\log(1-x)dx=\frac{1}{108}(37-3\pi^2)$$ we get rid of the 2: $$ -\int_0^1(216x^2-216x+37)\log{x}\log(1-x)dx=\frac{\pi^2}{6}$$ $\endgroup$ Jan 26, 2020 at 8:39
3
$\begingroup$

For $0< x<1$, \begin{align}\text{Li}_2(x)+\text{Li}_2(1-x)+\ln x\ln(1-x)=\dfrac{\pi^2}{6}\end{align}

And, for $0\leq x\leq 1$, $\displaystyle \text{Li}_2(x)=\sum_{n=1}^\infty \dfrac{x^n}{n^2}$

$\endgroup$
3
$\begingroup$

In physics, $\pi^2/6$ appears as a proportionality constant between a metal's internal energy (or at least the contribution of the electrons to that energy) on the one hand and the density of states $\times$ the Fermi temperature on the other hand. It appears there as another manifestation of the identity $\zeta(2)=\pi^2/6$, i.e. its derivation has not really an independent character here compared to the installments of the other answers.

$\endgroup$
3
$\begingroup$

I was quite pleasantly surprised to derive the continued fraction identity $$\frac{\pi^2}{6}=-1+\cfrac{1}{1-\cfrac{1}{2-\cfrac{1}{5-\cfrac{16}{13-\cfrac{81}{25-\cfrac{256}{41-\cdots}}}}}},$$ which comes from the identity for the polylogarithm $\mathrm{Li}_s(z)$: $$\mathrm{Li}_s(z)+1=\cfrac{1}{1-\cfrac{z}{1+z-\cfrac{z}{2^s+z-\cfrac{2^{2s}z}{3^s+2^sz-\cfrac{3^{2s}z}{4^s+3^sz-\cfrac{4^{2s}z}{5^s+4^sz-\cdots}}}}}},$$ which is valid for $\Re(s)>1$ and $|z|\le1$.

$\endgroup$
3
$\begingroup$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ With Abel-Plana Formula: \begin{align} {\pi^{2} \over 6\phantom{^{2}}} & = {3 \over 2} + 2\int_{0}^{\infty}{\sin\pars{2\arctan\pars{t}} \over \pars{1 + t^{2}}\pars{\expo{2\pi{\large t}} - 1}}\,\dd t \end{align}

$\endgroup$
2
$\begingroup$

Background

For a certain project, I collected a considerable number of definite integrals whose evaluations amounted to numbers of the form $P :=a+b \pi^{2} $, where $a, b \in \mathbb{Q}$. I won't list them all here (the document comprises 34 pages, including references), but state the ones that stood out to me most in terms of bringing an element of surprise.

Integrals of Elementary Functions

  1. By [1] we have $$\int_{0}^{1} \left( \frac{\arctan(x)}{1+x^{2}} \right) dx = \frac{\pi^{2}}{32} . \tag{1} $$
  2. In the same source, we find $$\int_{0}^{\ln(\phi)} x \coth(x) dx = \frac{\pi^{2}}{20}, \tag{2} $$ where $\phi = \frac{1+\sqrt{5}}{2} $ is the golden ratio.
  3. In [2] we find $$\int_{0}^{\pi/2} \frac{\ln(\sec(x))}{\tan(x)} = \frac{\pi^{2}}{24}. \tag{3} $$
  4. Another logarithmic integral yet quite different from the previous one is [3] $$\int_{0}^{1} \frac{\ln(1+x+x^{2}+ x^{3} + x^{4} + x^{5} + x^{6})}{x} dx = \frac{\pi^{2}}{7}. \tag{4} $$
  5. By Theorem 1 of [4] we retrieve $$\int_{0}^{\infty} \Big{(} \sinh^{-1} (\cosh(u)) - u \Big{)} du = \frac{\pi^{2}}{16}. \tag{5} $$
  6. On p. 375, entry 3.521 of [5] we read $$\int_{0}^{\infty} \frac{ x }{\sinh(ax) }dx = \frac{\pi^{2}}{4a^{2}} \tag{6} .$$
  7. By p. 389, entry 3.557.6 of [5] we have $$\int_{0}^{\infty} x \frac{(e^{-x} -1 ) }{ (\cosh(x) - 1 ) } dx = - \frac{\pi^{2}}{3} . \tag{7} $$
  8. We have by p. 568, entry 4.317.11 of [5] that $$\int_{-\infty}^{\infty} \ln \left| \frac{1+2 \sqrt{1+x^{2}} }{1-2 \sqrt{1+x^{2}} } \right| \frac{dx}{(1+x^{2})^{1/2}} = \frac{\pi^{2}}{3}. \tag{8} $$
  9. By p. 569, entry 4.323.3 of [5] we learn that the following identity holds: $$\int_{0}^{\infty} \ln \bigg{(} \frac{1+\tan(x) }{1-\tan(x)} \bigg{)}^{2} \frac{dx}{x} = \frac{\pi^{2}}{2}. \tag{9} $$
  10. Another integral from [5]: on p. 582, entry 4.384.12, it is stated that $$\int_{0}^{\pi/2} \ln \big{(} \sin(x) \big{)} \tan(x) dx = - \frac{\pi^{2}}{24} . \tag{10}$$
  11. In [6] we find a long list of integrals. Here, [7] is relevant, among others. We find $$ \int_{0}^{\infty} \operatorname{arctanh}\big{(} e^{-ax} \big{)} dx = \frac{\pi^{2}}{8a}, \tag{11} $$
  12. and we also find $$\int_{0}^{\infty} x \operatorname{csch}(ax) dx = \frac{\pi^{2}}{4a^2} \tag{12} $$ when we set $n=2 $ in [8].
  13. On p. 544 of entry 2.6.36.3 of [9] we find $$\int_{0}^{\pi/2} \frac{ \ln \Big{(}1+\frac{\sin(x)}{2} \Big{)} }{\sin(x) } dx = \frac{5 \pi^{2}}{72} \tag{13} $$ when we set $a = 1/2 $.

We now move to the second half of the answer.

Integrals of Special Functions

To me, this is the most fun part.

  1. On p. 28 of [10], we find $$\int_{1}^{\infty} [-W_{0}(-xe^{-x})]^{\alpha} x^{-\alpha} dx = \alpha \psi_{1}(\alpha) -1, $$ where $W_{0}$ is the principal branch of the Lambert W-function and $\psi_{1}(\cdot) $ is the trigamma function and $\alpha>1$. If we take into account the special value of the trigamma function at $\alpha = 2 $, we obtain $$\int_{1}^{\infty} [-W_{0}(-xe^{-x})]^{2} x^{-2} dx = \frac{\pi^{2}}{3} -3. \tag{14}$$ Something similar holds for another branch of the Lambert W function: $$I_{3} := \int_{0}^{1} [-W_{-1}(-xe^{-x})]^{\alpha} x^{-\alpha} dx = \alpha \psi_{1}(1-\alpha)+ 1,$$ where in this case $|\alpha|<1$.
  2. On p. 120 of [11] we find $$\int_{0}^{1} x \Big{ \{ }\frac{1}{x} \Big{ \} } \Big{ \lfloor } \frac{1}{x} \Big{ \rfloor } dx = \frac{1}{2} \Big{(} 1 - \frac{\pi^{2}}{6} \Big{)} . \tag{15}$$ Here, $\big{ \lfloor }\cdot \big{ \rfloor } $ is the floor function, and $\{ \cdot \}$ is the fractional part function.
  3. Moreover, we obtain via [12] that $$ \int_{0}^{1} \frac{1}{ \Big{ \lfloor } \frac{1}{x} \Big{ \rfloor } } dx = \frac{\pi^{2}}{6} -1 . \tag{16} $$
  4. By [13] we have $$ \int_{0}^{\infty} x K_{0}(x) K_{1}(x) dx = \frac{\pi^{2}}{8}, \tag{17}$$ where $K_{0}(\cdot) $ and $K_{1}(\cdot) $ are the modified Bessel functions of the second kind of the zeroth and first order. Other definite integrals of Bessel functions involving $\pi^{2}$ can be found in the OEIS entries of the decimal expansions of $\frac{\pi^{2}}{4}$ and $\frac{\pi^{2}}{16}$.
  5. By p. 632, entry 6.141.2 of [5] we have $$ \int_{0}^{1} \Big{(} \textbf{K} (k') \Big{)} dk = \frac{\pi^{2}}{4} , \tag{18}$$ where $\textbf{K}(\cdot) $ is the Complete Elliptic Integral of the first kind, and $k' := \sqrt{1-k^{2}}$ is its complementary modulus.
  6. By p. 885, entry 8.219.1 of [5] we obtain $$ \int_{0}^{\infty} \operatorname{Ei}^{2}(x) e^{-2x} dx = \frac{\pi^{2}}{4}. \tag{19} $$ Here, $\operatorname{Ei}(\cdot)$ is the exponential integral function. Other, similar integrals can be found in [16], for instance on p. 203.
  7. Define the Rogers dilogarithm function as follows: $$L(x) := - \frac{1}{2} \int_{0}^{x} \bigg{(} \frac{\ln(1-y)}{y} + \frac{\ln(y)}{1-y} \bigg{)} dy. $$ By p. 10 of [14] we have $$L \Big{(} \frac{\sqrt{5}-1}{2} \Big{)} = \frac{\pi^{2}}{10}, \quad \text{and} \quad L \Big{(} \frac{3-\sqrt{5}}{2} \Big{)} = \frac{\pi^{2}}{15} . \tag{20}$$
  8. Let $|q|<1$ and define the $q$-shifted factorials as $$(a;q)_{n} := \prod_{i=0}^{n-1} \left(1-a q^{i} \right) ,$$ with the limiting cases $$\left(a;q \right)_{0} = 1 $$ and $$\left( a;q \right)_{\infty} = \prod_{i=0}^{\infty} \left(1-a q^{i} \right) .$$ By equations $(12)$ and $(13)$ of [15], we find $$ \int_{- \frac{1}{2} \ln(q) }^{ \frac{1}{2} \ln(q) } \ln \left( q e^{\pm 2x} ; q^{2} \right)_{\infty} dx = \frac{\zeta(2)}{2} = \frac{\pi^{2}}{12}. \tag{21} $$
  9. Finally, define the logarithmic integral function as $$\operatorname{li}(x) := \int_{0}^{x} \frac{dt}{\ln(t)} .$$ By [17] we have $$\int_{0}^{1} \left( \frac{\operatorname{li}(x)}{x} \right)^{2} dx = \frac{\pi^{2}}{6} .$$

Sources

[1] N. J. A. Sloane, J. F. Alcover. OEIS A002388, 2022, 2013. Online Encyclopedia for Integer Sequences, link

[2] N. J. A. Sloane. OEIS A222171, 2022. Online Encyclopedia for Integer Sequences, link

[3] N. J. A. Sloane. OEIS A195056, 2022. Online Encyclopedia for Integer Sequences, link

[4] F. M. S. Lima. New definite integrals and a two-term dilogarithm identity. Indagationes Mathematicae, 2012, link

[5] I.S. Gradshteyn, I.M. Ryzhik. Table of Integrals, Series, and Products, seventh edition. Academic Press, 2007, link

[6] A. Dieckmann, Table of Integrals, Elsa Physik, Physikalisches Institut der Uni Bonn, 14 July 2022, link

[7] A. Dieckmann, Table of Integrals, Elsa Physik, Physikalisches Institut der Uni Bonn, 14 July 2022, specific integral link

[8] A. Dieckmann, Table of Integrals, Elsa Physik, Physikalisches Institut der Uni Bonn, 14 July 2022, specific integral link

[9] A. P. Prudnikov and Yu. A. Brychkov and O. I. Marichev. Integrals and Series, Volume 1: Elementary Functions. Gordon and Breach Science Publishers, 1986, link

[10] W. Gautschi. The Lambert W-functions and some of their integrals: a case study of high-precision computation. Numerical Algorithms, 2011, link

[11] O. Furdui. Limits, Series, and Fractional Part Integrals. Springer, 2012, link

[12] jh235, Tintarn. A Funny Integral. Art of Problem Solving, 2015, link

[13] N. J. A. Sloane. OEIS A111003, 2022. Online Encyclopedia for Integer Sequences, link

[14] A. N. Krillov. Dilogarithm Identities. ArXiv, 1994, link

[15] M. E. Bachraoui. Short proofs for $q$-Raabe formula and integrals for Jacobi theta functions. Journal of Number Theory, April 2017, link

[16] E. Geller, E. W. Ng. A Table of Integrals of the Exponential Integral. JOURNAL OF RESEARCH of the National Bureau af Standards, 1969, link

[17] Erik Satie, ComplexYetTrivial. prove that $\int_{0}^{1}\Big(\frac{\operatorname{li}(x)}{x}\Big)^2dx= \frac{\pi^2}{6}$. Math Stackexchange, 2020, link

$\endgroup$
1
$\begingroup$

You have $$\int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{+\infty}\frac{\sin(x^2+y^2)}{x^2+y^2}dxdy=\frac{\pi^2}{2}.$$ Of course, dividing by 3 you have the expected value $\pi^2/6$.

$\endgroup$
1
$\begingroup$

$$\sum_{n=0}^{\infty}\frac{C_n^5}{2^{10n}}\left(\frac{n+1}{2n-1}\right)^4(4n-1)\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)[4k^2-2k+1]}{(2k-1)(2k+1)}=\frac{4}{\pi^2}$$

Where $C_n$; Catalan numbers

$\endgroup$
2
  • 1
    $\begingroup$ Sibawayh, Where's $\frac{\pi^2}{6}$ here? $\endgroup$ May 16, 2021 at 8:58
  • $\begingroup$ @BooleanWick I guess Sibawayh means it's equal to $$\frac{2}{3}\left(\frac{\pi^2}{6}\right)^{-1}$$ $\endgroup$ Jun 11, 2021 at 10:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.