Unexpected appearances of $\pi^2 /~6$.

Question

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

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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! :$)$

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

Answer 1

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 $\frac{\pi^2}6$.

Answer 2

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.

Answer 3

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.

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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.

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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 $$j(a)=\frac12(f(a)+f(-a))=\int_0^\infty\frac{(x^2+1)\arctan x}{x^4+2(1-2a^2)x^2+1}dx=\frac{\pi^2}{8\sqrt{1-a^2}}.\tag1$$ Hence $$j\left(\sqrt{1-\sqrt{3}/2}\right)=\int_0^\infty\frac{(x^2+1)\arctan x}{x^4+2(\sqrt3-1)x^2+1}dx=\frac{\pi^2}{6}.$$

On a side note, upon integrating $(1)$ over $a\in[0,t]$ with $|t|\le1$, we get

$$\int_0^\infty\frac{\arctan x}{x}\log\left(\frac{1+2tx+x^2}{1-2tx+x^2}\right)dx=\frac{\pi^2}{2}\arcsin t$$

Then integration by parts yields $$\int_0^\infty \frac{\mathrm{Ti}_2(x)(x^2-1)}{x^4-2(2t^2-1)x^2+1}dx=\frac{\pi^2}{8t}\arcsin t,$$ where $\mathrm{Ti}_2(x)$ is the inverse tangent integral $$\mathrm{Ti}_2(x)=\int_0^x\frac{\arctan y}{y}dy.$$

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Another cool integral is $$\int_0^\infty\left\{x-\log(2\sinh x)\right\}\cos(x)\cos(x/n)^ndx=\\ =\frac{1}{2^{n+1}}\left[\color{red}{\frac{\pi^2}{6}}+\frac{1}{2}\sum_{k=1}^n \frac{n^2}{k^2}{n\choose k}\left(\frac{\pi k}{n}\coth\frac{\pi k}{n}-1\right)\right]$$

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Expect more nice examples as I gather the best ones.

Answer 4

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.

Answer 5

The simplest

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

Answer 6

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}

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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}

$\endgroup$

Answer 7

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 $.
14. As pointed out by user Black Emperor in the comment section of this question [19], we have $$ \int_{0}^{\pi/4} \frac{(\sin(x)+\cos(x)) \arctan( \sin(x) + \cos(x) ) }{2 - \sin(2x)} dx = \frac{7}{96} \pi^{2}. \tag{14} $$
15. In MSE question [25], user Zacky proves that $$\int_0^{\pi/3}\arccos(2\sin^2 x-\cos x)\mathrm dx=\frac{\pi^2}{5}.$$

We now move to the second third 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 $$\int_{1}^{\infty} [-W_{0}(-xe^{-x})]^{2} x^{-2} dx = \frac{\pi^{2}}{3} -3. \tag{15}$$ 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 $$\int_{0}^{1} x \Big{ \{ }\frac{1}{x} \Big{ \} } \Big{ \lfloor } \frac{1}{x} \Big{ \rfloor } dx = \frac{1}{2} \Big{(} 1 - \frac{\pi^{2}}{6} \Big{)} . \tag{16}$$ Here, $\big{ \lfloor }\cdot \big{ \rfloor } $ is the floor function, and $\{ \cdot \}$ is the fractional part function.
3. Moreover, we obtain via [12] that $$ \int_{0}^{1} \frac{1}{ \Big{ \lfloor } \frac{1}{x} \Big{ \rfloor } } dx = \frac{\pi^{2}}{6} -1 . \tag{17} $$
4. By [13] we have $$ \int_{0}^{\infty} x K_{0}(x) K_{1}(x) dx = \frac{\pi^{2}}{8}, \tag{18}$$ 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 $$ \int_{0}^{1} \Big{(} \textbf{K} (k') \Big{)} dk = \frac{\pi^{2}}{4} , \tag{19}$$ 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 $$ \int_{0}^{\infty} \operatorname{Ei}^{2}(x) e^{-2x} dx = \frac{\pi^{2}}{4}. \tag{20} $$ 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 $$L \Big{(} \frac{\sqrt{5}-1}{2} \Big{)} = \frac{\pi^{2}}{10}, \quad \text{and} \quad L \Big{(} \frac{3-\sqrt{5}}{2} \Big{)} = \frac{\pi^{2}}{15} . \tag{21}$$
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 $$ \int_{- \frac{1}{2} \ln(q) }^{ \frac{1}{2} \ln(q) } \ln \left( q e^{\pm 2x} ; q^{2} \right)_{\infty} dx = \frac{\zeta(2)}{2} = \frac{\pi^{2}}{12}. \tag{22} $$
9. Now, define the logarithmic integral function as $$\operatorname{li}(x) := \int_{0}^{x} \frac{dt}{\ln(t)} .$$ By [17] we have $$\int_{0}^{1} \left( \frac{\operatorname{li}(x)}{x} \right)^{2} dx = \frac{\pi^{2}}{6} . \tag{23} $$
10. Finally, we have by [20] that the following holds: $$ \int_1^{\infty} \frac{\operatorname{li}(x)^2 (x - 1)}{x^4} dx = \frac{5 \pi^2}{36}. \tag{24} $$

Sums

In the final third part of the answer, we delve into relevant sums.

1. Let $t_n$ be the $n$'th term of the Thue-Morse Sequence. Then by the following 2022 paper by László Tóth [18], we have: $$ \sum_{n=1}^{\infty} \frac{5 t_{n-1} + 3 t_{n}}{n^2} = \frac{2 \pi^2}{3} . \tag{25}$$
2. In the comment section of this MSE question [21], user Ethan Splaver points out that $$ \sum_{n=1}^{\infty} \frac{H_{n}}{n 2^{n}} = \frac{\pi^{2}}{12} . \tag{26}$$
3. On p. 9 (eq. 1.73) of [22] we also see - among many other, related formulas - the following identity: $$ \sum_{k=1}^{\infty} \frac{28k^2 + 31k + 8}{(2k+1)^2 k^3 \binom{2k}{k}^3 } = \frac{\pi^2-8}{2}. \tag{27} $$
4. In [23], the author obtains the following identity: $$\sum_{k=1}^{\infty} \frac{(3k-1)16^k }{k^{3} \binom{2k}{k}^3} = \frac{\pi^2}{2}. \tag{28} $$
5. In the MSE question [24], user Ty. inquires about the series identity $$\sum_{n=1}^{\infty} \arctan\left(\frac{1}{F_n}\right) \arctan\left(\frac{1}{F_{n+1}}\right)=\frac{\pi^2}{8}. \tag{29}$$ Here, $F_{n}$ represents the $n$'th Fibonacci number. The question was answered by user metamorphy.

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

[18] L. Tóth, Linear Combinations of Dirichlet Series associated with the Thue-Morse Sequence. Integers, 2022, link

[19] OnTheWay, Black Emperor. Evaluate $\int\limits_0^{\sqrt 2 } {\frac{1}{{3{a^2} + 2}}\frac{{\arctan \left( {\sqrt {{a^2} + 1} } \right)}}{{\sqrt {{a^2} + 1} }}da}$. Math Stackexchange, September 2023, link

[20] mick, metamorphy, Marco Cantarini. Show that $ \int_1^{\infty} \frac{\operatorname{li}(x)^2 (x - 1)}{x^4} dx = \frac{5 \pi^2}{36} $. Math Stackexchange, April 2023, link

[21] OlegK, mick, Ethan Splaver. Infinite Series $\sum_{n=1}^\infty\frac{H_n}{n^32^n}$. August 2014, link

[22] Z.-W. Sun. New Series for powers of $\pi$ and related Congruences. Electronic Research Archive and ArXiv, 2020 link

[23] J. Guillera. Hypergeometric identities for 10 extended Ramanujan-type series. The Ramanujan Journal, 2008, link

[24] Ty., metamorphy. Prove $\sum_{n=1}^{\infty} \arctan\left(\frac{1}{F_n}\right) \arctan\left(\frac{1}{F_{n+1}}\right)=\frac{\pi^2}{8}$. Math Stackexchange, April 2021, link

[25] Dan, Zacky. $\int_0^{\pi/3}\arccos(2\sin^2 x-\cos x)\mathrm dx=\frac{\pi^2}{5}$. Math Stackexchange, January 2024, link

Answer 8

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

Answer 9

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

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

Answer 10

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 $\frac{1}{\zeta(2)}$.

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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 $k$.

Answer 11

Consider the following picture:, 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 $\pi\zeta(3).$

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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 $\frac{1}{2}\pi\zeta(2).$

Answer 12

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

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

Answer 13

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

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

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Using inverse hyperbolic functions:

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

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

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

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

Answer 14

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.

Answer 15

This I found neat

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

Answer 16

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

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

Answer 17

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

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

Answer 18

How about

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

Answer 19

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

Answer 20

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

Answer 21

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.

Answer 22

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

Answer 23

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 $|z|\le1$.

Answer 24

Draw line segments from the centre of a unit circle, to two uniformly random points on the circle.

The expected product of the areas is $\frac{1}{\pi}\int_0^\pi\frac{1}{2}x\left(\pi-\frac{1}{2}x\right)dx=\frac{\pi^2}{6}$.

This question asks whether this result can be used to solve the Basel problem.

Answer 25

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 $\pi^2/6$.

Answer 26

If $a_n$ is $1$ or $-1$ with equal probability, then

$$\mathbb{E}\left[\left(\sum\limits_{n=1}^\infty\frac{a_n}{n}\right)^2\right]=\frac{\pi^2}{6}$$

Source: last page of this article.

Answer 27

On a circle of area $\sqrt{n}\pi$, draw $3n$ evenly spaced points. Draw $n$ triangles using the points as vertices, as shown in this example with $n=4$.

As $n\to\infty$, the sum of squares of areas of the triangles approaches $\dfrac{\pi^2}{6}$.

Proof:

Using the fact that the area of a triangle is $\frac12 ab\sin C$, the limit is:

$\begin{align} L&=\lim\limits_{n\to\infty}\sum\limits_{k=1}^n \left[\frac12\sqrt{n}\cdot 2\sin\left(\frac{(3k-2)\pi}{3n}\right)2\sin\left(\frac{(3k-1)\pi}{3n}\right)\sin\left(\frac{\pi}{3n}\right)\right]^2\\ \end{align}$

Use the compound angle formula for sine, and use $\cos\theta\to 1$ and $\sin\theta\to 0$ as $\theta\to 0$.

$\begin{align} L&=\lim\limits_{n\to\infty}\sum\limits_{k=1}^n \left[2\sqrt{n} \sin\left(\frac{k\pi}{n}\right)\sin\left(\frac{k\pi}{n}\right)\sin\left(\frac{\pi}{3n}\right)\right]^2\\ &=\lim\limits_{n\to\infty}\sum\limits_{k=1}^n \left[2\sqrt{n}\frac{\pi}{3n}\sin\left(\frac{k\pi}{n}\right)\sin\left(\frac{k\pi}{n}\right)\frac{\sin\left(\frac{\pi}{3n}\right)}{\frac{\pi}{3n}}\right]^2\\ &=\frac{4\pi^2}{9}\lim\limits_{n\to\infty}\frac1n\sum\limits_{k=1}^n \sin^4\left(\frac{k\pi}{n}\right)\\ &=\frac{4\pi^2}{9}\int_0^1\sin^4(x\pi)\mathrm dx\\ &=\frac{\pi^2}{6}\\ \end{align}$

Answer 28

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

Where $C_n$; Catalan numbers

Answer 29

A regular $n$-gon of side length $n$ is inscribed in a circle. As $n\to\infty$, the difference between their perimeters approaches $\frac{\pi^2}{6}$.

Answer 30

I might be late for the party, but I add these two:

$ 1) $ Let $ 0 < a < 1 $ and $ b > 0$ satisfy $ 4a^2 - b^2 = \frac{4}{3} $. Then

$ \int_{0}^{\pi} \arctan(a + b\cos\theta) \, d\theta = \frac{\pi^2}{6} $

Here the proof

$ 2) $ From here we know that $\int_0^1 \frac{\ln(1+x-x^2)}{x}dx = 2\ln^2\varphi $. It can be shown that $ \lim_{k \to ∞} \int_0^1 \frac{\ln(\sum\limits_{i=0}^{k-1}x^i - x^k)}{x}dx = \frac{\pi^2}{6} $

Answer 31

$$\int_0^{\pi/2} \cot^{-1}\frac{215+36\cos^2\theta}{88\sqrt{21}}\ d\theta =\frac{\pi^2}6$$

Answer 32

Start with 1 as your initial score, and one white ball with one black ball in a bag. You pick up at random one of them, and you do it twice (independently). If you get white in both cases, then you advance. Otherwise you stop and your score is 1. Everytime you advance, 1 is added up to your score and one new white ball is added into the bag, and you proceed as before, and so on, until you have to stop. The expected value of your final score is $\zeta(2)$.

I know, we don't get $\pi^2/6$ as requested, but I find it interesting that this number (anyhow we write it) is the expected value of the score in a game so easy to play.

Answer 33

Here is one. If we define the harmonic numbers $H_x$ to the positive real line using $$H_x=\sum_{k=1}^\infty\frac{x}{k(k+x)}$$Then $$H'_0=\frac{\pi^2}6$$And $$\lim_{k\rightarrow0}\frac{H_k}k=\frac{\pi^2}6$$

Answer 34

The harmonic series is the series with partial sums $$s_1=1\text{ and }s_n=s_{n-1}+\frac1n$$

Similarly, let's define harmonic product as the product with partial products $$p_1=1\text{ and }p_n=p_{n-1}\left(1+\frac1n\right)$$

Now let's combine these into the harmonic series-product, which has partial terms

$$u_1=1\text{ and }u_n=\left(u_{n-1}+\frac1n\right)\left(1+\frac1n\right)$$

Then we have:

$$\lim_{n\to\infty}\frac{u_n}{n}=\frac{\pi^2}{6}-\frac12$$

Explanation:

$u_n=\left(\frac{n+1}{n}\right)\left(\frac1n+u_{n-1}\right)$

$\dfrac{u_n}{n}$
$=\frac1n\left(\frac{n+1}{n}\right)\left(\frac1n+u_{n-1}\right)$
$=\frac1n\left(\frac{n+1}{n}\right)\left(\frac1n+\left(\frac{n}{n-1}\right)\left(\frac{1}{n-1}+u_{n-2}\right)\right)$
$=\frac1n\left(\frac{n+1}{n}\right)\left(\frac1n+\left(\frac{n}{n-1}\right)\left(\frac{1}{n-1}+\left(\frac{n-1}{n-2}\right)\left(\frac{1}{n-2}+u_{n-3}\right)\right)\right)$
$=\cdots$
$=\left(\frac{n+1}{n}\right)\left(\frac{1}{n^2}+\frac{1}{(n-1)^2}+\frac{1}{(n-2)^2}+\cdots+\frac{1}{2^2}+\frac{1}{2}\right)$
$\therefore \lim\limits_{n\to\infty}\dfrac{u_n}{n}=\dfrac{\pi^2}{6}-\dfrac12$

Answer 35

I got this fascinating result from this previous answer of mine

$$I(a,b) = \int_{-\infty}^{\infty} \frac{\mathrm{arccoth}(a\sqrt{x^2+b^2})}{\sqrt{x^2+b^2}}dx = \pi \cdot \arcsin \left(\frac{1}{ab}\right)$$

Setting $a=1$ and $b=2$, we get:

$$I(1,2) = \int_{-\infty}^{\infty} \frac{\mathrm{arccoth}(\sqrt{x^2+4})}{\sqrt{x^2+4}}dx = \pi \cdot \arcsin(\frac{1}{2}) = \pi \cdot \frac{\pi}{6} = \frac{\pi^2}{6} = \zeta(2)$$

Also

$$I(2,1) = I(1,2) = \int_{-\infty}^{\infty} \frac{\mathrm{arccoth}(2\sqrt{x^2+1})}{\sqrt{x^2+1}}dx = \frac{\pi^2}{6} = \zeta(2)$$

This holds for any combination of positive $a,b$ such that $ab =2$.

Thus we can write in general, for positive $a$:

$$I(a, \frac{2}{a}) = I(\frac{2}{a}, a) = \int_{-\infty}^{\infty} \frac{\mathrm{arccoth}\left(a\sqrt{x^2+\frac{4}{a^2}}\right)}{\sqrt{x^2+\frac{4}{a^2}}}dx = \int_{-\infty}^{\infty} \frac{\mathrm{arccoth}\left(\frac{2}{a}\cdot \sqrt{x^2+a^2}\right)}{\sqrt{x^2+a^2}}dx = \frac{\pi^2}{6} = \zeta(2)$$

Other interesting results which follow from this result that involve $\frac{\pi^2}{6}$ are:

$$I(1,1) = \int_{-\infty}^{\infty} \frac{\mathrm{arccoth}(\sqrt{x^2+1})}{\sqrt{x^2+1}}dx = \frac{\pi^2}{2} = 3\zeta(2)$$

$$I(1,\sqrt2) = \int_{-\infty}^{\infty} \frac{\mathrm{arccoth}(\sqrt{x^2+2})}{\sqrt{x^2+2}}dx = \frac{\pi^2}{4} = \frac{3}{2} \zeta(2)$$ $$I(1,\sqrt3) = \int_{-\infty}^{\infty} \frac{\mathrm{arccoth}(\sqrt{x^2+3})}{\sqrt{x^2+3}}dx = \frac{\pi^2}{3} = 2\zeta(2)$$

Answer 36

In one of my papers in Journal of Number Theory: https://www.sciencedirect.com/science/article/abs/pii/S0022314X1930277X

A remark after Corollary 8.1 implies the result: $$ \lim_{x\rightarrow\infty}\frac1x\sum_{p\leq x} \frac{\tau(p-1)\phi(p-1)}{p-1}=\frac6{\pi^2}. $$ Here, $\tau(n)$ is the number of divisors of $n$, and $\phi(n)$ is the Euler's totient function. The sum is taken over primes at most $x$.