# Basel Problem

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## Theorem

$\displaystyle \zeta \left({2}\right) = \sum_{n \mathop = 1}^{\infty} {\frac 1 {n^2}} = \frac {\pi^2} 6$

where $\zeta$ denotes the Riemann zeta function.

## Proof 1

Consider the expression:

$\displaystyle \int_0^1 \int_0^1 \frac 1 {1 - x y} \mathrm d A$

WLOG, the Riemann Double Sum of this multi-variable function can be set such that $0 < x y < 1$.

$\displaystyle \int_0^1 \int_0^1 \frac 1 {1 - x y} \mathrm d A = \int_0^1 \int_0^1 \sum_{n \mathop = 0}^\infty \left({x y}\right)^n \mathrm d A$
$\displaystyle \int_0^1 \int_0^1 \frac 1 {1 - x y} \mathrm d A = \sum_{n \mathop = 0}^\infty \left({\int_0^1 x^n \mathrm d x \int_0^1 y^n \mathrm d y}\right)$

Calculating both integrals via Fundamental Theorem of Calculus:

$\displaystyle \int_0^1 \int_0^1 \frac 1 {1 - x y} \mathrm d A = \sum_{n \mathop = 0}^\infty \left({\frac 1 {n + 1} \frac 1 {n + 1} }\right)$

Correcting the sum shows the relationship between the first expression and the value to calculate.

$\displaystyle \zeta \left({2}\right) = \sum_{n \mathop = 1}^\infty \frac 1 {n^2} = \int_0^1 \int_0^1 \frac 1 {1 - x y}\mathrm d A$

Let $\displaystyle \left({u, v}\right) = \left({\frac{x + y} 2, \frac{y - x} 2}\right)$ so that:

$\left({x, y}\right) = \left({u - v, \ u + v}\right)$

Let:

$\left|{J}\right| = \left\vert{\dfrac{\partial \left({x, y}\right)} {\partial \left({u, v}\right)} }\right\vert = 2$
$\displaystyle \zeta \left({2}\right) = 2 \iint \limits_S \frac{\mathrm d u \ \mathrm d v} {1 - u^2 + v^2}$

where $S$ is the square defined by the coordinates:

$\left({0, 0}\right), \ \left({\dfrac 1 2, -\dfrac 1 2}\right), \ \left({1, 0}\right), \ \left({\dfrac 1 2, \dfrac 1 2}\right)$

Exploiting the symmetry of the square and the function over the $u$-axis, we have:

$\displaystyle \zeta \left({2}\right) = 4 \left({\int_0^{\frac 1 2} \! \int_0^u \frac {\mathrm d v \ \mathrm d u} {1 - u^2 + v^2} + \int_{\frac 1 2}^1 \! \int_0^{1 - u} \frac {\mathrm d v \ \mathrm d u} {1 - u^2 + v^2}}\right)$

Factoring $1-u^2$ gives us:

$\displaystyle \zeta \left({2}\right) = 4 \left({\int_0^{\frac 1 2} \! \int_0^u \frac 1 {1 - u^2} \ \frac {\mathrm d v \ \mathrm d u} {\frac {v^2} {1 - u^2} + 1} + \int_{\frac 1 2}^1 \! \int_0^{1 - u} \frac 1 {1 - u^2} \frac {\mathrm d v \ \mathrm d u} {\frac {v^2} {1 - u^2} + 1}}\right)$

and letting:

$s = \dfrac v {\sqrt{1 - u^2} }, \mathrm d s = \dfrac 1 {\sqrt{1 - u^2} }$

allows us to make a substitution into each integral, giving:

$\displaystyle \zeta \left({2}\right) = 4 \left({\int_0^{\frac 1 2} \frac 1 {\sqrt{1 - u^2} } \arctan \left({\frac u {\sqrt{1 - u^2} } }\right) \mathrm d u + \int_{\frac 1 2}^1 \frac 1 {\sqrt{1 - u^2} } \arctan \left({\frac{1 - u} {\sqrt{1 - u^2} } }\right) \mathrm d u}\right)$

Consider the right triangle with sides $1$, $x$ and $\sqrt{1 - x^2}$.

From it arises the identity:

$\arcsin x = \arctan \dfrac x {\sqrt{1 - x^2} }$

Let:

$\theta = \arctan \left({\dfrac{1 - u} {\sqrt{1 - u^2} } }\right)$

Then:

$\tan^2 \theta = \dfrac{1 - u} {1 + u}$

and from Sum of Squares of Sine and Cosine/Corollary 1, we have that:

$\sec^2 \theta = \dfrac 2 {1 + u}$

Solving for $u$, and by the Double Angle Formulas/Cosine/Corollary 1:

$\cos 2 \theta = u$

Taking arccosines at both sides, and using Sine of Complement equals Cosine:

$\theta = \dfrac 1 2 \arccos u = \dfrac \pi 4 - \dfrac{\arcsin u} 2$

This allows us to convert the arctans from the integrals into arcsines:

$\displaystyle \zeta \left({2}\right) = 4 \left({\int_0^{\frac 1 2} {\frac {\arcsin u} {\sqrt{1 - u^2} }\, \mathrm d u} + \int_{\frac 1 2}^1 {\frac 1 {\sqrt{1 - u^2} } \left({\frac \pi 4 - \frac {\arcsin u} 2}\right)\,\mathrm d u} }\right)$

Substituting:

$s = \arcsin u$, $\mathrm d s = \dfrac 1 {\sqrt{1 - u^2} }$

into the arcsines, and splitting the second integral:

$\zeta \left({2}\right) = 4 \left({\dfrac {\pi^2} {72} + \dfrac {\pi^2} {36} }\right) = \dfrac {\pi^2} 6$

$\blacksquare$

## Proof 2

Let:

$\displaystyle P_k = x \prod_{n \mathop = 1}^k \left({1 - \frac {x^2} {n^2 \pi^2}}\right)$

We note that:

 $$\displaystyle P_k - P_{k - 1}$$ $$=$$ $$\displaystyle \left({- \frac {x^3} {k^2 \pi^2} }\right) \prod_{n \mathop = 1}^{k - 1} \left({1 - \frac {x^2} {n^2 \pi^2} }\right)$$ $$\displaystyle$$ $$=$$ $$\displaystyle - \frac {x^3} {k^2 \pi ^2} + O \left({x^5}\right)$$ Big-O Notation

By Telescoping Series we find that the coefficient of $x^3$ in $P_k$ is given by:

$(1): \quad \displaystyle - \frac 1 {\pi^2} \sum_{i \mathop = 1}^k \frac 1 {i^2}$
$\displaystyle \sin x = x \prod_{n \mathop = 1}^\infty \left({1 - \frac {x^2} {n^2 \pi^2}}\right)$

So by taking the limit as $k \to \infty$ in $(1)$ and equating with the coefficient of $x^3$ in the Power Series Expansion for Sine Function, we can deduce:

$\displaystyle - \frac 1 {\pi^2} \sum_{i \mathop = 1}^{\infty} \frac 1 {i^2} = - \frac 1 {3!}$

hence:

$\displaystyle \sum_{i \mathop = 1}^{\infty} \frac 1 {i^2} = \frac {\pi^2} 6$

$\blacksquare$

## Proof 3

Let $x \in \left({0 \,.\,.\, \dfrac \pi 2}\right)$ and let $n$ be a non-negative integer.

 $$\displaystyle \frac {\cos \left({2n + 1}\right) x + i \sin \left({2n + 1}\right) x} {\sin^{2n + 1} x}$$ $$=$$ $$\displaystyle \frac {\left({\cos x + i \sin x}\right)^{2n + 1} } {\sin^{2n + 1} x}$$ De Moivre's Formula $$\displaystyle$$ $$=$$ $$\displaystyle \left({\cot x + i}\right)^{2n + 1}$$ $$\displaystyle$$ $$=$$ $$\displaystyle \sum_{r \mathop = 0}^{2n + 1} \binom{2n + 1} r i^r \cot^{2n + 1 - r} x$$ Binomial Theorem $$\displaystyle \implies$$ $$\displaystyle \frac {\sin \left({2n + 1}\right) x} {\sin^{2n + 1} x}$$ $$=$$ $$\displaystyle \sum_{r \mathop = 0}^n \binom {2n + 1} {2r + 1} \left({-1}\right)^r \cot^{2 \left({n - r}\right)} x$$ equating imaginary parts

Let $x_k = \dfrac {k \pi} {2n + 1}$ for $k = 1, 2, \ldots, n$.

Then:

$\sin \left({2n + 1}\right) x_k = 0$

So we have:

$\displaystyle \sum_{r \mathop = 0}^n \binom {2n + 1} {2r + 1} \left({-1}\right)^r \cot^{2 \left({n - r}\right)} x_k = 0$

for $k = 1, 2, \ldots, n$.

The numbers $x_k$ are all distinct and in the interval $\left({0 \,.\,.\, \dfrac \pi 2}\right)$.

By Shape of Cotangent Function, $\cot x$ is positive and injective in the interval $\left({0 \,.\,.\, \dfrac \pi 2}\right)$.

Therefore $\cot^2 x$ is also injective in this interval.

Hence the numbers $c_k = \cot^2 x_k$ are distinct for $k = 1, 2, \ldots, n$.

These numbers are the $n$ distinct roots of the $n$th degree polynomial:

$\displaystyle f \left({c}\right) := \sum_{r \mathop = 0}^n \binom {2n + 1} {2r + 1} (-1)^r c^{n - r}$

By Viète's Formulas, we can calculate the sum of the roots:

$\displaystyle \sum_{k \mathop = 1}^n \cot^2 x_k = \frac {\binom {2n + 1} 3} {\binom {2n + 1} 1} = \frac {2 n \left({2 n - 1}\right)} 6$

From the Difference of Squares of Cosecant and Cotangent $\cot^2 x = \csc^2 x - 1$ we can similarly deduce:

$\displaystyle \sum_{k \mathop = 1}^n \csc^2 x_k = \frac {2 n \left({2 n - 1}\right)} 6 + n = \frac {2 n \left({2 n + 2}\right)} 6$

By Shape of Sine Function, $\sin x$ is positive in the interval $\left({0 \,.\,.\, \dfrac \pi 2}\right)$.

So from the Tangent Inequality $\sin x < x < \tan x$ for $x \in \left({0 \,.\,.\, \dfrac \pi 2}\right)$, we can deduce that $\cot^2 x < 1 / x^2 < \csc^2 x$ in the same interval.

Summing this inequality from $x_1$ to $x_n$ gives:

 $$\displaystyle \frac {2 n \left({2 n - 1}\right)} 6$$ $$<$$ $$\displaystyle \sum_{k \mathop = 1}^n \left({\frac {2 n + 1} {k \pi} }\right)^2$$ $$\displaystyle$$ $$<$$ $$\displaystyle \frac {2 n \left({2 n + 2}\right)} 6$$

or equivalently:

 $$\displaystyle \frac {\pi^2} 6 \frac {2n \left({2n - 1}\right)} {\left({2n + 1}\right)^2}$$ $$<$$ $$\displaystyle \sum_{k \mathop = 1}^n \frac 1 {k^2}$$ $$\displaystyle$$ $$<$$ $$\displaystyle \frac {\pi^2} 6 \frac {2n \left({2n + 2}\right)} {\left({2n + 1}\right)^2}$$

By Combination Theorem for Limits of Functions, the left and right hand sides tend to $\dfrac {\pi^2} 6$ as $n$ tends to infinity.

Therefore by the Squeeze Theorem:

$\displaystyle \zeta \left({2}\right) = \sum_{k \mathop = 1}^\infty \frac 1 {k^2} = \frac {\pi^2} 6$

$\blacksquare$

## Proof 4

Let $n$ be a positive integer.

 $$\displaystyle \int_0^1 \frac {x^{2n + 1} } {\sqrt{1 - x^2} } \mathrm d x$$ $$=$$ $$\displaystyle \int_0^{\pi / 2} \sin^{2n + 1} x \ \mathrm d x$$ by substituting $x \to \sin x$ $$\displaystyle (1):$$ $$\displaystyle$$ $$=$$ $$\displaystyle \frac {2^{2n} \left({n!}\right)^2} {\left({2n + 1}\right)!}$$ by Reduction Formula for Definite Integral of Power of Sine

We have:

 $$\displaystyle \int_0^1 \frac {\arcsin x} {\sqrt {1 - x^2} } \ \mathrm d x$$ $$=$$ $$\displaystyle \int_0^1 \sum_{n \mathop = 0}^\infty \frac {\left({2n}\right)!} {2^{2n} \left({n!}\right)^2 \left({2n + 1}\right)} \frac {x^{2n + 1} } {\sqrt {1 - x^2} } \mathrm d x$$ by Taylor Series of Arcsine Function $$\displaystyle$$ $$=$$ $$\displaystyle \sum_{n \mathop = 0}^\infty \frac {\left({2n}\right)!} {2^{2n} \left({n!}\right)^2 \left({2n + 1}\right)} \int_0^1 \frac {x^{2n + 1} } {\sqrt {1 - x^2} } \mathrm d x$$ interchange of sum and integral is valid by Tonelli's Theorem $$\displaystyle$$ $$=$$ $$\displaystyle \sum_{n \mathop = 0}^\infty \frac {\left({2n}\right)!} {2^{2n} \left({n!}\right)^2 \left({2n + 1}\right)} \frac {2^{2n} \left({n!}\right)^2} {\left({2n + 1}\right)!}$$ by $(1)$ $$\displaystyle$$ $$=$$ $$\displaystyle \sum_{n \mathop = 0}^\infty \frac 1 {\left({2n + 1}\right)^2}$$

We also have:

 $$\displaystyle \int_0^1 \frac {\arcsin x} {\sqrt {1 - x^2} } \ \mathrm d x$$ $$=$$ $$\displaystyle \left[{\frac 1 2 \left({\arcsin x}\right)^2}\right]_0^1$$ by Fundamental Theorem of Calculus $$\displaystyle$$ $$=$$ $$\displaystyle \frac {\pi^2} 8$$

So we can deduce:

$\displaystyle \sum_{n \mathop = 0}^\infty \frac 1 {\left({2n + 1}\right)^2} = \frac {\pi^2} 8$

Finally note that:

 $$\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {n^2}$$ $$=$$ $$\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {\left({2n}\right)^2} + \sum_{n \mathop = 0}^\infty \frac 1 {\left({2n + 1}\right)^2}$$ $$\displaystyle$$ $$=$$ $$\displaystyle \frac 1 4 \sum_{n \mathop = 1}^\infty \frac 1 {n^2} + \frac {\pi^2} 8$$ $$\displaystyle \implies$$ $$\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {n^2}$$ $$=$$ $$\displaystyle \frac {\pi^2} 6$$

$\blacksquare$

## Historical Note

Although often credited to Euler, it has been suggested that it was in fact solved by Nicolaus I Bernoulli.