# Basel problem (I)

One may find quite amazing to notice how relative time is ( check WBW’s post ). Let’s consider Poincaré conjecture, the only solved Millennium Problem. It took circa 100 years to solve it,  which as we know is not that long.  The Fermat’s Last Theorem remained  unsolved for over 358 years ( conjectured  in 1637 (!) ), which is almost nothing in comparison with more than 2000 years that took for proving inability of angle’s trisection construction.

The Basel problem which asks for the precise sum of reciprocals of the squares of the natural numbers, was posed in 1644 by Petro Mengoli and solved in 1735 by Leonard Euler. 91 year seems quite a short period if you consider the ones mentioned above. What’s interesting about the problem which absorbed the ranks of mathematicians ( especially Bernoulli’s family ) is that nowadays even the freshmen maths students can arrive at its solution. It highlights the development in mathematics and the progress which was made in it. I will emphasis on the proof announced in 1741, fourth and the last published Euler’s paper about the Basel problem.

Theorem: $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$
Proof: Let $t = \arcsin{x}$ and $\mathrm{d}t = \frac{1}{\sqrt{(1-x^2)}}$. We will integrate their product:

$\int_0^1 \! \arcsin{x} \, \frac{\mathrm{d}x}{\sqrt{(1-x^2)}} = \int_0^{\frac{\pi}{2}} \! t \ \mathrm{d}t = \arcsin^2{\frac{\pi}{2}} - \arcsin^2{0} = \frac{\pi^2}{8}.$

At this point, expand $\mathrm{d}t$ using generalized binomial theorem and integrate to receive $t$ series:
$(1 + (-x^2) )^{-\frac{1}{2}} = \sum_{k=0} \binom{-\frac{1}{2}} {k} = 1 + \frac{1}{2} x^2 + \frac{1\cdot 3}{2\cdot 4} x^4 +\frac{1\cdot 3 \cdot 5}{2\cdot 4 \cdot 6} x^6 + \frac{1\cdot 3 \cdot 5 \cdot 7}{2\cdot 4 \cdot 6 \cdot 8} x^8 + \ldots$
$t = \int \mathrm{d}t = x + \frac{1}{2 \cdot 3} x^3 + \frac{1\cdot 3}{2\cdot 4\cdot 5} x^5 +\frac{1\cdot 3 \cdot 5}{2\cdot 4 \cdot 6 \cdot 7} x^7 + \frac{1\cdot 3 \cdot 5 \cdot 7}{2\cdot 4 \cdot 6 \cdot 8 \cdot 9} x^9 + \ldots$

Multiplying $t$ series by $\mathrm{d}t$ gives first-equation integrand:
$\frac{x}{\sqrt{(1-x^2)}} + \frac{1}{2 \cdot 3} \frac{x^3}{\sqrt{(1-x^2)}} + \frac{1\cdot 3}{2\cdot 4\cdot 5} \frac{x^5}{\sqrt{(1-x^2)}} +\frac{1\cdot 3 \cdot 5}{2\cdot 4 \cdot 6 \cdot 7} \frac{x^7}{\sqrt{(1-x^2)}} + \frac{1\cdot 3 \cdot 5 \cdot 7}{2\cdot 4 \cdot 6 \cdot 8 \cdot 9} \frac{x^9}{\sqrt{(1-x^2)}} + \ldots$
As we consider $\frac{(n)!!}{(n+1)!! \cdot(n+2)}\frac{x^{n+2}}{\sqrt{(1-x^2)}}$ for $n = 1,3,5,\cdots$ as the formula of following elements, it may be easy to find recurrence relation.

$I_n = \int \frac{x^{n} \ \mathrm{d}x}{\sqrt{(1-x^2)}}$
$L_n = I_n - I_{n+2} = \int x^n \sqrt{1-x^2} \ \mathrm{d}x$

After application of integration by parts, it turns out:
$L_n = \sqrt{1-x^2}\frac{x^{n+1}}{n+1}+\frac{1}{n+1}I_{n+2}$.
After substituting $J_n$ with $I_n - I_{n+2}$
$I_{n+2} = \frac{n+1}{n+2} I_n - \frac{x^{n+1}}{n+2} \sqrt{1-x^2}$
The second addend can be eliminated as we reflect integral range $[0,1]$, thus:
$I_{n+2} = \frac{n+1}{n+2} I_n$

$\int_0^1 \frac{x \ \mathrm{d}x}{\sqrt{(1-x^2)}} = 1 \\ \int_0^1 \frac{x^3 \ \mathrm{d}x}{\sqrt{(1-x^2)}} = \frac{2}{3} \int_0^1 \frac{x \ \mathrm{d}x}{\sqrt{(1-x^2)}} = \frac{2}{3} \\ \int_0^1 \frac{x^5 \ \mathrm{d}x}{\sqrt{(1-x^2)}} = \frac{4}{5} \int_0^1 \frac{x^3 \ \mathrm{d}x}{\sqrt{(1-x^2)}} = \frac{2\cdot4}{3\cdot5} \\ \int_0^1 \frac{x^7 \ \mathrm{d}x}{\sqrt{(1-x^2)}} = \frac{6}{7} \int_0^1 \frac{x^5 \ \mathrm{d}x}{\sqrt{(1-x^2)}} = \frac{2\cdot4\cdot6}{3\cdot5\cdot7}$

According to the fact above and substituting integrals with values, we have:
$\frac{\pi^2}{8} = \int t \ \mathrm{d}t = 1 + \frac{1}{2 \cdot 3} \cdot \frac{2}{3} + \frac{1\cdot 3}{2\cdot 4\cdot 5} \cdot \frac{2\cdot4}{3\cdot5} +\frac{1\cdot 3 \cdot 5}{2\cdot 4 \cdot 6 \cdot 7} \cdot \frac{2\cdot4\cdot6}{3\cdot5\cdot7} + \ldots$
$\frac{\pi^2}{8} = 1 + \frac{1}{3\cdot3} + \frac{1}{5\cdot5} + \frac{1}{7\cdot7} + \ldots = \sum_{n=1}^\infty \frac{1}{(2n-1)^2}$
And obviously:
$\sum_{n=1}^\infty \frac{1}{n^2} = \sum_{n=1}^\infty \frac{1}{(2n-1)^2}+ \sum_{n=1}^\infty \frac{1}{(2n)^2} = \frac{\pi^2}{8} \cdot \frac{1}{1-\frac{1}{4}} = \frac{\pi^2}{6}. \square$