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Undid revision 929555438 by Contribute.Math (talk) WP:WINARS |
I found the book where Euler wrote his second proof. I mentionned that the proof shown here is almost the same. This proof may seem more familiar to some readers because it uses the famous Wallis' integrals. |
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Using the [[Weierstrass factorization theorem]], it can also be shown that the left-hand side is the product of linear factors given by its roots, just as we do for finite polynomials (which Euler assumed as a [[heuristic]] for expanding an infinite degree [[polynomial]] in terms of its roots, but is in general not always true for general <math>P(x)</math>):<ref>A priori, since the left-hand-side is a [[polynomial]] (of infinite degree) we can write it as a product of its roots as
:<math>\begin{align}
\sin(x) & = x (x^2-\pi^2)(x^2-4\pi^2)(x^2-9\pi^2) \cdots \\
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and this completes the proof.
== Euler's second proof ==
The first accepted proof of the theorem was made by Euler in 1741<ref>{{Cite book|title=Opera Omnia|last=Euler|first=Leonhard|publisher=|year=1741|isbn=|___location=|pages=177-186}}</ref>, six years after his first proof. The rigor of the latter was challenged at that time because the Weierstrass factorization theorem had not been discovered yet. The following proof is almost the same as Euler's proof. The integration by substitution <math>\ u = sin(t)</math> links the two proofs.
First of all, <math>\displaystyle \int ^{\frac{\pi }{2}}_{0} \operatorname{Arcsin}( \sin\ t) \ dt\ =\int ^{\frac{\pi }{2}}_{0} t\ dt\
=\frac{\pi ^{2}}{8}
</math>
Now one can use the [[List of mathematical series|power series of Arcsin]] :
<math>\begin{align}
& \int ^{\frac{\pi }{2}}_{0} \ (\sum ^{+\infty }_{n=0}\frac{( 2n) !}{\left( n!\ 2^{n}\right)^{2}( 2n+1) \ } \ \sin( t)^{2n+1}) dt\ & = & \ \ \sum ^{+\infty }_{n=0}\frac{( 2n) !}{\left( n!\ 2^{n}\right)^{2}( 2n+1) \ } \ \int ^{\frac{\pi }{2}}_{0} \sin( t)^{2n+1} dt\\
\\
& & = & \ \ \sum ^{+\infty }_{n=0}\frac{( 2n) !}{\left( n!\ 2^{n}\right)^{2}( 2n+1) \ } \ \frac{\left( n!\ 2^{n}\right)^{2}}{( 2n+1) !}\\
& & = & \ \ \sum ^{+\infty }_{n=0}\frac{1}{( 2n+1)^{2}} \\
& & = & \ \ \frac{\pi ^{2}}{8}
\end{align}
</math>
The sum and the integral signs were interchanged thanks to Beppo Levi's [[monotone convergence theorem]] for Lebesgue integral. Then, [[Wallis' integrals]] enabled us to integrate the powers of sine.
One can separate even and odd numbers :
<math>\displaystyle \sum ^{+\infty }_{n=1}\frac{1}{n^{2}} =\sum ^{+\infty }_{n=1}\frac{1}{( 2n)^{2}} \ +\sum ^{+\infty }_{n=0}\frac{1}{( 2n+1)^{2}} \
=\frac{1}{4}\sum ^{+\infty }_{n=1}\frac{1}{n^{2}} +\displaystyle \ \frac{\pi ^{2}}{8}
</math>
Finally,
<math>\displaystyle \sum ^{+\infty }_{n=1}\frac{1}{n^{2}}
=\frac{4}{3}\frac{\pi ^{2}}{8} \ =\frac{\pi ^{2}}{6}
</math>
==Other identities==
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