# Why do authors claim that Euler gave no proof to his “$\sin(\pi x)= \pi x\prod\limits_{k=1}^{\infty}\left(1-\frac{x^2}{k^2} \right )$” when…

When he proved the relation between $\pi \cot(\pi x)$ and the harmonic series in "Introductio in analysin infinitorum" which states that $$\pi \cot(\pi x)=\sum_{k \to \infty}^{\infty} \frac{1}{x+k}=\frac{1}{x}+\sum_{k=1}^{\infty}\left(\frac{1}{x+n}+\frac{1}{x-n} \right) \text{ for } x\in \mathbb{R} \backslash \mathbb{Z}.$$ It doesn't take a genius to transform this into an infinite product just by knowing the fact that $$\int \pi \cot(\pi x) = \log\left(\sin(\pi x) \right)+C.$$

So my question is, why does every historian/author claim that Euler's first proof of $\displaystyle \zeta (2)=\frac{\pi^2}{6}$ was not rigourous at all because Euler didn't prove his famous infinite product in his lifetime when the proof of the relation between the cotangent and the harmonic series implies directly his infinite product?

EDIT:I'm sorry for doing this but shameless self bump.

I got no answers and once again, I'm sorry.

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I am afraid that in Euler's days integrating those infinite series would be even more striking example of an argument that is not rigorous. Even Cauchy got it wrong the first time round. :) – J.H. Feb 19 '13 at 19:46
Thank you Asaf Karagila. – Kobesky Feb 19 '13 at 19:54
J.H.: I agree but the difference between no rigorous justification of the infinite product and not considering uniform convergence when integrating term by term is a big one. Authors don't even say he proved the product in any way. – Kobesky Feb 19 '13 at 20:37
Do we have a definition for a "rigorous proof"? – Mhenni Benghorbal Feb 19 '13 at 20:50
That's just shifting the semantic burden from the words "rigorous proof" to the words "completely justified". Do you have a definition for "completely justified"? – Greg Martin Feb 20 '13 at 1:23

why does every historian/author claim that Euler's first proof of $\displaystyle \zeta (2)=\frac{\pi^2}{6}$ was not rigorous at all