I have a question about the sine product formulae.
As far as I know, Euler thinks the function as a polynomial of order, and writes it by using its roots as
for some constants .
From the fact that
the constant in (1) is computed to be .
My question comes at this point.
How I can be sure that the right-hand side of (2) gives , not ?
Anyway, it is not elementary to make Euler's argument rigorous (note that if you multiply by where is an entire function, the function is still analytic and has the same zeroes and same orders). As far as I know, the proof that is closest (in intuition, not technique...) to Euler's proof would go along Weierstrass Factorization Theorem (which gives the existence of an analytic function with given zeroes, and its expression as an infinite product). The difficulty (besides the proof of the theorem...) is then to prove that the above entire function is a constant; you can find a full proof of the sine product formula along these lines in "Functions of one complex variable I" by John B. Conway (Springer, Graduate texts in mathematics), p. 175 in 2nd edition. (His proof uses an expansion of the cotangent function that can be obtained from Residue Theorem. )
Actually, this is not how SPF (sine product formula) is obtained, it is rather in this way if you check the proof
and as you have mentioned the proof makes use of Weierstrass' Gamma function expansion
where is the Euler–Mascheroni constant defined by
This is where my interest to the SPF originates.
If you truly understand this proof, then you'll notice the key to its being resolved is proving that a function defined to be the second derivative of , \;where \;\Phi(.)[/tex] as defined above, is a CONSTANT. Key in concluding that is also determining that is periodic! is not periodic. Once you notice this, a contradiction can be constructed. I finally got around to looking at the proof today.
There is a proof starting from the expansion (proved by one mean or another), then noting that so that if we can integrate term by term, we get a series expansion of which is equivalent to the sine product formula. A ref (also gives another proof)
There is also a fairly elementary proof I learned from "Analysis by its history" by Hairer and Wanner, the French edition. I looked it up, it is p.64 of the English edition, which you can probably find in any good library (The justification of a convergence step is partly left as an exercise, with a hint) By the way, this book is nice reading; rather simple maths (undergraduate analysis), but put in its historical context. Of course, they mention how Euler came up with the formula!