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Show that for $0<t<1$, $$\log\sin (t\pi)=\log(t\pi) + \sum_{n=1}^\infty\log\left(1-\frac{t^2}{n^2}\right)$$


So I derived the following Fourier series: $$\frac{\pi\cos(t\pi)}{\sin{t\pi}}=\frac1t+\sum_{n=1}^\infty\frac{2t}{t^2-n^2},\quad t\notin\mathbb{Z}$$

Then the next natural step is to integrate. But how do I get the constant $\log\pi$ on the RHS? I try to set $t=1/2$ but that didn't help too much. By the way, I'm not suppose to use the Infinite Product Formula for Sine, since this question is supposed to derive the formula. Thanks.

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@Alyosha "I'm not suppose to use the Infinite Product Formula for Sine" –  Cortizol Dec 9 '13 at 17:42
    
Sorry, didn't read the last part! –  Alyosha Dec 9 '13 at 17:43

1 Answer 1

up vote 2 down vote accepted

If one knows that $$\log\sin (t\pi)=\log(ct) + \sum_{n=1}^\infty\log\left(1-\frac{t^2}{n^2}\right)$$ for some constant $c$ then considering the limit when $t\to0$, the sum on he RHS disappears and one is left with $$ \log\sin (t\pi)-\log(ct)=\log(\sin(t\pi)/(ct)). $$ Since $\sin(\pi t)\sim\pi t$, the RHS converges to $\log(\pi/c)$. By identification this limit should be $0$ hence $c=\pi$.

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That's cool. By the way, how I should justify integrating term by term in the infinite series? Thanks. –  Christmas Bunny Dec 10 '13 at 5:10
    
Uniform convergence on every interval $[0,T]$ with $T\lt1$. –  Did Dec 10 '13 at 7:27

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