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

Question

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.

Answer 1

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 $c=\pi$.