How to show that $\ln\left(\sin(\pi x)/(\pi x)\right)=\sum_{n=1}^{+\infty}\ln\left(1-x^2/n^2\right)$

Question

I wanted to show for $x \in \left]-1,1\right]$ the beautiful relation

$$\ln\left(\frac{\sin\left(\pi x\right)}{\left(\pi x\right)}\right)=\sum_{n=1}^{+\infty}\ln\left(1-\frac{x^2}{n^2}\right)$$

I want to use it in order to prove the complement formula about $\Gamma$ function, so I would like to know if I could show this with differential equation, or partial sums but without fourier series. I know that it is a well known product result if we take the exponential.

Answer 1

HINT

Recall that the following holds

$$\sin(x) = x\prod_{n=1}^\infty \left(1-\frac{x^2}{n^2\pi^2}\right)$$

and note that

$$\sum_{n=1}^{+\infty}\ln\left(1-\frac{x^2}{n^2}\right)=\ln\left(\prod_{n=1}^{+\infty}\left(1-\frac{x^2}{n^2}\right)\right)$$

Answer 2

It's pretty straightforward if you are familiar with the product expansion of $\sin\theta$

$$\frac {\sin\theta}{\theta}=\prod\limits_{n\geq1}\left(1-\frac {\theta^2}{n^2\pi^2}\right)$$

Take the natural log of both sides to immediately see that$$\log\left(\frac {\sin\theta}{\theta}\right)=\sum\limits_{n\geq1}\log\left(1-\frac {\theta^2}{n^2\pi^2}\right)$$

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Proof: The proof isn't very simple. Start off with the basic identity$$\sin \theta=2\sin\left(\frac {\theta}2\right)\cos\left(\frac {\theta}2\right)=2\sin\left(\frac {\theta}2\right)\sin\left(\frac {\pi}2+\frac {\theta}2\right)\tag1$$

Now, let $\theta\mapsto\frac {\theta}2$ to derive

$$\sin\left(\frac {\theta}2\right)=2\sin\left(\frac {\theta}4\right)\sin\left(\frac {2\pi}4+\frac {\theta}4\right)$$

And similarly, let $\theta\mapsto\frac {\pi}2+\frac {\theta}2$ so that

$$\sin\left(\frac {\pi}2+\frac {\theta}2\right)=2\sin\left(\frac {\pi}4+\frac {\theta}4\right)\sin\left(\frac {3\pi}4+\frac {\theta}4\right)$$Replacing the right-hand side of (1) gives us$$\sin\theta=2^3\sin\left(\frac {\theta}{4}\right)\sin\left(\frac {\pi}4+\frac {\theta}4\right)\sin\left(\frac {2\pi}{4}+\frac {\theta}4\right)+\cdots\tag2$$In fact, continuing indefinitely (try it yourself!), gives us$$\sin\theta=2^{n-1}\sin\left(\frac {\theta}n\right)\sin\left(\frac {\pi}n+\frac {\theta}n\right)\sin\left(\frac {2\pi}n+\frac {\theta}n\right)+\cdots+\sin\left(\frac {\pi(n-1)}n+\frac {\theta}n\right)\tag3$$The last factor of (3) can be simplified to give us what we desire. Using the basic identity for sine that we all learned in first grade, most notably$$\sin(\pi-\theta)=\sin\theta$$we get$$\sin\theta=2^{n-1}\sin\left(\frac {\theta}n\right)\sin\left(\frac {\pi}n+\frac {\theta}n\right)\sin\left(\frac {2\pi}n+\frac {\theta}n\right)\cdots\sin\left(\frac {2\pi}n-\frac {\theta}n\right)\sin\left(\frac {\pi}n-\frac {\theta}n\right)$$And thus, selectivly multiplying the "conjugates" leaves$$\begin{multline}\sin\theta=2^{n-1}\sin\left(\frac {\theta}n\right)\cos\left(\frac {\theta}n\right)\left[\sin^2\left(\frac {\pi}n\right)-\sin^2\left(\frac {\theta}n\right)\right]\left[\sin^2\left(\frac {2\pi}n\right)-\sin^2\left(\frac {\theta}n\right)\right]\\\cdots\left[\sin^2\left(\frac {\pi(\tfrac n2-1)}n\right)-\sin^2\left(\frac {\theta}n\right)\right]\end{multline}\tag4$$Finally, starting with (4), divide both sides by $\sin\left(\frac {\theta}n\right)$ and take the limit as $n\to0$ to isolate $n$ on the left-hand side. Hence$$n=2^{n-1}\sin^2\left(\frac {\pi}n\right)\sin^2\left(\frac {2\pi}n\right)\cdots\sin^2\left(\frac {\pi\left(\tfrac n2-1\right)}n\right)\tag5$$Dividing (4) by (5) and taking the limit as $n\to\infty$ gives$$\begin{align*}\sin\theta & =n\sin\left(\frac {\theta}n\right)\left[1-\frac {\sin^2\left(\frac {\theta}n\right)}{\sin^2\left(\frac {\pi}n\right)}\right]\left[1-\frac {\sin^2\left(\frac {\theta}n\right)}{\sin^2\left(\frac {2\pi}n\right)}\right]\cdots\left[1-\frac {\sin^2\left(\frac {\theta}n\right)}{\sin^2\left(\frac {\pi\left(\frac {n}2-1\right)}n\right)}\right]\end{align*}$$As $n\to\infty$, we are left with$$\sin\theta=\theta\prod\limits_{k\geq1}\left(1-\frac {\theta^2}{k^2\pi^2}\right)$$Hence, the identity has been proven. Man this was a long post!