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I found this way of integrating $\frac{x}{\sin x}$ using infinite products and fraction decomposition.

$$I=\int\frac{xdx}{\sin x}=\int\frac{xdx}{x\prod_{n\geq1}(1-\frac{x^2}{\pi^2n^2})}\\I=\int\prod_{n\geq1}\frac{1}{1-\frac{x^2}{\pi^2n^2}}\ dx$$ Fraction decomposition:

Let $a_n=\frac1{\pi n}$. Suppose $$\prod_{n\geq1}\frac1{1-a_n^2x^2}=\sum_{n\geq1}\frac{b_n}{1-a_n^2x^2}$$ $$\therefore \prod_{n\geq1}\frac1{1-a_n^2x^2}=\frac{\sum_{n\geq1}b_n\prod_{n\neq i\in\Bbb N}(1-a_i^2x^2)}{\prod_{k\geq1}(1-a_k^2x^2)}$$ $$\therefore 1=\sum_{n\geq1}b_n\prod_{n\neq i\in\Bbb N}(1-a_i^2x^2)$$ $$\therefore 1=b_n\prod_{n\neq i\in\Bbb N}\bigg(1-\frac{a_i^2}{a_n^2}\bigg)$$ $$\therefore b_n=\prod_{n\neq i\in\Bbb N}\frac1{1-\frac{a_i^2}{a_n^2}}$$ Which gives $$I=\sum_{n\geq1}\bigg(\prod_{n\neq i\in\Bbb N}\frac1{1-\frac{a_i^2}{a_n^2}}\bigg)\int\frac{dx}{1-a_n^2x^2}$$ Now the integral $$G_n=\int\frac{dx}{1-a_n^2x^2}$$ $a_nx=\sin u$: $$G_n=\frac1{a_n}\int\sec u\ du$$ $$G_n=\frac1{a_n}\log\bigg|\frac{1+a_nx}{\sqrt{1-a_n^2x^2}}\bigg|$$ Thus $$I=C+\sum_{n\geq1}\frac1{a_n}\log\bigg|\frac{1+a_nx}{\sqrt{1-a_n^2x^2}}\bigg|\prod_{n\neq i\in\Bbb N}\frac1{1-\frac{a_i^2}{a_n^2}}$$ Question: Is this valid/does this work? Can similar techniques be employed to find similar integrals? Thanks.

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    $\begingroup$ $\frac{1}{\sin x} = \lim_{N \to \infty} \sum_{n=-N}^N \frac{(-1)^n}{x-\pi n}$, this is quite equivalent to the infinite product for $\sin x $ $\endgroup$ – reuns Nov 5 '18 at 23:02
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We have that, for a sufficient function $f(x)$, $$f(x)=\prod_{f(\omega)=0}(x-\omega)$$ so that $$\frac{1}{f(x)}=\prod_{f(\omega)=0}\frac{1}{x-\omega}.$$ We then assume that we may write $$\frac{1}{f(x)}=\sum_{f(\omega)=0}\frac{b(\omega)}{x-\omega}.$$ We can show that $$b(\omega)=\prod_{f(r)=0\ ,\ r\ne\omega}\frac{1}{\omega-r}\ .$$ But on the other hand, $$\ln f(x)=\sum_{f(\omega)=0}\ln(x-\omega)$$ so that $$f'(x)=\sum_{f(\omega)=0}\frac{f(x)}{x-\omega}=\sum_{f(\omega)=0}\ \prod_{f(r)=0\ ,\ r\ne\omega}(x-r)\ .$$ Thus, for any $q$ with $f(q)=0$, we plug in $x=q$ to get $$f'(q)=\prod_{f(r)=0\ ,\ r\ne q}(q-r)=\frac{1}{b(q)}\ .$$ Which gives $$\frac{1}{f(x)}=\sum_{f(\omega)=0}\frac{1}{(x-\omega)f'(\omega)}\ .$$ Choosing $f(x)=\sin x$ implies that $$\frac{\omega}{\pi}\in\Bbb Z$$ and $$f'(\omega)=\cos\omega=(-1)^k,$$ so we end up with $$\frac{1}{\sin x}=\sum_{k\in\Bbb Z}\frac{(-1)^k}{x+\pi k}$$ for $x/\pi\not\in\Bbb Z$. This expression is equivalent to $$\frac{x}{\sin x}=\sum_{k\in \Bbb N}\frac{\pi^2k^2}{\pi^2k^2-x^2}\prod_{k\ne j\in\Bbb N}\frac{j^2}{j^2-k^2}\ .$$

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