# closed form for $\sum_{n\geq1}\pi n\log\frac{4n+1}{4n-1}\prod_{k\geq1\\k\neq n}\frac{k^2}{k^2-n^2}$?

Look at what I found! $$\int_{-\pi/4}^{\pi/4}x\csc x\ \mathrm dx=\sum_{n\geq1}\pi n\log\bigg(\frac{4n+1}{4n-1}\bigg)\prod_{k\geq1\\k\neq n}\frac{k^2}{k^2-n^2}$$ I want to know if it is valid, and if there is a nice closed form to go with my series.

Here's my proof.

Recall the famous product representation $$\sin x=x\prod_{k\geq1}\bigg(1-\frac{x^2}{\pi^2k^2}\bigg)$$ Hence we have that $$I=\int_{-\pi/4}^{\pi/4}x\csc x\ \mathrm dx=\int_{-\pi/4}^{\pi/4}\prod_{k\geq1}\frac{\pi^2k^2}{\pi^2k^2-x^2}\mathrm dx$$ We may now preform a fraction decomposition, and start with $$\prod_{k\geq1}\frac{\pi^2k^2}{\pi^2k^2-x^2}=\sum_{n\geq1}\frac{\pi^2n^2}{\pi^2n^2-x^2}b_n$$ Multiplying both sides by $$\prod_{k\geq1}\frac{\pi^2k^2-x^2}{\pi^2k^2}$$, $$1=\sum_{n\geq1}\frac{\pi^2n^2}{\pi^2n^2-x^2}b_n\prod_{k\geq1}\frac{\pi^2k^2-x^2}{\pi^2k^2}$$ $$1=\sum_{n\geq1}b_n\prod_{k\geq1\\k\neq n}\frac{\pi^2k^2-x^2}{\pi^2k^2}$$ Which in turn implies that, for any $$m\in\Bbb N$$, $$1=b_m\prod_{k\geq1\\k\neq m}\frac{\pi^2k^2-\pi^2m^2}{\pi^2k^2}$$ $$b_m=\prod_{k\geq1\\k\neq m}\frac{k^2}{k^2-m^2}$$ Hence we can begin the process of integration: $$I=\int_{-\pi/4}^{\pi/4}\sum_{n\geq1}\frac{\pi^2n^2}{\pi^2n^2-x^2}b_n\mathrm dx$$ $$I=\sum_{n\geq1}\pi^2n^2b_n\int_{-\pi/4}^{\pi/4}\frac{\mathrm dx}{\pi^2n^2-x^2}$$ This final integral evaluates to $$\int_{-\pi/4}^{\pi/4}\frac{\mathrm dx}{\pi^2n^2-x^2}=\frac1{\pi n}\log\bigg(\frac{4n+1}{4n-1}\bigg)$$ So finally we have $$I=\sum_{n\geq1}\pi n\log\bigg(\frac{4n+1}{4n-1}\bigg)\prod_{k\geq1\\k\neq n}\frac{k^2}{k^2-n^2}$$ Does this work? Is there a closed form value for $$I$$? Is there another way to prove this?

Update:

Integrating form $$0$$ to $$\pi/2$$ instead of $$-\pi/4$$ to $$\pi/4$$ gives, according to Wolfram Alpha, $$\sum_{n\geq1}\frac{\pi n}2\log\bigg(\frac{2n+1}{2n-1}\bigg)\prod_{k\geq1\\k\neq n}\frac{k^2}{k^2-n^2}=2G$$ Where $$G$$ is Catalan's constant. Isn't that cool?! That's crazy cool!

• Yes it is 'crazy-cool' ... & there are lirerally thousands of crazy-cool results like that! I can't get enough of 'em. – AmbretteOrrisey Dec 4 '18 at 22:00