Product $\left(\frac31\right)^{1/2}\cdot\left(\frac75\right)^{1/6}\cdot \left(\frac{11}9\right)^{1/10}\cdot \left(\frac{15}{13}\right)^{1/14}\cdots$

Recently I rediscovered by a new method a family of infinite products I obtained years ago, and one of the examples you may find below,

$$\left(\frac{3}{1}\right)^{1/2}\cdot\left(\frac{7}{5}\right)^{1/6}\cdot \left(\frac{11}{9}\right)^{1/10}\cdot \left(\frac{15}{13}\right)^{1/14}\cdots=\exp\bigg(\frac{\pi}{4}\int_0^{\pi/4}\frac{\tan(x)}{x}\textrm{d}{x}\biggr).$$

I wonder if such results are known in the mathematical literature, or anything similar to the example stated above. If yes, I'd be glad to receive information about the related literature.

Supplementary question: Find the infinite product representations of $$\exp\bigg(\frac{\pi}{4}\int_0^{\pi/4}\frac{\tan^2(x)}{x}\textrm{d}{x}\biggr), \ \exp\bigg(\frac{\pi}{4}\int_0^{\pi/4}\frac{\tan^3(x)}{x}\textrm{d}{x}\biggr), \ \exp\bigg(\frac{\pi}{4}\int_0^{\pi/4}\frac{\tan^4(x)}{x}\textrm{d}{x}\biggr)$$

The central function for the development (which I prefer here) of all integrals mentioned above into products of the type of your example is the Digamma function $\,\psi\,$ .

Literature: Any useful formula collection, which contains the Digamma function; e.g. please look at https://en.wikipedia.org/wiki/Digamma_function , the relevant sections are Taylor series, series formula, reflection formula and Recurrence formula and characterization .

In part $\,(C)\,$ of my answer I've listed the formulas which I've used here.

$(A)\,$: $\enspace$ About the given product.

Your case comes from

$\displaystyle f(z):=\sum\limits_{k=0}^\infty \frac{\ln(1+\frac{z}{2k+1})}{2k+1} = \int\limits_0^z \frac{\gamma + \psi(1+x)}{x}dx - \frac{1}{2}\int\limits_0^{z/2} \frac{\gamma + \psi(1+x)}{x}dx $

using $\,\displaystyle z=-\frac{1}{2}\,$ and $\,\displaystyle z=+\frac{1}{2}\,$ .

It’s

$\displaystyle \sum\limits_{k=0}^\infty \frac{\ln(4k+3)-\ln(4k+1)}{4k+2} = \frac{1}{2} \left(f(\frac{1}{2})-f(-\frac{1}{2})\right) = $

$\displaystyle =\frac{1}{2}\int\limits_0^{1/2} \frac{\psi(1+x)-\psi(1-x)}{x}dx - \frac{1}{4}\int\limits_0^{1/4} \frac{\psi(1+x)-\psi(1-x)}{x}dx $

$\displaystyle = \frac{\pi}{4} \int\limits_0^{1/4}\frac{\cot(\pi x)-2\cot(2\pi x)}{x} dx = \frac{\pi}{4} \int\limits_0^{\pi/4}\frac{\tan x}{x} dx \,$ .

One type of generalization of the initial product is with $\,|z|<1\,$ :

$\displaystyle \sum\limits_{k=0}^\infty \frac{\ln(2k+1+z)-\ln(2k+1-z)}{2k+1} = \frac{\pi}{2}\int\limits_0^{z\pi/2}\frac{\tan x}{x} dx$

Note: The derivation with respect to $\,z\,$ shows us an alternative way to the result.

$(B)\,$: $\enspace$ Example for the 'supplementary question'.

We can take the calculations of $(A)$, modify and reverse them.

$\displaystyle g(z):=\sum\limits_{k=1}^\infty \frac{\ln(1+\frac{z}{k})}{k^2} = -\int\limits_0^z \frac{\psi(1+x)+\gamma-\zeta(2)x}{x^2}dx $

With

$\displaystyle \frac{\pi^2}{4} \int\limits_0^{\pi/4} \frac{\tan^2 x}{x}dx = \pi - \frac{\pi^2}{4} + \frac{\pi^2}{4} \int\limits_0^{\pi/4} \frac{\tan x - x}{x^2}dx $

and

$\displaystyle \frac{\pi^2}{4} \int\limits_0^{\pi/4} \frac{\tan x - x}{x^2}dx =$

$\displaystyle =\frac{1}{4} \int\limits_0^{1/4} \frac{1 - \frac{1}{3} (2\pi x)^2 - 2\pi x \cot(2\pi x)}{x^3}dx - \frac{1}{4} \int\limits_0^{1/4} \frac{1 - \frac{1}{3} (\pi x)^2 - \pi x \cot(\pi x)}{x^3}dx $

$\displaystyle = \int\limits_0^{1/2} \frac{- \frac{1}{3} \pi^2 x - \psi(1-x) + \psi(1+x)}{x^2}dx - \frac{1}{4} \int\limits_0^{1/4} \frac{- \frac{1}{3} \pi^2 x - \psi(1-x) + \psi(1+x)}{x^2}dx $

$\displaystyle =\frac{1}{4} \left(g(\frac{1}{4})+g(-\frac{1}{4})\right) - \left(g(\frac{1}{2})+g(-\frac{1}{2})\right) $

we get

$\displaystyle \frac{\pi^2}{4} \int\limits_0^{\pi/4} \frac{\tan^2 x}{x}dx = \pi - \frac{\pi^2}{4} +\frac{\pi^2}{6}\ln 2 + \frac{1}{4} \sum\limits_{k=1}^\infty \frac{1}{k^2}\ln\frac{k^6 ((4k)^2-1)}{((2k)^2-1)^4}$

and with that a product.

Note:

$\displaystyle \int\limits_0^z \frac{\tan^3 x}{x}dx = \frac{\tan^2 z}{2z} + \frac{\tan z - z}{2 z^2} - \int\limits_0^z \frac{\tan x}{x}dx + \int\limits_0^z \frac{\tan x - x}{x^3}dx $

$\displaystyle \int\limits_0^z \frac{\tan^4 x}{x}dx =\frac{\tan^3 z}{3z} +\frac{\tan^2 z - z^2}{6 z^2} - \frac{\tan z - z}{z} + \frac{\tan z - z - \frac{1}{3}z^3}{3 z^3} $ $\hspace{3cm}\displaystyle -\frac{4}{3}\int\limits_0^z \frac{\tan x - x}{x^2}dx + \int\limits_0^z \frac{\tan x - x - \frac{1}{3}x^3}{x^4}dx $

So, for the second integral of the 'supplementary question' it's necessary to evaluate

$\displaystyle\int\limits_0^{\pi/4}\frac{\tan x -x}{x^3}dx \,$ and for the third integral $\,\displaystyle\int\limits_0^{\pi/4} \frac{\tan x - x - \frac{1}{3}x^3}{x^4}dx $

which can be done with the same method as used in $(A)$ and $(B)$ .

And so on ... .

$(C)\,$: $\enspace$ Used formulas.

$\displaystyle \psi(1+x) = \psi(x) + \frac{1}{x}\,$ , $\enspace \psi(1-x) - \psi(x) = \pi \cot(\pi x)\,$ , $\enspace \tan x = \cot x - 2 \cot(2x)$

$\displaystyle \gamma+\psi(1+x) =\sum\limits_{n=1}^\infty (-1)^{n-1} \zeta(n+1) x^n \,$ , $\enspace \displaystyle \ln(1+x)=\sum\limits_{n=1}^\infty (-1)^{n-1} \frac{x^n}{n} $

$\displaystyle \cot x = \frac{1}{x} + \sum\limits_{n=1}^\infty \frac{(-4)^n B_{2n}}{(2n)!}x^{2n-1}\enspace$ where $\,B_k\,$ is defined by $\enspace\displaystyle \frac{x}{e^x - 1} = \sum\limits_{k=0}^\infty \frac{x^k}{k!}B_k $