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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 $$\exp\bigg(\frac{\pi}{4}\int_0^{\pi/4}\frac{\tan^2(x)}{x}\textrm{d}{x}\biggr), \ \exp\bigg(\frac{\pi}{4}\int_0^{\pi/4}\frac{\tan^3(x)}{x}\textrm{d}{x}\biggr), \ \exp\bigg(\frac{\pi}{4}\int_0^{\pi/4}\frac{\tan^4(x)}{x}\textrm{d}{x}\biggr)$$

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    $\begingroup$ What do you mean by the infinite product representations of $\ldots$ ? They are just three numbers, I guess they admit a lot of representations as infinite products. How a canonical infinite product is defined here? $\endgroup$ Sep 26, 2017 at 16:23
  • $\begingroup$ Let $a_0,a_1,a_2,\ldots,a_k,\ldots $ a a.p. $a_k+a_0+k\cdot r$. $\sigma(0)=1,\sigma(1)=3,\sigma(2)=5,\ldots , \sigma(k)=1+k\cdot 2$ Then $\prod_{k=0}{\infty}(\frac{a_{\sigma(k+1}}{a_{\sigma(k)}})^{1/\sigma(k}+1}}=?$ $\endgroup$ Sep 27, 2017 at 11:45
  • $\begingroup$ @JackD'Aurizio I had in mind very similar ways of posing the products as in the first example, but avoided on purpose to give more details. I'll possibly add my solution in the future. $\endgroup$ Sep 28, 2017 at 13:53
  • $\begingroup$ similar home.earthlink.net/~jsondow $\endgroup$ Oct 2, 2017 at 21:19
  • $\begingroup$ @123 : Not similar but somehow the same category. There are indeed a lot of number relations. (Srinivasa Ramanujan had shown that more than hundred years ago.) $\endgroup$
    – user90369
    Oct 3, 2017 at 14:17

1 Answer 1

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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 $

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  • $\begingroup$ Nice observation. +1 $\endgroup$ Sep 26, 2017 at 17:44
  • $\begingroup$ @RandomVariable : Thanks! :-) $\endgroup$
    – user90369
    Sep 26, 2017 at 18:41
  • $\begingroup$ @user90369 Hi, I was wondering if you could say what $\psi(x)$ represented? Thanks :) $\endgroup$
    – Jam
    Sep 27, 2017 at 13:26
  • $\begingroup$ @Jam : Digamma function, e.g. en.wikipedia.org/wiki/Digamma_function . $\endgroup$
    – user90369
    Sep 27, 2017 at 15:02
  • $\begingroup$ @user90369 (+1) You did a great job. $\endgroup$ Sep 28, 2017 at 13:55

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