On the product $\displaystyle\prod^{n=\infty}_{n=0}\left[1+\left(\frac{x}{a+nd}\right)^3\right]$
Journal of the Indian Mathematical Society, VII, 1915, 209 – 211

1. Let

$$\phi (\alpha, \beta) = \left\{1+ \left(\frac{\alpha+\beta}{1+\alpha}\right)^3\right\} \left\{1+ \left(\frac{\alpha+\beta}{2+\alpha}\right)^3\right\}.$$
It is easy to see that
\begin{align} & \left\{1+\left(\frac{\alpha+\beta}{n+\alpha}\right)^3\right\} \left\{1+\left(\frac{\alpha +\beta}{n+\beta}\right)^3\right\}\notag\\ & \hspace{1cm} = \frac{\left(1+\frac{\alpha+2\beta}{n}\right) \left(1+\frac{\beta+2\alpha}{n}\right)}{\left(1+\frac{\alpha}{n}\right)^3 \left(1+\frac{\beta}{n}\right)^3} \left[1-\left\{\frac{(\alpha - \beta) + i(\alpha + \beta)\sqrt{3}}{2n}\right\}^2 \right]\notag\\ & \hspace{1.7cm} \times \left[1- \left\{\frac{(\alpha-\beta) - i(\alpha+\beta)\sqrt{3}}{2n}\right\}^2\right]; \end{align}
$$\prod^{n=\infty}_{n=1} \left\{\frac{\left(1+\frac{\alpha + 2 \beta}{n}\right) \left(1+\frac{\beta+2\alpha}{n}\right)}{\left(1+\frac{\alpha}{n}\right)^3 \left(1+\frac{\beta}{n}\right)^3}\right\} = \frac{\{\Gamma (1+\alpha) \Gamma(1+\beta)\}^3}{\Gamma(1+\alpha + 2 \beta) \Gamma(1+\beta+2\alpha)};$$
and
\begin{align} \prod^{n=\infty}_{n=1} \left[1-\left\{\frac{(\alpha -\beta)+i(\alpha+\beta)\sqrt{3}}{2n}\right\}^2\right] & \left[ 1-\left\{\frac{(\alpha-\beta)-i(\alpha+\beta)\sqrt{3}}{2n}\right\}^2\right]\notag\\ & = \frac{\cosh \pi(\alpha+\beta) \sqrt{3}-\cos \pi (\alpha-\beta)}{2 \pi^2 (\alpha^2 +\alpha \beta +\beta^2)}. \end{align}

It follows from (1) - (4) that

$$\phi (\alpha, \beta) \phi (\beta, \alpha) = \frac{\{\Gamma(1+\alpha) \Gamma(1+\beta)\}^3}{\Gamma(1+\alpha + 2\beta)\Gamma(1+\beta+2\alpha)} \left\{\frac{\cosh \pi(\alpha+\beta)\sqrt{3}-\cos \pi (\alpha-\beta)}{2\pi^2(\alpha^2 +\alpha\beta +\beta^2)} \right\}.$$

But it is evident that, if $\alpha-\beta$ be any integer, then $\phi (\alpha,\beta)/\phi (\beta, \alpha)$ can be expressed in finite terms. From this and (5) it follows that $\phi(\alpha, \beta)$ can be expressed in finite terms, if $\alpha-\beta$ be any integer. That is to say $$\left\{1+\left(\frac{x}{a}\right)^3\right\} \left\{1+\left(\frac{x}{a+d}\right)^3\right\} \left\{1+\left(\frac{x}{a+2d}\right)^3\right\}\cdots$$ can be expressed in finite terms if $x-2a$ be a multiple of $d$.

2. Suppose now that $\alpha=\beta$ in (5). We obtain

$$\left\{1+\left(\frac{2\alpha}{1+\alpha}\right)^3\right\} \left\{1+\left(\frac{2\alpha}{2+\alpha}\right)^3\right\} \left\{1+\left(\frac{2\alpha}{3+\alpha}\right)^3\right\}\cdots = \frac{\{\Gamma(1+\alpha)\}^3} {\Gamma (1+3\alpha)} \frac{\sinh \pi \alpha \sqrt{3}}{\pi \alpha \sqrt{3}}.$$
Similarly, putting $\beta = \alpha+1$ in (5), we obtain
$$\left\{1+\left(\frac{2\alpha+1}{1+\alpha}\right)^3\right\} \left\{1+\left(\frac{2\alpha+1}{2+\alpha}\right)^3\right\} \cdots =\frac{\{\Gamma(1+\alpha)\}^3}{\Gamma(2+3\alpha)} \frac{\cosh \pi (\half + \alpha)\sqrt{3}}{\pi}.$$
Again, since $$\left\{1+\left(\frac{\alpha}{n}\right)^3\right\} \left\{1+3\left(\frac{\alpha}{2n+\alpha}\right)^2\right\} = \frac{\left(1+\frac{\alpha}{n}\right) \left(1+\frac{\alpha^2}{n^2} + \frac{\alpha^4}{n^4}\right)}{\left(1+\frac{\alpha}{2n}\right)^2},$$ it is easy to see that
\begin{align} \left[\left (1+\frac{\alpha^3}{1^3}\right)\left(1+\frac{\alpha^3}{2^3}\right)\cdots\right] & \left[\left\{1+3\left(\frac{\alpha}{2+\alpha}\right)^2\right\} \left\{1+3\left(\frac{\alpha}{4+\alpha}\right)^2\right\}\ldots\right]\notag\\ &\quad = \frac{\Gamma (\half \alpha)}{\Gamma\{\half (1+\alpha)\}} \left(\frac{\cosh \pi \alpha \sqrt{3} - \cos \pi \alpha}{2^{\alpha+2} \pi \alpha \sqrt{\pi}} \right). \end{align}

3. It is known that, if the real part of $\alpha$ is positive, then

$$\log \Gamma (\alpha) = (\alpha - \half ) \log \alpha - \alpha + \half \log 2 \pi + 2 \int\limits^\infty_0 \frac{\tan^{-1} (x/\alpha)}{e^{2 \pi x} - 1} dx.$$
From this we can shew that, if the real part of $\alpha$ is positive, then
\begin{align} & \half \log 2 \pi \alpha + \frac{\pi \alpha}{\sqrt{3}} + \log \left\{\left(1+\frac{\alpha^3}{1^3}\right) \left(1+\frac{\alpha^3}{2^3}\right) \left(1+\frac{\alpha^3}{3^3}\right)\ldots\right\}\notag\\ &\qquad = \log \left(\frac{\cosh \pi \alpha \sqrt{3} - \cos \pi \alpha}{\pi \alpha}\right) + 2 \int\limits^\infty_0 \frac{\tan^{-1} (x/\alpha)^3}{e^{2 \pi x} - 1} dx. \end{align}
From this and the previous section it follows that $$\int\limits^\infty_0 \frac{\tan^{-1} x^3}{e^{2 \pi n x}-1} dx$$ can be expressed in finite terms if $n$ is a positive integer. Thus, for example,
$$\int\limits^\infty_0 \frac{\tan^{-1} x^3}{e^{2\pi x}-1} dx = \frac{1}{4} \log 2 \pi - \frac{\pi}{4\sqrt{3}} - \half \log (1+e^{-\pi \sqrt{3}});$$
$$\int\limits^\infty_0 \frac{\tan^{-1} x^3}{e^{4\pi x}-1} dx = \frac{1}{8} \log 12 \pi - \frac{\pi}{4\sqrt{3}} - \frac{1}{4} \log (1- e^{-2 \pi \sqrt{3}});$$
and so on.

It is also easy to see that

\begin{align} & \frac{1^2}{1^3 + n^3} - \frac{2^2}{2^3+n^3} + \frac{3^2}{3^3+n^3} - \frac{4^2}{4^3+n^3} + \cdots\notag\\ &\quad = \frac{1}{3} \left(\frac{1}{1+n} - \frac{1}{2+n} + \frac{1}{3+n} - \frac{1}{4+n} + \ldots \right)\notag\\ &\qquad + \frac{4}{3} \left\{\frac{2-n}{(2-n)^2+3n^2} - \frac{4-n}{(4-n)^2 + 3n^2} + \frac{6-n}{(6-n)^2+3n^2} - \ldots \right\}. \end{align}
Since $$\frac{\pi}{4 \cosh \half \pi x} = \frac{1}{1^2 + x^2} - \frac{3}{3^2+x^2} + \frac{5}{5^2 + x^2} - \ldots,$$ it is clear that the left-hand side of (13) can be expressed in finite terms if $n$ is any odd integer. For example,
$$\frac{1^2}{1^3+1} - \frac{2^2}{2^3 +1} + \frac{3^2}{3^3+1} - \frac{4^2}{4^3+1} + \ldots = \frac{1}{3} (1-\log2 + \pi \mbox{ sech } \half \pi \sqrt{3}).$$
The corresponding integral in this case is
\begin{eqnarray} \int\limits^\infty_0 \frac{x^5}{\sinh \pi x} \frac{dx}{n^6+x^6} = \frac{2}{\pi} \int\limits^\infty_0 \left\{\frac{1}{2x^2} + \sum^{\nu=\infty}_{\nu=1} \frac{(-1)^\nu}{\nu^2+x^2}\right\} \frac{x^6dx}{n^6+x^6}\nonumber\\ = \frac{1}{3} \left(\frac{1}{n} - \frac{1}{n+1} + \frac{1}{n+2} - \frac{1}{n+3}+\ldots \right) \nonumber\\ -\frac{4}{3} \left\{\frac{n+2}{(n+2)^2 + 3n^2} - \frac{n+4}{(n+4)^2+3n^2} + \frac{n+6}{(n+6)^2 +3n^2}-\ldots \right\}; \end{eqnarray}
and so the integral on the left-hand side of (15) can be expressed in finite terms if $n$ is any odd integer. For example,
$$\int\limits^\infty_0 \frac{x^5}{\sinh \pi x} \frac{dx}{1+x^6} = \frac{1}{3} (\log 2 - 1 + \pi \mbox{ sech } \half \pi \sqrt{3}).$$