On the integral $\int\limits^x_0\frac{\tan^{-1} t}{t} dt$
Journal of the Indian Mathematical Society, VII,1915, 93 – 96

Let

$$\phi(x) = \int\limits^x_0 \frac{\tan^{-1}t}{t} \ dt.$$
Then it is easy to see that
$$\phi (x)+\phi(-x)=0;$$
and that
$$\phi(x) =\frac{x}{1^2} -\frac{x^3}{3^2} + \frac{x^5}{5^2}-\frac{x^7}{7^2}+\ldots$$
provided that $|x|\leq 1.$ Changing $t$ into $1/t$ in (1), we obtain
$$\phi(x) - \phi \left(\frac{1}{x}\right) =\frac{1}{2} \pi \log x,$$
provided that the real part of $x$ is positive. The results in the following two sections can be very easily proved by differentiating both sides with respect to $x$.

2. If $0 \lt x \lt \frac{1}{2} \pi,$ then

$$\frac{\sin2x}{1^2} + \frac{\sin6x}{3^2} + \frac{\sin 10x}{5^2} + \ldots = \phi (\tan x) - x \log (\tan x).$$
If, in particular, we put $x=\frac{1}{8} \pi$ and $\frac{1}{12} \pi$ in (5), we obtain
$$\frac{1}{1^2}+\frac{1}{3^2} - \frac{1}{5^2}-\frac{1}{7^2} + \frac{1}{9^2} + \cdots = {\sqrt{2}} \phi (\sqrt{2}-1) + \frac{\pi}{4\sqrt{2}} \log (1+\sqrt{2});$$
and
$$2\phi(1) = 3\phi (2- \sqrt{3}) + \frac{1}{4} \pi \log (2+\sqrt{3}).$$

If $-\frac{1}{2} \pi \lt x \lt \frac{1}{2} \pi,$ then

$$2\phi \left(\tan \frac{x}{2}\right) = \sin x +\frac{2}{3} \frac{\sin^3 x}{3} + \frac{2\cdot4}{3\cdot5} \frac{\sin^5 x}{5}+ \cdots .$$
If $0 \lt x \lt \frac{1}{2} \pi,$ then
\begin{eqnarray} \frac{\sin x}{1^2} \cos x + \frac{\sin 2x}{2^2} \cos^2 x &+& \frac{\sin 3x}{3^2} \cos^3 x + \cdots\nonumber\\ &=& \phi (\tan x) + \frac{1}{2} \pi \log \cos x - x \log \sin x; \end{eqnarray}
and
\begin{eqnarray} \frac{\cos x+\sin x}{1^2} &+& \frac{1}{2} \frac{\cos^3 x+\sin^3 x}{3^2} + \frac{1\cdot3}{2\cdot4} \frac{\cos^5 x + \sin^5 x}{5^2} + \cdots\nonumber\\ &=& \phi (\tan x)+\frac{1}{2} \pi \log (2 \cos x). \end{eqnarray}
If $- \frac{1}{2} \pi \lt x \lt \frac{1}{2} \pi$ and $a$ be any number such that $$|(1-a) \sin x | \leq 1, \left| \left(1-\frac{1}{a}\right) \cos x\right| \leq 1,$$ then
\begin{eqnarray} \frac{\sin x}{1^2} \left(1- \frac{1}{a}\right) \cos x + \frac{\sin 2x}{2^2} \left(1- \frac{1}{a} \right)^2 \cos^2 x &+& \frac{\sin 3x}{3^2} \left(1- \frac{1}{a}\right)^3 \cos^3 x + \cdots \nonumber\\ + \frac{\sin (x +\frac{1}{2} \pi)}{1^2} (1-a) \sin x &-& \frac{\sin 2 (x+\frac{1}{2} \pi)}{2^2} (1-a)^2 \sin^2 x + \cdots\nonumber \\ &=& \phi (\tan x) - \phi (a \tan x) + x \log a. \end{eqnarray}

3. Let $R(x)$ and $I(x)$ denote the real and the imaginary parts of $x$ respectively. Then, if $-1 \lt R (x) \lt 1,$

\begin{eqnarray} \log \left(1-\frac{x^2}{1^2}\right) &-& 3 \log \left(1-\frac{x^2}{3^2}\right) + 5 \log \left(1-\frac{x^2}{5^2}\right) - \cdots\nonumber \\ &=& \frac{4}{\pi} [\phi (1)-\phi \{\tan \frac{1}{4} \pi (1-x)\}] + \log \tan \frac{1}{4} \pi (1-x). \end{eqnarray}
Putting $x=\frac{2}{3}$ in (12) and using (7), we obtain $${\left(1-\frac{4}{3^2}\right) \left(1-\frac{4}{9^2}\right)^{-3} \left(1-\frac{4}{15^2}\right)^5 \left(1-\frac{4}{21^2}\right)^{-7} \left(1-\frac{4}{27^2}\right)^9 \cdots=(2-\sqrt{3})^{\frac{2}{3}} e^n,}$$ where
$$n= \frac{4}{3 \pi} \phi (1)$$
Again, subtracting $\log (1-x)$ from both sides in (12) and making $x \to 1$, we obtain
$$\left(1-\frac{1}{3^2}\right)^{-3} \left(1-\frac{1}{5^2}\right)^5 \left(1-\frac{1}{7^2}\right)^{-7} \left(1-\frac{1}{9^2}\right)^9 \cdots = \frac{\pi}{8} e^{3n} .$$
If $-1 \lt I(x)\lt 1,$ then
\begin{eqnarray} \log \left(1+\frac{x^2}{1^2}\right) &-& 3 \log \left(1+\frac{x^2}{3^2}\right) + 5 \log \left(1+ \frac{x^2}{5^2}\right) - \cdots\nonumber\\ &=& \frac{4}{\pi} \{\phi (1) - \phi (e^{-\frac{1}{2} \pi x})\} - 2x \tan^{-1} e^{-\frac{1}{2} \pi x}. \end{eqnarray}
From this and (7) we see that, if $\frac{1}{2} \pi x = \log (2+\sqrt{3}),$ then
$$\left(1+\frac{x^2}{1^2}\right) \left(1+\frac{x^2}{3^2}\right)^{-3} \left(1+\frac{x^2}{5^2}\right)^5 \left(1+\frac{x^2}{7^2}\right)^{-7} \cdots = e^n,$$
where $n$ is the same as in (13). It follows at once from (12) and (15) that, if $-1 \lt R (\beta) \lt 1, -1 \lt I (\alpha) \lt 1$, then
$$e^{\frac{1}{2} \pi\alpha \beta} = \left(\frac{1^2 + \alpha^2}{1^2 - \beta^2}\right) \left(\frac{3^2 - \beta^2}{3^2 + \alpha^2}\right)^3 \left(\frac{5^2 + \alpha^2}{5^2 - \beta^2}\right)^5 \left(\frac{7^2 - \beta^2}{7^2 + \alpha^2}\right)^7 \cdots,$$
provided that $\cosh \frac{1}{2}\pi \alpha = \sec \frac{1}{2} \pi \beta.$

4. Now changing $x$ into $2x (1+i)$ in (15), we have \begin{eqnarray*} \log \left(1+\frac{8ix^2}{1^2} \right) &-& 3 \log \left(1+ \frac{8ix^2}{3^2}\right) + 5 \log \left(1+ \frac{8ix^2}{5^2}\right)-\cdots \\ &=& \frac{4}{\pi} \phi (1) - 4x (1+i) \tan^{-1} e^{-\pi x(1+i)} - \frac{4}{\pi} \left\{\frac{1}{1^2} e^{-\pi x (1+i)} - \frac{1}{3^2} e^{-3\pi x(1+i)} + \ldots \right\}. \end{eqnarray*} Equating real and imaginary parts we see that, if $x$ is positive, then

\begin{eqnarray} \log \left(1+\frac{64x^4}{1^4}\right) &-& 3 \log \left(1+\frac{64x^4}{3^4}\right) + 5 \log \left(1+\frac{64x^4}{5^4}\right) - \cdots\nonumber\\ &=& \frac{8}{\pi} \phi (1) - 2x \log \left(\frac{\cosh \pi x + \sin \pi x}{\cosh \pi x - \sin \pi x}\right) - 4x \tan^{-1} \left(\frac{\cos \pi x}{\sinh \pi x}\right)\nonumber\\ &&\quad\quad\quad -\frac{8}{\pi} \left\{ \frac{\cos \pi x}{1^2} e^{-\pi x} - \frac{\cos 3 \pi x}{3^2} e^{-3\pi x} + \frac{\cos 5 \pi x}{5^2} e^{-5 \pi x} - \cdots \right\}; \end{eqnarray}
and
\begin{eqnarray} \tan^{-1} \frac{8x^2}{1^2} &-& 3 \tan^{-1} \frac{8x^2}{3^2} + 5 \tan^{-1} \frac{8x^2}{5^2} - \cdots\nonumber \\ &=& x \log \left(\frac{\cosh \pi x + \sin \pi x}{\cosh \pi x - \sin \pi x} \right) - 2x \tan^{-1} \left(\frac{\cos \pi x}{\sinh \pi x}\right)\nonumber\\ &+& \frac{4}{\pi} \left\{\frac{\sin \pi x}{1^2} e^{-\pi x} - \frac{\sin 3 \pi x}{3^2} e^{-3 \pi x} + \frac{\sin 5 \pi x}{5^2} e^{-5 \pi x} - \cdots \right\}. \end{eqnarray}
It follows from (18) that, if $n$ is a positive odd integer, then
\begin{eqnarray} \left(1+\frac{4n^4}{1^4}\right) \left(1 + \frac{4n^4}{3^4}\right)^{-3} && \left(1+\frac{4n^4}{5^4}\right)^5 \left(1+\frac{4n^4}{7^4}\right)^{-7}\cdots\nonumber\\ &=& e^{\frac{8}{\pi} \phi (1)} \left(\frac{1-e^{-\frac{1}{2}\pi n}}{1+e^{-\frac{1}{2} \pi n}}\right)^{2 n \cos \frac{1}{2} (n-1)\pi}, \end{eqnarray}
and, if $n$ is any even integer, then
\begin{eqnarray} \left(1+\frac{4n^4}{1^4} \right) \left(1+\frac{4n^4}{3^4}\right)^{-3} && \left(1+\frac{4n^4}{5^4}\right)^5 \left(1+\frac{4n^4}{7^4}\right)^{-7}\cdots\nonumber\\ &=& \exp \left\{ \frac{8}{\pi} \phi (1) - \frac{8}{\pi} (-1)^{\frac{1}{2} n} [\phi (e^{-\frac{1}{2}\pi n}) + \frac{1}{2} \pi n \tan^{-1} e^{-\frac{1}{2} \pi n}] \right\}. \end{eqnarray}
Similarly from (19) we see that, if $n$ is any positive odd integer, then
\begin{eqnarray} \tan^{-1} \frac{2n^2}{1^2} &-& 3 tan^{-1} \frac{2n^2}{3^2} + 5 \tan^{-1} \frac{2n^2}{5^2} - \cdots \nonumber\\ &=& \frac{4}{\pi} (-1)^{\frac{1}{2}(n-1)} \left\{\frac{\pi n}{4} \log \left(\frac{1+e^{-\frac{1}{2} \pi n}}{1-e^{-\frac{1}{2} \pi n}}\right) + \frac{1}{1^2} e^{-\frac{1}{2} \pi n} + \frac{1}{3^2} e^{-\frac{3}{2} \pi n} + \frac{1}{5^2} e^{-\frac{5}{2} \pi n}+\cdots\right\}; \end{eqnarray}
and, if $n$ is a positive even integer, then
$$\tan^{-1} \frac{2n^2}{1^2} - 3 \tan^{-1} \frac{2n^2}{3^2} + 5 \tan^{-1} \frac{2n^2}{5^2} - \ldots = 2n (-1)^{\frac{1}{2} n-1} \tan^{-1} e^{-\frac{1}{2} \pi n}.$$
In this connection it may be interesting to note that
$$\tan^{-1} e^{-\frac{1}{2} \pi n} = \frac{\pi}{4} -\left(\tan^{-1} \frac{n}{1} - \tan^{-1} \frac{n}{3} + \tan^{-1} \frac{n}{5} - \ldots \right)$$
for all real values of $n$.

5. Remembering that $\ds\frac{\pi}{4 \cosh \pi x} = \frac{1}{1^2 + 4x^2} - \frac{3}{3^2 + 4x^2} + \frac{5}{5^2 + 4x^2} - \ldots$ we have

\begin{eqnarray} \frac{\pi}{4} \sum^\infty_1 \frac{1}{n^2\cosh \pi n x} &=& \sum^{n=\infty}_{n=1} \left\{ \frac{1}{n^2 (1^2 + 4n^2 x^2)} - \frac{3}{n^2 (3^2+4n^2 x^2)} + \cdots \right\}\nonumber\\ &=& \frac{\pi^3}{8} \left(\frac{1}{3} + \frac{x^2}{2}\right) - \pi x \left(\frac{\coth \frac{\pi}{2x}}{1^2} - \frac{ \coth \frac{3\pi}{2x}}{3^2} + \frac{\coth \frac{5 \pi}{2x}}{5^2} - \cdots \right). \end{eqnarray}
That is to say, if $\alpha$ and $\beta$ are real and $\alpha \beta = \pi^2$, then
\begin{eqnarray} \phi (1)&+& 2 \phi (e^{-\alpha}) + 2 \phi (e^{-2\alpha}) + 2\phi (e^{-3\alpha}) + \cdots \nonumber\\ &=& \frac{\pi}{8} \left(\frac{\alpha}{3} + \frac{\beta}{2}\right) - \frac{\pi}{4 \beta} \left\{\frac{1}{1^2 \cosh \beta} + \frac{1}{2^2 \cosh 2 \beta} + \cdots \right\}. \end{eqnarray}
If, in particular, we put $\alpha = \beta = \pi$ in (26), we obtain
\begin{eqnarray} \phi (1) = \frac{5 \pi^2}{48} &-& 2 \left\{\frac{1}{1^2(e^\pi-1)} - \frac{1}{3^2(e^{3 \pi} -1)} + \frac{1}{5^2 (e^{5 \pi} -1)} \ldots \right\}\nonumber\\ &-& \frac{1}{2} \left\{\frac{1}{(1^2e^\pi + e^{-\pi})} + \frac{1}{2^2(e^{2 \pi} + e^{-2 \pi})} + \frac{1}{3^2 (e^{3 \pi} + e^{-3 \pi})} + \ldots \right\}\nonumber\\ &=& .9159655942 , \end{eqnarray}
approximately.