It is known that
\begin{equation*}
\tan x=\sum_{k=1}^{\infty}\frac{2^{2k}\bigl(2^{2k}-1\bigr)}{(2k)!}|B_{2k}|x^{2k-1}, \quad |x|<\frac{\pi}{2}
\end{equation*}
and
\begin{equation*}
\ln\tan x=\ln x+\sum_{k=1}^{\infty}\frac{2^{2k}\bigl(2^{2k-1}-1\bigr)}{k(2k)!}|B_{2k}|x^{2k}, \quad 0<x<\frac{\pi}{2},
\end{equation*}
where $B_{2k}$ denotes the classical Bernoulli numbers. From the second series expansion, we acquire
\begin{equation*}
\ln\frac{\tan x}{x}=\sum_{k=1}^{\infty}\frac{2^{2k}\bigl(2^{2k-1}-1\bigr)}{k(2k)!}|B_{2k}|x^{2k}, \quad |x|<\frac{\pi}{2}.
\end{equation*}
My question is:
What and where is the series expansion of the function $\ln\bigl(\frac{\tan x}{x}-1\bigr)$ or $\ln(\tan x-x)$ around $x=0$?
References
- I. S. Gradshteyn and I. M. Ryzhik, Table of Integrals, Series, and Products, Translated from the Russian, Translation edited and with a preface by Daniel Zwillinger and Victor Moll, Eighth edition, Revised from the seventh edition, Elsevier/Academic Press, Amsterdam, 2015; available online at https://doi.org/10.1016/B978-0-12-384933-5.00013-8.
Best Answer
Another approach is obtained by introducing the Bessel functions to express \begin{align} f(x)&=\ln\left(\frac{\tan x-x}{x^3}\right)\\ &=\ln\left( \frac{\sin x-x\cos x}{x^3\cos x} \right)\\ &=\ln\left( \frac{J_{3/2}(x)}{x^2J_{-1/2}(x)} \right) \end{align} In a paper by Dickinson a uniformly convergent series expansion is given for the logarithm of a Bessel function: \begin{equation} \ln J_{\nu}(x)=\ln \left(\frac{(x / 2)^{\nu}}{\Gamma(\nu+1)}\right)+\sum_{n=1}^{\infty} \ln \left(1-\frac{x^2}{j_{\nu, n}}\right) \end{equation} when $\nu>-1$ and $0<x<j_{\nu ,1}$ ($j_{\nu,n}$ is the $n$-th zero of the Bessel function of order $\nu$). Then, \begin{align} \ln J_{\nu}(x)&=\ln \left(\frac{(x / 2)^{\nu}}{\Gamma(\nu+1)}\right)-\sum_{n=1}^{\infty}\sum_{k=0}^{\infty} \frac{1}{k+1}\frac{x^{2k+2}}{j_{\nu,n}^{2k+2}}\\ &=\ln \left(\frac{(x / 2)^{\nu}}{\Gamma(\nu+1)}\right)-\sum_{k=0}^{\infty} \frac{x^{2k+2}}{k+1}\sum_{n=1}^{\infty}\frac{1}{j_{\nu,n}^{2k+2}} \end{align} (all the summands in the convergent double series are positive and we may interchange the order of summation). Thus \begin{align} f(x)&=-2\ln x+\ln J_{3/2}(x)-\ln J_{-1/2}(x)\\ &=-\ln3 +\sum_{k=0}^{\infty} \frac{x^{2k+2}}{k+1}\left(\sum_{n=1}^{\infty} \frac{1}{j_{-1/2,n}^{2k+2}}-\sum_{n=1}^{\infty}\frac{1}{j_{3/2,n}^{2k+2}} \right) \end{align}
The sum of inverse even powers of zeros of Bessel functions appears in problems involving the diffusion equation and a broad literature exists on the topic. The quantity \begin{equation} \sigma(p,\nu)=\sum_{n=1}^\infty\frac{1}{j_{\nu,n}^{2p}} \end{equation} is discussed here. A recent analysis can be found in a paper by Jorge L. deLyra where a general formula for certain linear combinations of these sums is given ; it can be used to derive the formulas for $\sigma(p, \nu)$ by purely linear-algebraic means. The first values for $\nu=3/2$ are \begin{align} &\sigma(1,3/2)=\frac{1}{10}\;;\;\sigma(2,3/2)=\frac{1}{350}\;;\;\sigma(3,3/2)=\frac{1}{7875}\\ &\sigma(4,3/2)=\frac{37}{6063750}\;;\;\sigma(5,3/2)=\frac{59}{197071875} \end{align} while $j_{-1/2,n}=(2n-1)\pi/2$. Then \begin{align} \sigma(k+1,-1/2)&=\sum_{n=1}^\infty\frac{1}{j_{-1/2,n}^{2k+2}}\\ &=\sum_{n=1}^\infty\left(\frac{2}{(2n-1)\pi} \right)^{2k+2}\\ &=\left( \frac{2}{\pi} \right)^{2k+2}(2^{2k+2}-1)\zeta(2k+2)\\ &=(2^{2k+2}-1)\frac{2^{2k+1}}{(2k+2)!}|B_{2k+2}| \end{align} Latter identity is obtained expressing the Zeta function at even values in terms of the Bernoulli numbers. Finally, \begin{align} f(x)&=-\ln3 +\sum_{k=0}^{\infty} \frac{x^{2k+2}}{k+1}\left(\sigma(k+1,-1/2)-\sigma(k+1,3/2)\right)\\ &=-\ln3 +\sum_{k=0}^{\infty} \frac{x^{2k+2}}{k+1}\left((2^{2k+2}-1)\frac{2^{2k+1}}{(2k+2)!}|B_{2k+2}|-\sigma(k+1,3/2)\right) \end{align} which leads to the same first terms as in the expansion given by other users.