Series expansion of $\log\frac{\sin z}{z}$

Question

What is the series expansion of $f(z)= \log\frac{\sin z}{z}$ upto $z^6$ terms? Since $f(z)$ is an even function,so the coefficients of all the odd powers of z must be $0$ and I calculated the coefficient of $z^2 $ to be $-1/4$ (am I correct) but what will be the expansion till $z^6$?

Answer 1

Using steps, start with $$\log \left(\frac{\sin (x)}{x}\right)=\log \left(1+\left(\frac{\sin (x)}{x}-1\right)\right)$$ Let $t=\frac{\sin (x)}{x}-1$ and use $$\log(1+t)=t-\frac{t^2}{2}+\frac{t^3}{3}-\frac{t^4}{4}+\frac{t^5}{5}-\frac{t^6}{6}+\frac{t^7}{ 7}+O\left(t^8\right)$$ Now $$t=\frac{\sin (x)}{x}-1=-\frac{x^2}{6}+\frac{x^4}{120}-\frac{x^6}{5040}+O\left(x^8\right)$$ I suppose that the binomila expansion will be simple since you are not asked for too many terms.

Answer 2

I will post another approach with Bernoulli numbers. This answer provides the general terms of the coefficients. $$\ln\frac{\sin x}x=\int_0^x\left(\cot t-\frac1t\right)dt\\ =\int_0^x\sum_{n\ge 1}\frac{(-4)^nB_{2n}}{(2n)!}t^{2n-1}dt\\ =\sum_{n\ge 1}\frac{(-4)^nB_{2n}}{(2n)!(2n)}x^{2n}\\ =-\frac16x^2-\frac1{180}x^4-\frac1{2835}x^6-\cdots$$