What's the $n$-th derivative of $\ln(\sin(x))$?

Question

I want to find the $n$-th derivative of $\ln(\sin x)$, i.e. $$ \frac{d^n\ln(\sin x)}{dx^n} $$ where $x\in (0,\pi/2)$ such that $\sin x>0$. To make the problem definitely, $x=\pi/4$ is assumed. In wolframalpha, we know that for $n=1,2,\cdots$, we have $1,-2,4,-16,80,-512,\cdots$. So, what's the general behavior of the derivative w.r.t. $n$?

Answer 1

Hint. One may observe that $$ \frac{d\ln(\sin x)}{dx}= \cot x \tag1 $$ gives $$ \left.\frac{d^n\ln(\sin x)}{dx^n}\right|_{\large x=\frac{\pi}4}=\left.\frac{d^{n-1}}{dx^{n-1}}\left( \cot x\right)\right|_{\large x=\frac{\pi}4}\tag2 $$ An approach is to use the reflection formula for the digamma function: $$ \cot{ x}=\frac1\pi\left(\psi\left(1-\frac{x}{\pi}\right)-\psi\left(\frac{x}{\pi}\right)\right)\tag3 $$

Then, for $x$ sufficiently near $\dfrac{\pi}4$, we have $$ \begin{align} \cot{x} &=\frac1\pi\left(\psi\left(1-\frac{x}{\pi}\right)-\psi\left(\frac{x}{\pi}\right)\right)\\ &=\frac1\pi\int^1_0\frac{t^{x/\pi-1}-t^{-x/\pi}}{1-t}\:{\rm d}t\\ &=\frac1\pi\sum_{k=0}^\infty\int^1_0\left(t^{x/\pi+k-1}-t^{-x/\pi+k}\right){\rm d}t\\ &=\sum_{k=0}^\infty\left(\frac{1}{x+k\pi}+\frac{1}{x-(k+1)\pi}\right)\\ &=\sum_{k=0}^\infty\left(\frac{1}{x-\pi/4+(k+1/4)\pi}+\frac{1}{x-\pi/4-(k+3/4)\pi}\right)\\ &=\sum^\infty_{k=0}\sum^\infty_{n=0}\left(\frac{(-1)^n}{\pi^{n+1}(k+1/4)^{n+1}}-\frac{1}{\pi^{n+1}(k+3/4)^{n+1}}\right)\left(x-\frac{\pi}4\right)^n\\ &=1+\sum^\infty_{n=1}\frac{(-1)^n\zeta(n+1,1/4)-\zeta(n+1,3/4)}{\pi^{n+1}}\left(x-\frac{\pi}4\right)^n \end{align} $$ where $\zeta(\cdot,\cdot)$ is the Hurwitz zeta function.

Thus, by the Taylor series expansion formula, one gets, for $n=2,3,4,\cdots$,

$$ \left.\frac{d^n\ln(\sin x)}{dx^n}\right|_{\large x=\frac{\pi}4}=(n-1)!\:\frac{(-1)^{n-1}\zeta(n,1/4)-\zeta(n,3/4)}{\pi^{n}}\tag4 $$

which one may eventually rewrite as

$$ \left.\frac{d^n\ln(\sin x)}{dx^n}\right|_{\large x=\frac{\pi}4}=(-1)^{n(n-1)/2}\left(2^{n-1}E_{n-1}+2^{2n-1}\left(2^n-1\right)\frac{B_n}n\right) \tag5 $$

where $E_n$ and $B_n$ are Euler and Bernoulli numbers respectively.