Show that $ ix+\frac{2ix}{e^{2ix}-1} = 1+\sum_{j=1}^\infty \frac{(-1)^jB_{2j}}{2j!}(2x)^{2j}$

Question

I proved that $x\cot x = ix+2ix/(e^{2ix}-1)$, now I need to show that

$ ix+\frac{2ix}{e^{2ix}-1} =x\cot x = 1+\sum_{j=1}^\infty \frac{(-1)^jB_{2j}}{2j!}(2x)^{2j}$.

I know that $z/(e^z-1)=\sum_{k=0}^\infty(B_k/k!)z^k$ for all complex numbers $z$ such that $|z| < 2\pi$

So if I let $z=2ix$ I get $x\cot x=ix+\frac{2ix}{e^{2ix}-1}\overset{sub}{=}z/2+z/(e^z-1) = z/2+\sum\limits_{k=0}^\infty\frac{B_k}{k!}z^k$ but I don't know how to get from there to here:$1+\sum_{j=1}^\infty \frac{(-1)^jB_{2j}}{2j!}(2x)^{2j}$

Answer 1

Hint:

As $$g(x)={x\over e^x-1}=\sum_{j=0}^\infty B_j{x^j\over j!}$$ and $B_0=1$, your RHS is $$ \sum_{j=0}^\infty B_{2j}{(2ix)^{2j}\over (2j)!}={1\over2}(g(2ix)+g(-2ix))\\ $$