# Proving $\int\limits^{\infty}_{-\infty}\frac{\sin(2n\arctan(x))}{\left(x^2+1\right)^n\left( e^{x\pi}+1\right)}dx=\eta(2n)-\zeta(2n)$

For quite some time now I've been stuck on proving the following:

$$\int\limits^{\infty}_{-\infty}\frac{\sin(2n\arctan(x))}{\left(x^2+1\right)^n\left( e^{x\pi}+1\right)}dx=\eta(2n)-\zeta(2n),$$

where $$\eta(z)$$ is the Dirichlet Eta function and $$\zeta(z)$$ is the Riemann Zeta function. I would like to prove this for $$n>1$$, but showing it to be true only for integer $$n$$ would be fine too. For the case of integer $$n$$, I have found an expansion for the numerator that may be of use.

$$\sin(2n\arctan(x))=\frac{x}{(x^2+1)^n}\sum ^{\left\lfloor \frac{2n-1}{2}\right\rfloor }_{k=0}\left(\begin{matrix}2n\\2k+1\end{matrix}\right)(-x^2)^k.$$

Other than that, I really haven't been able to do anything to this integral.

First note that $$\frac{\sin (2n\arctan x)}{(1+x^2)^n}=\Im(1-ix)^{-2n}$$ Letting $$f(z)=\frac{(1-ix)^{-2n}}{e^{x\pi}+1}$$ Integrating $$f(z)$$ over the semicircle in the upper half plane and avoiding the poles in the integrating path, the original integral turns into $$\Im 2\pi i\sum_{k \text{ positive and odd}}\operatorname{Res}_{z=ki}f(z)=-\sum_{k=1}^\infty\frac{(2k)^{-2n}2\pi}{\pi}=-2^{1-2n}\zeta(2n)=\eta(2n)-\zeta(2n)$$