About Mellin transform and harmonic series

Question

Let $\mathfrak{M}\left(*\right)$ the Mellin transform. We know that holds this identity$$\mathfrak{M}\left(\underset{k\geq1}{\sum}\lambda_{k}g\left(\mu_{k}x\right),\, s\right)=\underset{k\geq1}{\sum}\frac{\lambda_{k}}{\mu_{k}^{s}}\mathfrak{M}\left(g\left(x\right),s\right)$$ so for example if we want a closed form of$$\underset{k\geq1}{\sum}\frac{\sin\left(nx\right)}{n^{3}}=x^{3}\underset{k\geq1}{\sum}\frac{\sin\left(nx\right)}{n^{3}x^{3}}\,\,(1)$$ we are in the case $$\lambda_{k}=1,\thinspace\mu_{k}=k,\, g\left(x\right)=\frac{\sin\left(x\right)}{x^{3}}$$ and so our Mellin transform is $$\zeta\left(s\right)\mathfrak{M}\left(g\left(x\right),s\right)$$ and then we find the sum of $(1)$ by the residue theorem. I tried to use this argument for this sum$$\underset{k\geq1}{\sum}\frac{\left(-1\right)^{k}}{k\left(e^{k\pi}+1\right)}=\pi\underset{k\geq1}{\sum}\frac{\left(-1\right)^{k}}{k\pi\left(e^{k\pi}+1\right)}\,\,(2)$$ so we are in the case $$\lambda_{k}=\left(-1\right)^{k},\thinspace\mu_{k}=k,\, g\left(x\right)=\frac{1}{x\left(e^{x}+1\right)}$$ and so the Mellin transform of $(2)$ , using $$\underset{k\geq1}{\sum}\frac{\left(-1\right)^{k}}{k^{s}}=-\left(1-2^{1-s}\right)\zeta\left(s\right)$$ is$$\left(1-2^{1-s}\right)\zeta\left(s\right)\Gamma\left(s-1\right)\left(-1+2^{2-s}\right)\zeta\left(s-1\right)$$ hence, with $x=\pi$ $$\underset{k\geq1}{\sum}\frac{\left(-1\right)^{k}}{k\left(e^{k\pi}+1\right)}=\frac{1}{2\pi i}\int_{\mathbb{C}}\left(1-2^{1-s}\right)\zeta\left(s\right)\Gamma\left(s-1\right)\left(-1+2^{2-s}\right)\zeta\left(s-1\right)\pi^{1-s}ds=\frac{1}{2\pi i}\int_{\mathbb{C}}Q\left(s\right)ds.$$ Now we note that there are only two poles, because $\zeta\left(s\right)$ cancels poles of Gamma at negative even integer and $\zeta\left(s-1\right)$ cancels poles of Gamma at negative odd integers. So the compute is$$\underset{s=0}{\textrm{Res}}\,Q\left(s\right)=\frac{1}{8}\pi$$ $$\underset{s=1}{\textrm{Res}}\,Q\left(s\right)=-\frac{1}{2}\log\left(2\right)$$ so$$\underset{k\geq1}{\sum}\frac{\left(-1\right)^{k}}{k\left(e^{k\pi}+1\right)}=\frac{1}{8}\pi-\frac{1}{2}\log\left(2\right)$$ but this is wrong because the correct result is $$\frac{1}{8}\pi-\frac{5}{8}\log\left(2\right).$$ I don't understand why in this case this argument doesn't work. Where is the error? Thank you.