i was reading about the mellin transform ans i found the following $$\sum _{k=1}^{\infty } \left(\frac{e^{-k x}}{e^{-2 k x}+1}-\frac{\pi \text{sech}\left(\frac{\pi ^2 k}{x}\right)}{2 x}\right)=\frac{\pi }{4 x}-\frac{1}{4}$$ but i do not how to show? it seem interesting i try firt ti get the mellin transform it is equivalent to $$\sum _{k=1}^{\infty } \left(\frac{1}{2} \text{sech}(k x)-\frac{\pi \text{sech}\left(\frac{\pi ^2 k}{x}\right)}{2 x}\right)=\frac{\pi }{4 x}-\frac{1}{4}$$ get by ramanujan I found a similar formula $$\sum _{k=1}^{\infty } \frac{\pi (-1)^{k+1} \coth \left(\frac{\pi ^2 (2 k-1)}{2 x}\right)}{2 x}-\frac{1}{4}=\sum _{k=1}^{\infty } \frac{1}{2} \text{sech}(k x)$$ could you show it? Also i like to include $$\sum _{k=1}^{\infty } -\frac{16 x^2 \left(\pi \coth \left(\frac{\pi ^2 (2 k-1)}{2 x}\right)-\pi \right)}{\pi ^5 (2 k-1)^5}-\frac{1}{180} \left(x \left(\pi ^2-6 x^2\right) \left(x^2+\pi ^2\right)-90 \zeta (5)\right)=\sum _{k=1}^{\infty } \frac{1}{k^5 \left(e^{2 k x}+1\right)}$$ and $$\sum _{k=1}^{\infty } \frac{4 x^2 \left(\pi \coth \left(\frac{\pi ^2 (2 k-1)}{2 x}\right)-\pi \right)}{\pi ^3 (2 k-1)^3}-\frac{1}{12} \left(x^3+\pi ^2 x-6 \zeta (3)\right)=\sum _{k=1}^{\infty } \frac{1}{k^3 \left(e^{2 k x}+1\right)}$$
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