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Could you show that $$\sum _{k=0}^{\infty } \frac{k}{e^{\frac{\pi k}{2}}+1}=\frac{7 \pi ^2+6 \left(\psi _{e^{2 \pi }}^{(1)}(1)+\psi _{e^{2 \pi }}^{(1)}\left(1-\frac{i}{2}\right)-\psi _{e^{2 \pi }}^{(1)}\left(1+\frac{i}{4}\right)-\psi _{e^{2 \pi }}^{(1)}\left(1-\frac{i}{4}\right)\right)}{24 \pi ^2}$$ using mellin tranformthe Mellin transform?

Could you show that $$\sum _{k=0}^{\infty } \frac{k}{e^{\frac{\pi k}{2}}+1}=\frac{7 \pi ^2+6 \left(\psi _{e^{2 \pi }}^{(1)}(1)+\psi _{e^{2 \pi }}^{(1)}\left(1-\frac{i}{2}\right)-\psi _{e^{2 \pi }}^{(1)}\left(1+\frac{i}{4}\right)-\psi _{e^{2 \pi }}^{(1)}\left(1-\frac{i}{4}\right)\right)}{24 \pi ^2}$$ using mellin tranform

Could you show that $$\sum _{k=0}^{\infty } \frac{k}{e^{\frac{\pi k}{2}}+1}=\frac{7 \pi ^2+6 \left(\psi _{e^{2 \pi }}^{(1)}(1)+\psi _{e^{2 \pi }}^{(1)}\left(1-\frac{i}{2}\right)-\psi _{e^{2 \pi }}^{(1)}\left(1+\frac{i}{4}\right)-\psi _{e^{2 \pi }}^{(1)}\left(1-\frac{i}{4}\right)\right)}{24 \pi ^2}$$ using the Mellin transform?

ramanujan type Ramanujan-type sum

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ramanujan type sum

Could you show that $$\sum _{k=0}^{\infty } \frac{k}{e^{\frac{\pi k}{2}}+1}=\frac{7 \pi ^2+6 \left(\psi _{e^{2 \pi }}^{(1)}(1)+\psi _{e^{2 \pi }}^{(1)}\left(1-\frac{i}{2}\right)-\psi _{e^{2 \pi }}^{(1)}\left(1+\frac{i}{4}\right)-\psi _{e^{2 \pi }}^{(1)}\left(1-\frac{i}{4}\right)\right)}{24 \pi ^2}$$ using mellin tranform