I'm still working on new identities, which arise immediately from stocha equation (2):

$$\log (2)+2 \int_0^{\infty } (\psi (1+i\,z)+\psi (1-i\,z))\, \text{sech}(2 \pi \,z) \, dz-\int_0^{\infty } (\psi (1+i\,z)+\psi (1-i\,z))\, \text{sech}(\pi \,z) \, dz=\frac{\pi }{2}-1$$ $(1)$

Inspired by Oloa's integral Oloa and User, the Laplace transform of the complex valued psi function is: $$L_{C}(a)=\int_0^{\infty } e^{-a z} \psi (1+i z) \, dz$$ $(2)$

similar to Dixit.

$$L_C (a)=2 a \sum _{n=1}^m \frac{\log (n)}{a^2-4 \pi ^2 n^2}+\frac{\pi \ (a+2\, i)}{4 \,a}-\frac{\log (a)+\gamma }{a}+\frac{i}{2}\, \log\left(\frac{a}{2\pi }\right)$$

$$\frac{i}{4} \, \left(\psi \left(1-\frac{a}{2 \pi }\right)+\psi \left(1+\frac{a}{2 \pi }\right)\right)+\frac{\log \left(\frac{2 \,i\, \pi}{a}\right) \left(a \cot \left(\frac{a}{2}\right)-2\right)}{2 a}$$ $(3)$

Our interest is to integrate

$$P_C=\int_0^{\infty } \text{sech}(2\pi \,z) \, \psi (1+i z) \, dz$$ $(4)$

with $$\text{sech}(z)=2 \sum _{k=0}^{\infty } (-1)^k e^{-(2 k+1)\, z}$$ $(5)$

Problems, which arise from the double sum and emerging problems with infinities can be avoided with the integral form of the sum (3):

$$\sum _{n=1}^{\infty } \frac{\text{Log}[n]}{a^2-4 n^2 \pi ^2}=\frac{\ \label{eq:thirtyeight} \left(2-a \cot \left(\frac{a}{2}\right)\right) \log \left(\frac{2 \pi \ }{a}\right)}{4a^2}-\int_0^{\infty } \frac{-2+x \coth \left(\frac{x}{2}\right)}{4 x \left(a^2+x^2\right)} \, dx$$ $(6)$

The Imaginary part of the integral is given in closed form:

$$\Im\left(\int_0^{\infty } \text{sech}(z) \psi (1+i z) \, dz\right)=\ \sum _{k=0}^{\infty } (-1)^k \left(\log \left(\frac{2 k+1}{2 \pi }\right)-\psi\left(\frac{2 k+1}{2 \pi }\right)\right)-\frac{\pi ^2}{4}$$ $(7)$

The sum on the right hand side

$$\sum _{k=1}^{\infty } \left(\psi (\alpha\, k+s)-\log (\alpha \,k+s)+\frac{1}{2\, (\alpha \, k+s)}\right)$$ $(8)$

for $\alpha =\frac{1}{\pi }, \,s=\frac{1}{2 \pi }$

• without $(-1)^k$ is done in Oloa.

We found the following closed forms: $$\Im\left(\int_0^{\infty } \text{sech}(2 \pi \, z) \,\psi (1+i z) \, dz\right)=\frac{\gamma }{2}+\log \left(\frac{\Gamma \left(\frac{3}{4}\right)}{2^{3/4}\,\Gamma \left(\frac{5}{4}\right)}\right)$$ $(9)$

and

$$\Im\left(\int_0^{\infty } \text{sech}(\pi\,z)\, \psi (1+i z) \, dz\right)=\frac{\gamma }{2}+\log \left(\frac{\Gamma \left(\frac{3}{4}\right)}{\sqrt{2}\,\Gamma \left(\frac{5}{4}\right)}\right)$$ $(10)$

where $\gamma$ is the Euler-Mascheroni constant.