# How to calculate $\Re\psi(\mathrm{i}y)=$? and $\Im\psi(\mathrm{i}y)=\frac{1}{2}y^{-1}+\frac{1}{2}\pi\coth{\pi y}$?

How to calculate $$\Re\psi(\mathrm{i}y)= ?$$ and how to proof $$\Im\psi(\mathrm{i}y)=\frac{1}{2}y^{-1}+\frac{1}{2}\pi\coth{\pi y}.$$ Here $\mathrm{i}^2=-1.\psi(s)$ is digamma function.

Can you help me with this problem ?

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The reflection formula is : $$\Gamma(z)\Gamma(1-z)=\frac {\pi}{\sin(\pi z)}$$ The logarithm of this becomes : $$\log(\Gamma(z))+\log(\Gamma(1-z))=\log(\pi)-\log(\sin(\pi z))$$ Using $\psi(x)=(\log(\Gamma(z))'$ we may write the derivative of this as : $$\psi(z)-\psi(1-z)=-\pi\cot(\pi z)$$ We may use the reccurence formula $\ \psi(1+x)=\psi(x)+\frac 1x\$ to rewrite this as : $$\psi(z)-\psi(-z)+\frac 1z=-\pi\cot(\pi z)$$ For $\ z:=iy\$ and $\ y\in \mathbb{R}\$ we get : $$\psi(iy)-\psi(-iy)+\frac 1{iy}=-\pi\cot(\pi iy)$$ But $\ \overline{\psi(z)}=\psi(\overline{z})\$ and $\ \cot(iu)=-i\coth(u)\$ getting : $$\psi(iy)-\overline{\psi(iy)}=-\frac 1{iy}+i\pi\coth(\pi y)$$ i.e. $$2i\,\Im(\psi(iy))=i\left(\frac 1y+\pi\coth(\pi y)\right)$$ and your result for the imaginary part.
But there are many other ways to write it (see at Wikipedia, MathWorld, DLMF and especially Wolfram functions) for example using : $$\psi(z)=-\frac 1z-\gamma+\sum_{k=2}^\infty(-1)^k\zeta(k)z^{k-1}$$ or rather this expression (from the DLMF link) where the odd powers of $z$ (giving the imaginary part when $z=iy$) and the even powers are neatly separated : $$\psi(z)=-\frac 1{2z}-\frac{\pi}2\cot(\pi z)+\frac 1{z^2-1}+1-\gamma-\sum_{k=1}^\infty (\zeta(2k+1)-1)z^{2k}\quad \text{for}\ |z|<2, z\not =0, z\not = \pm 1$$
i.e. $$\Re(\psi(iy))=-\frac 1{y^2+1}+1-\gamma-\sum_{k=1}^\infty (-1)^k(\zeta(2k+1)-1)y^{2k}\quad \text{for}\ |y|<2$$