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 ?
 A: 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.

Concerning the real part I fear that no such elementary relation ('closed form') exist. 
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.
$$\Re(\psi(iy))=-\frac 1{y^2+1}+1-\gamma-\sum_{k=1}^\infty (-1)^k(\zeta(2k+1)-1)y^{2k}\quad \text{for}\ |y|<2$$
