Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ?

share|improve this question

1 Answer 1

up vote 0 down vote accepted

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. $$\Re(\psi(iy))=-\frac 1{y^2+1}+1-\gamma-\sum_{k=1}^\infty (-1)^k(\zeta(2k+1)-1)y^{2k}\quad \text{for}\ |y|<2$$

share|improve this answer
    
@:Raymond Manzoni Thank you very mach! –  Dao yi Peng Nov 11 '12 at 15:29
    
@Dao: you are welcome! –  Raymond Manzoni Nov 11 '12 at 15:46

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.