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## digamma identities

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### digamma identities

Wed Jun 12, 2013 9:50 am
galactus
Global Moderator

Posts: 902
Here are a few cool identities I list for no particulalr other than their cool and may come in handy at some point.

$\displaystyle \Im \psi(1+ix)=\frac{\pi}{2}\coth(\pi x)-\frac{1}{2x}$

$\displaystyle \Re \psi(1+ix) =-\gamma + \sum_{k=1}^{\infty}\frac{x^{2}}{k(k^{2}+x^{2})}$

### Re: digamma identities

Wed Jun 12, 2013 3:34 pm
zaidalyafey Global Moderator

Posts: 357
For the first one :

$\displaystyle \psi(1-x)-\psi(x) = \pi \cot(\pi x)$---(1)

$\displaystyle \psi(1+x)-\psi(x) = \frac{1}{x}$----(2)

subtracting (1) from (2) we obtain

$\displaystyle \psi(1+x) - \psi(1-x) = \frac{1}{x} - \pi \cot(\pi x)$

$\displaystyle \psi(1+x) = \frac{1}{x} - \pi \cot(\pi x) + \psi(1-x)$

$\displaystyle \psi(1+ix) = -i\frac{1}{x} + i \pi\, \text{coth}(\pi x) + \psi(1-ix)$

But

$\displaystyle \Im \psi(1-ix) = - \Im \psi(1+ix)$

Because we can have

$\displaystyle \psi(1+x)= -\gamma -\sum_{k\geq 1}\zeta (k+1)\;(-x)^k$

$\displaystyle \psi(1+ix)= -\gamma -\sum_{k\geq 1} \zeta (k+1)\;(-ix)^k$

$\displaystyle \Im \psi(1+ix)= \sum_{k\geq 1} (-1)^k \zeta (k+1)\; x^{2k-1}$

$\displaystyle \Im \psi(1-ix)= -\sum_{k\geq 1} (-1)^k \zeta (k+1)\; x^{2k-1}$

Hence we can deduce

$\displaystyle \Im\psi(1+ix) = -\frac{1}{x} + \pi\, \text{coth}(\pi x) -\Im \psi(1-ix)$

$\displaystyle 2 \Im\psi(1+ix) = \pi\, \text{coth}(\pi x)- \frac{1}{x} \,\,\, \square$

For the second one

We know that

$\displaystyle \psi(1+x) = -\gamma + \sum_{n\geq 1} \frac{x}{n(n+x)}$

$\displaystyle \psi(1+ix) = -\gamma + \sum_{n\geq 0} \frac{ix}{n(n+ix)}$

Multiplying by the conjugate we get

$\displaystyle \psi(1+ix) = -\gamma + \sum_{n\geq 1} \frac{ix(n-ix)}{n(n^2+x^2)}$

$\displaystyle \Re \psi(1+ix) = -\gamma + \sum_{n\geq 1} \frac{x^2}{n(n^2+x^2)}\,\,\, \square$
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