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digamma identities

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Post Wed Jun 12, 2013 9:50 am
galactus User avatar
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Here are a few cool identities I list for no particulalr other than their cool and may come in handy at some point.

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

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

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