Digamma Function

The digamma function usually denoted by \(\psi\) is defined as the logarithmic derivative of the gamma function.

\[\psi(z) = \dfrac{\mathrm{d}}{\mathrm{d}z} \ln \Gamma(z) = \dfrac{\Gamma^\prime (z)}{\Gamma(z)}.\]

\[\psi(s+1)=\psi(s)+\dfrac{1}{s}\]

Consider \[\Gamma(s+1)=s\Gamma(s)\] taking log we have \[\ln(\Gamma(s+1))=\ln(\Gamma(s))+\ln(s)\] differentiating with respect to s we get \[\psi(s+1)=\psi(s)+\dfrac{1}{s}\]

\[\psi(s+1)=-\gamma+\sum_{k=1}^\infty \left(\dfrac{1}{k}-\dfrac{1}{k+s}\right)\]

Consider the weierstrass representation of the gamma function: \[\Gamma(s)=\dfrac{e^{-\gamma s}}{s} \prod_{k=1}^\infty e^{s/k} \left(1+\dfrac{s}{k}\right)^{-1} \] Taking log \[\ln(\Gamma(s))=-\gamma s-\ln(s)+\sum_{k=1}^\infty \left(\dfrac{s}{k}-\ln \left(1+\dfrac{s}{k}\right)\right)\] differentiating with respect to s we have \[\psi(s)=-\gamma-\dfrac{1}{s}+\sum_{k=1}^\infty \left(\dfrac{1}{k}-\dfrac{1}{k+s}\right)=-\gamma+\sum_{k=1}^\infty \left(\dfrac{1}{k}-\dfrac{1}{k+s-1}\right)\] So, changing s with s+1 \[\psi(s+1)=-\gamma+\sum_{k=1}^\infty \left(\dfrac{1}{k}-\dfrac{1}{k+s}\right)\]

\[\psi(s+1) = -\gamma + \int_0^1 \dfrac{1-x^s}{1-x} dx\]

Consider \[\psi(s)=-\gamma+\sum_{n=1}^\infty \left(\dfrac{1}{n}-\dfrac{1}{n+s}\right)\] \[\psi(s)=-\gamma+\sum_{n=1}^\infty \int_0^1 (x^{n-1}-x^{n+s-1})dx=-\gamma+ \int_0^1\sum_{n=1}^\infty (x^{n-1}-x^{n+s-1})dx\] By applying geometric progression sum we have \[\psi(s+1) = -\gamma + \int_0^1 \dfrac{1-x^s}{1-x} dx\]

also by the integral representation of harmonic numbers , \[\psi(s+1) = -\gamma + H_s\]

By Euler's reflection formula, we have the following relation

\[\psi(1-z) - \psi(z) = \pi \cot \pi z\]

Euler's reflection formula is as follows:

\[\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin \pi z}\]

Taking natural logarithm and differentiating the above expression, we observe that

\[\log(\Gamma(z))+\log(\Gamma(1-z)) = \log \pi - \log \sin \pi z\]

On differentiating, we get

\[\dfrac{\Gamma^{\prime}(z)}{\Gamma(z)}-\dfrac{\Gamma^{\prime}(1-z)}{\Gamma(1-z)} = - \dfrac{\pi \cos \pi z}{\sin \pi z}\]

\[\Rightarrow \psi(z) - \psi(1-z) = - \dfrac{\pi \cos \pi z}{\sin \pi z}\]

\[\Rightarrow \psi(1-z) - \psi(z) = \pi \cot \pi z\]

where \(\psi\) denotes the Digamma function, which is the logarithmic derivative of the Gamma function.

\[2\psi(2s)=2\ln(2)+\psi(s)+\psi(s+1/2)\]

Consider \[\sqrt{\pi} \ \Gamma(2s)=2^{2s-1} \Gamma(s)\Gamma(s+1/2)\] Taking log \[\ln(\sqrt{\pi})+\ln(\Gamma(2s))=(2s-1)\ln(2)+\ln(\Gamma(s))+\ln(\Gamma(s+1/2))\] differentiating with respect to s we have \[2\psi(2s)=2\ln(2)+\psi(s)+\psi(s+1/2)\]

Prove

\[\sum_{n=0}^\infty \dfrac{1}{n^2+1} =\dfrac{\pi+1}{2} + \dfrac{\pi}{e^{2\pi}-1}\]

---

\[ S = \sum_{n=0}^{\infty} \frac{1}{n^2+1} = \frac{1}{2i} \sum_{n=0}^{\infty} \left( \frac{1}{n-i} - \frac{1}{n+i} \right) \] We can write the summation as \[2iS= \sum_{n=1}^{\infty} \left( \frac{1}{n-1-i} - \frac{1}{n-1+i} \right)=\sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n-1+i} \right)-\sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n-1-i} \right)\] By the series representation of the digamma function, this is just: \[2iS=\psi(i)-\psi(-i)=\psi(i)-\psi(1-i)-\dfrac{1}{i}=-\pi \cot(i\pi)+i=\pi i \coth(\pi)+i\] so \[S= \dfrac{1+\pi\coth(\pi)}{2}\] We can simplify this further to get \[S=\dfrac{\pi+1}{2} + \dfrac{\pi}{e^{2\pi}-1}\]

The nth polygamma function is given by \[\psi_n (s)= \dfrac{d^n}{ds^n} \psi(s)=\psi^{(n)}(s)\]

From this we can also get the taylor series of the digamma function \[\psi(s)=\sum_{n=0}^\infty \dfrac{\psi^{(n)}(1)(s-1)^n}{n!}=-\gamma+\sum_{n=1}^\infty \dfrac{\psi_n(1)(s-1)^n}{n!}=-\gamma-\sum_{n=1}^\infty (-1)^n\zeta(n+1)(s-1)^n\\ \psi(s+1)=-\gamma-\sum_{n=1}^\infty \zeta(n+1)(-s)^n\] We can differentiate the integral representation n times to get \[\psi_n(s+1)=\int_0^1 \dfrac{\ln^n(x) x^s}{x-1}dx\] We can also do this to the functional equation to get \[\psi_n(s+1)=\psi_n(s)+(-1)^nn! z^{-n-1}\]

The digamma functions can be implemented in Mathematica as follows

1

PolyGamma[z]

or

1

PolyGamma[0,z]