Digamma Function
The digamma function usually denoted by \(\psi\) is defined as the logarithmic derivative of the gamma function.
Contents
Definition
\[\psi(z) = \dfrac{\mathrm{d}}{\mathrm{d}z} \ln \Gamma(z) = \dfrac{\Gamma^\prime (z)}{\Gamma(z)}.\]
Functional equation
\[\psi(s+1)=\psi(s)+\dfrac{1}{s}\]
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 \[\psi(s+1)=\psi(s)+\dfrac{1}{s}\]
Series Representation
\[\psi(s+1)=\gamma+\sum_{k=1}^\infty \left(\dfrac{1}{k}\dfrac{1}{k+s}\right)\]
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 \[\psi(s)=\gamma\dfrac{1}{s}+\sum_{k=1}^\infty \left(\dfrac{1}{k}\dfrac{1}{k+s}\right)=\gamma+\sum_{k=1}^\infty \left(\dfrac{1}{k}\dfrac{1}{k+s1}\right)\] So, changing s with s+1 \[\psi(s+1)=\gamma+\sum_{k=1}^\infty \left(\dfrac{1}{k}\dfrac{1}{k+s}\right)\]
Integral Representation
\[\psi(s+1) = \gamma + \int_0^1 \dfrac{1x^s}{1x} dx\]
Consider \[\psi(s)=\gamma+\sum_{n=1}^\infty \left(\dfrac{1}{n}\dfrac{1}{n+s}\right)\] \[\psi(s)=\gamma+\sum_{n=1}^\infty \int_0^1 (x^{n1}x^{n+s1})dx=\gamma+ \int_0^1\sum_{n=1}^\infty (x^{n1}x^{n+s1})dx\] By applying geometric progression sum we have \[\psi(s+1) = \gamma + \int_0^1 \dfrac{1x^s}{1x} dx\]From this we can find specific values of the digamma function easily, for example putting s=0 we get \(\psi(1)=\gamma\)
also by the integral representation of harmonic numbers , \[\psi(s+1) = \gamma + H_s\]
Euler's Reflection Formula
By Euler's reflection formula, we have the following relation
\[\psi(1z)  \psi(z) = \pi \cot \pi z\]
Euler's reflection formula is as follows:
\[\Gamma(z)\Gamma(1z) = \frac{\pi}{\sin \pi z}\]
Taking natural logarithm and differentiating the above expression, we observe that
\[\log(\Gamma(z))+\log(\Gamma(1z)) = \log \pi  \log \sin \pi z\]
On differentiating, we get
\[\dfrac{\Gamma^{\prime}(z)}{\Gamma(z)}\dfrac{\Gamma^{\prime}(1z)}{\Gamma(1z)} =  \dfrac{\pi \cos \pi z}{\sin \pi z}\]
\[\Rightarrow \psi(z)  \psi(1z) =  \dfrac{\pi \cos \pi z}{\sin \pi z}\]
\[\Rightarrow \psi(1z)  \psi(z) = \pi \cot \pi z\]
where \(\psi\) denotes the Digamma function, which is the logarithmic derivative of the Gamma function.
Legendre duplication formula
\[2\psi(2s)=2\ln(2)+\psi(s)+\psi(s+1/2)\]
Consider \[\sqrt{\pi} \ \Gamma(2s)=2^{2s1} \Gamma(s)\Gamma(s+1/2)\] Taking log \[\ln(\sqrt{\pi})+\ln(\Gamma(2s))=(2s1)\ln(2)+\ln(\Gamma(s))+\ln(\Gamma(s+1/2))\] differentiating with respect to s we have \[2\psi(2s)=2\ln(2)+\psi(s)+\psi(s+1/2)\]
Applications in summations
Prove
\[\sum_{n=0}^\infty \dfrac{1}{n^2+1} =\dfrac{\pi+1}{2} + \dfrac{\pi}{e^{2\pi}1}\]
\[ S = \sum_{n=0}^{\infty} \frac{1}{n^2+1} = \frac{1}{2i} \sum_{n=0}^{\infty} \left( \frac{1}{ni}  \frac{1}{n+i} \right) \] We can write the summation as \[2iS= \sum_{n=1}^{\infty} \left( \frac{1}{n1i}  \frac{1}{n1+i} \right)=\sum_{n=1}^{\infty} \left( \frac{1}{n}  \frac{1}{n1+i} \right)\sum_{n=1}^{\infty} \left( \frac{1}{n}  \frac{1}{n1i} \right)\] By the series representation of the digamma function, this is just: \[2iS=\psi(i)\psi(i)=\psi(i)\psi(1i)\dfrac{1}{i}=\pi \cot(i\pi)+i=\pi i \coth(\pi)+i\] so \[S= \dfrac{1+\pi\coth(\pi)}{2}\] We can simplify this further to get \[S=\dfrac{\pi+1}{2} + \dfrac{\pi}{e^{2\pi}1}\]
Polygamma Functions
We can get many properties from this, for example by differentiating the series representation n times we have \[\psi_n(s)=(1)^{n+1}n! \sum_{k=1}^\infty \dfrac{1}{(k+s1)^{n+1}}=(1)^{n+1}n! \sum_{k=0}^\infty \dfrac{1}{(k+s)^{n+1}}=(1)^{n+1} n!\zeta(n+1,s)\] Where \(\zeta(n+1,s)\) is the hurwitz zeta function. putting s=1 we can get \(\psi_n(1)=(1)^{n+1} n!\zeta(n+1)\)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 \[\psi(s)=\sum_{n=0}^\infty \dfrac{\psi^{(n)}(1)(s1)^n}{n!}=\gamma+\sum_{n=1}^\infty \dfrac{\psi_n(1)(s1)^n}{n!}=\gamma\sum_{n=1}^\infty (1)^n\zeta(n+1)(s1)^n\\ \psi(s+1)=\gamma\sum_{n=1}^\infty \zeta(n+1)(s)^n\] We can differentiate the integral representation n times to get \[\psi_n(s+1)=\int_0^1 \dfrac{\ln^n(x) x^s}{x1}dx\] We can also do this to the functional equation to get \[\psi_n(s+1)=\psi_n(s)+(1)^nn! z^{n1}\]
Example Problems
Implementation in Mathematica
The digamma functions can be implemented in Mathematica as follows
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or
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