The polygamma function of order \( (m-1) \) is defined as the \( m \)-th derivative of \( \ln \Gamma(x) \):
\[ \psi^{(m-1)}(x) = \frac{d^m}{dx^m} \ln \Gamma(x) \]Reflection Relation
The reflection relation between polygamma functions is given by:
\[ (-1)^m \psi^{(m)}(1-z) - \psi^{(m)}(z) = \pi \frac{\mathrm{d}^m}{\mathrm{d} z^m} \cot(\pi z) \]It can be easily derived by differentiating the reflection relation of the gamma function \( m \) times:
\[ \frac{d^m}{dz^m} \ln\left(\Gamma(z) \Gamma(1-z)\right) = \frac{d^m}{dz^m} \ln\left(\frac{\pi}{\sin(\pi z)}\right) \]Dirichlet Beta Function
The Dirichlet beta function is defined as:
\[ \beta(s) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^s} \]If we separate terms like this:
\[ \beta(s) = \sum_{n=0}^\infty \frac{1}{(4n + 1)^s} - \sum_{n=0}^\infty \frac{1}{(4n + 3)^s} \]Then, it can be expressed in terms of the Hurwitz zeta function and polygamma function as well:
\[ \beta(s) = 4^{-s} \left( \zeta\left(s,\frac{1}{4}\right)-\zeta\left(s,\frac{3}{4}\right) \right) \]Polygamma Series Expansion
For the polygamma function, let us recall(derive) the series expansion:
\[ \psi^{(m)}(x) = \frac{d^{(m+1)}}{dx^{(m+1)}} \ln \Gamma(x) \]differentiating:
\[ \psi^{(m)}(x) = \frac{d^{(m+1)}}{dx^{(m+1)}} \sum_{k=1}^{x} \frac{1}{k} \]adding and subtracting \( \sum_{k=x+1}^{\infty} \frac{1}{k} \) of eqn:
\[ \psi^{(m)}(x) = \frac{d^{(m+1)}}{dx^{(m+1)}} \left(\sum_{k=1}^{x} \frac{1}{k}+ \sum_{k=x+1}^{\infty} \frac{1}{k} - \sum_{k=x+1}^{\infty} \frac{1}{k}\right) \]now:
\[ \psi^{(m)}(x) =\frac{d^{(m+1)}}{dx^{(m+1)}} \left(\sum_{k=1}^{\infty} \frac{1}{k} + \sum_{k=1}^{\infty} \frac{1}{x+k}\right) \]now we can see that the first term inside brackets will be treated as constant because it's converging, and we can easily prove it by the integral test or also approximate. For more info, see:
By differentiating it \(m\) times, we get the following expression:
\[ \psi^{(m)}(z) = (-1)^{m+1}\, m! \sum_{k=0}^\infty \frac{1}{(z+k)^{m+1}} \]Now, considering specific values for \(z = \frac{1}{4}\) and \(z = \frac{3}{4}\):
\[ \psi^{(m)}\left(\frac{1}{4}\right) = (-1)^{m+1}\, m! \sum_{k=0}^\infty \frac{1}{\left(\frac{1}{4}+k\right)^{m+1}} \] \[ \psi^{(m)}\left(\frac{3}{4}\right) = (-1)^{m+1}\, m! \sum_{k=0}^\infty \frac{1}{\left(\frac{3}{4}+k\right)^{m+1}} \]Then, subtracting these two equations gives us the relation:
\[ \beta(s) = \frac{1}{2^s} \sum_{n=0}^\infty\frac{(-1)^{n}}{\left(n+\frac{1}{2}\right)^{s}} = \frac1{(-4)^s(s-1)!}\left[\psi^{(s-1)}\left(\frac{1}{4}\right)-\psi^{(s-1)}\left(\frac{3}{4}\right)\right] \]If \(s\) is odd, then:
\[ \beta(2m+1) = \frac{1}{2^{2m+1}} \sum_{n=0}^\infty\frac{(-1)^{n}}{\left(n+\frac{1}{2}\right)^{2m+1}} = \frac1{(-4)^{2m+1}(2m)!}\left[\psi^{(2m)}\left(\frac{1}{4}\right)-\psi^{(2m)}\left(\frac{3}{4}\right)\right] \]By using the reflection formula of the polygamma function, we obtain the value of:
\[ (-1)^m \psi^{(m)}\left(1-\frac{1}{4}\right) - \psi^{(m)}\left(\frac{1}{4}\right) = \left. \frac{\pi \, \mathrm{d}^{(m)}}{\mathrm{d} z^{(m)}} \cot(\pi z) \right|_{z=\frac{1}{4}} \]Which gives us the formula:
\[ \beta(2m+1) = \frac{1}{2^{2m+1}} \sum_{n=0}^\infty\frac{(-1)^{n}}{\left(n+\frac{1}{2}\right)^{2m+1}} = \frac{1}{(4)^{2m+1}(2m)!}\Bigg|_{z=\frac{1}{4}} \frac{\pi \, \mathrm{d}^{(2s)}}{\mathrm{d} z^{(2s)}} \cot(\pi z) \]Then:
\[ \beta(5) = \frac{5\pi^5}{2(2^5)(4)!} \]Booooooooooom!