Closed form answer by Wolfram Alpha

Question

I used Wolfram Alpha to estimate the sum of multiple Harmonic series. I was quite surprised that Wolfram Alpha came up with a closed form solution

$$ \psi\big(\tfrac{3}{16}\big) + \psi\big(\tfrac{5}{16}\big) + \psi\big(\tfrac{11}{16}\big) + \psi\big(\tfrac{13}{16}\big) - \psi\big(\tfrac{1}{16}\big) - \psi\big(\tfrac{7}{16}\big) - \psi\big(\tfrac{9}{16}\big) - \psi\big(\tfrac{15}{16}\big) = \\8 \sqrt{2} \log\big(\cot\big(\tfrac{\pi}{8}\big)\big) $$ where $\psi$ is the digamma function.

I am interested in knowing how the equality could be derived.

Answer 1

Using $$\psi(z)=-\gamma+\sum_{n=0}^\infty\left(\frac1{n+1}-\frac1{n+z}\right)$$ gives your expression as $$16\sum_{n=0}^\infty\left(\frac1{16n+1}-\frac1{16n+3}-\frac1{16n+5}+\frac1{16n+7}+\frac1{16n+9}-\frac1{16n+11}-\frac1{16n+13}+\frac1{16n+15}\right).$$ That is $\int_0^1f(x)\,dx$ where $$f(x)=16\sum_{n=0}^\infty(x^{16n}-x^{16n+2}-x^{16n+4}+x^{16n+6}+x^{16n+8}-x^{16n+10}-x^{16n+12}+x^{16n+14}) =16\frac{1-x^2-x^4+x^6}{1-x^8}.$$ One now just has to do the integral of a rational function. Computer algebra systems are good at this, even Wolfram's.

Answer 2

It is a straightfoward consequence of the Gauss Digamma theorem, which on its turn is a consequence of the reflection formula for the $\Gamma$ function: $$ \Gamma(s)\Gamma(1-s)=\frac{\pi}{\sin(\pi s)},\qquad \log\Gamma(s)+\log\Gamma(1-s) = \log \pi-\log\sin(\pi s), $$ $$\text{(Reflection formula for the Digamma function)}\qquad \psi(s) - \psi(1-s) = -\pi\cot(\pi s) . $$