TITLE: This formula is on Wolfram Mathworld but I cannot find it anywhere else online?
QUESTION [5 upvotes]: $$\sum_{k=1}^{\infty}\ln\left[ \frac{(4k+1)^{1/(4k+1)^{n}}}{(4k-1)^{1/(4k-1)^{n}}} \right] = -\beta'(n)$$. 
Where $\beta$ is the Dirichlet Beta Function and $n$ is a positive integer. 
I cannot find this cited anywhere nor values of the beta function derivative apart from at $-1,0,1$. How can I go about finding these things, I have searched googled and arxiv.

REPLY [5 votes]: It is known that (uniformly and absolutely)
$$
\beta(n)=\sum^{\infty}_{k=1}\frac{\chi_4(k)}{k^n}\textrm{, }Re(n)>1.
$$
Hence writing $1/k^n=e^{-n\log(k)}$, we have easily
$$
-\beta'(n)=\sum^{\infty}_{k=2}\frac{\chi_4(k)\log(k)}{k^n}.
$$
But 
$$
\chi_4(k)=\left\{
\begin{array}{cc}
   0\textrm{ if }k\equiv 0 (mod)4\\
 1\textrm{ if }k\equiv 1 (mod)4\\
 0\textrm{ if }k\equiv 2 (mod)4\\
 -1\textrm{ if }k\equiv 3 (mod)4
\end{array}
\right\}.
$$
Hence for $Re(n)>1$, we have
$$
-\beta'(n)=\sum^{\infty}_{k=1}\frac{\log(4k+1)}{(4k+1)^n}-\sum^{\infty}_{k=1}\frac{\log(4k-1)}{(4k-1)^n}=\sum^{\infty}_{k=1}\log\left(\frac{(4k+1)^{1/(4k+1)^{n}}}{(4k-1)^{1/(4k-1)^{n}}}\right).
$$
QED