TITLE: Is there a closed form for the series $\sum_{k=1}^\infty \frac{\ln(4k-3)}{(4k-3)}-\frac{\ln(4k-1)}{(4k-1)}?$
QUESTION [14 upvotes]: This is actually a question which was deleted by a previous user. I worked very hard on this question and became quite engrossed. The question was simply: Is there a closed form for the series $$\sum_{k=1}^\infty \frac{\ln(4k-3)}{(4k-3)}-\frac{\ln(4k-1)}{(4k-1)}?$$
I am submitting my work below as an answer and I'd like Math.SE to help me out here. There is likely something I've missed, and I would like some help on figuring out if there is a closed form.

REPLY [17 votes]: We can rewrite the given sum as
$$
\begin{align*}
- \sum_{n=0}^{\infty} (-1)^{n+1} \frac{\log(2n+1)}{2n+1} &= - \left. \sum_{n=0}^{\infty} (-1)^{n+1} \frac{\log(2n+1)}{(2n+1)^s} \right|_{s=1} \\
&= - \left. \frac{d}{ds} \right|_{s=1} \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^s} \\
&= - \beta'(1) \\
&= - \frac{\pi}{4} \Bigl\{\gamma + 2 \log 2 + 3 \log \pi - 4 \log\left[\Gamma(1/4)\right]\Bigr\},
\end{align*}
$$
where $\beta$ is the Dirichlet beta function. The last equality was found at MathWorld, which cites an OEIS entry (among other things), which in turn cites the MathWorld article.