Q:

Calculate the infinite sum $\sum_{1}^\infty \frac{\log{n}}{2n-1}$

I would like to calculate an asymptotic expansion for the following infinite sum:
$$\displaystyle \sum_{1}^N \frac{\log{n}}{2n-1}$$
when $N$ tends to $\infty$. I found that the asymptotic expansion for this partial sum is
$$ \displaystyle \frac{\log^2{N}}{4}+0.2282077...$$
and I would be interested in writing this constant term in an explicit way.  By similarity with other sums of the same type, I believe that an explicit expression should probably include $\displaystyle \gamma$ and the first Stieltjes constant $\displaystyle \gamma_1$, but I was not able to find it. 

A:

An incomplete answer but I hope it may clarify some things.  
Let us introduce
\begin{align*}
S_N&=\sum_{n=1}^N\frac{\ln n}{2n},\qquad 
\bar{S}_N=\sum_{n=1}^N\frac{\ln n}{2n-1}. 
\end{align*}
Now make two observations:

the sum $\displaystyle C_N:=\bar S_N-S_N=\sum_{n=1}^N\frac{\ln n}{2n\left(2n-1\right)}$ converges as $N\to \infty$.
the asymptotics of $S_N$ is known:
$$S_N=\frac14\ln^2N+\frac12\gamma_1+o\left(1\right).$$

Thus the constant we are looking for is nothing but
$$\frac12\gamma_1+C_{\infty}=\frac12\gamma_1+\sum_{n=1}^{\infty}\frac{\ln n}{2n\left(2n-1\right)}.$$
However the evaluation of the remaining infinite sum looks complicated (yet much simpler than the sum involving zeta values from another answer).