Summation $S=\sum_{k=1}^{\infty}\left(\frac{\left(-1\right)^{k-1}}{k+\frac{1}{2}}\ln\left(1-\frac{1}{\left(k+\frac{1}{2}\right)^{2}}\right)\right)$

Question

$$S=\sum_{k=1}^{\infty}\left(\frac{\left(-1\right)^{k-1}}{k+\frac{1}{2}}\ln\left(1-\frac{1}{\left(k+\frac{1}{2}\right)^{2}}\right)\right)$$

Let, $$f(a)=\sum_{k=1}^{\infty}\left(\frac{\left(-1\right)^{k}}{2k+1}\ln\left(2k+a\right)\right)$$ then S can be represented as follows (after expanding everything): $$S=4f(1)-2f(-1)-2f(3)$$

Although I am not sure, but $f(1)$ seems to have a closed form as follows: $$-\frac{1}{4}\gamma_{1}\left(\frac{3}{4}\right)+\frac{1}{4}\gamma_1\left(\frac{5}{4}\right)+\ln\left(2\right)\left(\frac{\pi}{2}-2\right)$$ where, $\gamma_n$ is the $n$th Stieltjes Constant.
I did get the answer for $f(1)$ without the Stieltjes Consants from one of my friends as follows: $$-\frac{\pi}{4}\left(\gamma+2\ln2+3\ln\left(\pi\right)-4\ln\left[\Gamma\left(\frac{1}{4}\right)\right]\right)$$ He mentioned it as "Dirichlet Beta Derivative Evaluated at 1".

As a side note, this would imply: $$\gamma_1\left(\frac{5}{4}\right)-\gamma_1\left(\frac{3}{4}\right)=-\pi \gamma-4\pi\ln2-3\pi\ln\left(\pi\right)+4\pi\ln\left[\Gamma\left(\frac{1}{4}\right)\right]+8\ln\left(2\right)$$

I have no idea about $f(-1)$ and $f(3)$.
Also, I have no reason to believe a Closed Form Exists.

Edit: Using another approach as follows:

Using Series Expansion for $\ln(1-x)$ and rearranging the Summation: $$S=\sum_{r=1}^{\infty}\frac{2^{1+2r}}{r}\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k}}{\left(2k+1\right)^{1+2r}}$$

And it seems that: Infinite Series $\sum\limits_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}$ $$\sum_{k=0}^{\infty}\frac{\left(-1\right)^{k}}{\left(2k+1\right)^{1+2r}}=\frac{\left(-1\right)^{r}E_{2r}\pi^{2r+1}}{2^{2r+2}\left(2r\right)!}$$ where, $E_{2r}$ are the Euler Numbers.

$$S=\lim_{\alpha\to\infty}\sum_{r=1}^{\alpha}\frac{2^{1+2r}}{r}\left(\frac{\left(-1\right)^{r}E_{2r}\pi^{2r+1}}{2^{2r+2}\left(2r\right)!}-1\right)$$ Though this probably made it tougher.

Answer 1

Too long for a comment

Using the Frullani's integral $\displaystyle\ln n=\int_0^\infty\frac{e^{-t}-e^{-nt}}tdt$

$$S=\sum_{k=1}^{\infty}\left(\frac{\left(-1\right)^{k-1}}{k+\frac{1}{2}}\ln\left(1-\frac{1}{\left(k+\frac{1}{2}\right)^{2}}\right)\right)=\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k+\frac12}\ln\frac{(k-\frac12)(k+\frac32)}{(k+\frac12)^2}$$ $$=\int_0^\infty\frac{2e^{-\frac t2}-e^{-\frac{3t}2}-e^{\frac t2}}t\left(\sum_{k=1}^\infty\frac{(-1)^{k-1}e^{-kt}}{k+\frac12}\right)dt$$ Now, using $$2\sum_{k=1}^\infty\frac{(-1)^{k-1}}{2k+1}e^{-kt}=2i\sum_{k=1}^\infty (i)^{2k+1}\frac{e^{-(2k+1)\frac t2}e^\frac t2}{2k+1}=2+ie^\frac t2\ln\frac{1+ie^{-\frac t2}}{1-ie^{-\frac t2}}$$ $$=2-2e^\frac t2\Im\ln(1+ie^{-\frac t2})=2-2e^\frac t2\Im\ln\left(\sqrt{1+e^{-t}}e^{i\arctan e^{-\frac t2}}\right)=2-2e^\frac t2\arctan e^{-\frac t2}$$ Therefore, the sum can be presented as an integral in the form $$S=2\int_0^\infty\frac{1-e^\frac t2\arctan e^{-\frac t2}}t\left(2e^{-\frac t2}-e^{-\frac{3t}2}-e^\frac t2\right)dt$$ $$=-8\int_0^\infty\frac{e^{-t}-\arctan(e^{-t})}t\sinh^2t\,dt$$

Answer 2

Starting from @Svyatoslav's solution $$S=-8\int_0^\infty\frac{e^{-t}-\arctan(e^{-t})}t\,\sinh^2(t)\,dt$$

Using $$\sinh^2(t)=\sum_{n=1}^\infty \frac{2^{2 n-1}}{(2 n)!}\, t^{2 n}$$ $$J_n=\int_0^\infty t^{2 n-1} \left(e^{-t}-\tan ^{-1}\left(e^{-t}\right)\right)\,dt$$

$$J_n=\frac{1}{2} (2-i \text{Li}_{2 n+1}(-i)+i \text{Li}_{2 n+1}(i))\, \Gamma (2 n)$$

$$S=-\sum_{n=1}^\infty \frac {4^n}n (2-i \text{Li}_{2 n+1}(-i)+i \text{Li}_{2 n+1}(i)) $$

But, using Dirichlet beta and eta functions $$i (\text{Li}_{2 n+1}(i)-\text{Li}_{2 n+1}(-i))=$$ $$2^{-(4 n+1)} \left(\zeta \left(2 n+1,\frac{3}{4}\right)-\zeta \left(2 n+1,\frac{1}{4}\right)\right)$$

$$\color{blue}{S=-\sum_{n=1}^\infty\frac{4^{2 n+1}+\zeta \left(2 n+1,\frac{3}{4}\right)-\zeta \left(2 n+1,\frac{1}{4}\right)}{4^{2 n+1}\, n }}$$

which converges very fast (the summand being more or less $e^{-(n+1)}$)

Now, I am stuck.

Edit

May be shorter but for the same result $$f(a)=\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k}}{2k+1}\log\left(2k+a\right)$$ $$f'(a)=\sum_{k=1}^{\infty}\frac{(-1)^k}{(2 k+1)(2k+a)}=\frac{2 \Phi \left(-1,1,\frac{a+2}{2}\right)-(4-\pi)}{4 (a-1)}$$