Explicit formula for the sum of an infinite series with Chebyshev's $U_k$ polynomials

Question

$$\sum _{k=0}^{\infty } \frac{U_k(\cos (\text{k1}))}{k+1}=\frac{1}{2} i \csc (\text{k1}) \left(\log \left(1-e^{i \text{k1}}\right)-\log (i \sin (\text{k1})-\cos (\text{k1})+1)\right)$$ where $U_k$ is the Chebyshev $U$ polynomial $$\sum _{k=0}^{\infty } \frac{U_k(\cos (\text{k1}))}{(k+2) \left(k-\frac{1}{2}\right)}=\frac{2 \left(\log \left(1-e^{i \text{k1}}\right)+2 \left(1-\sqrt{e^{-i \text{k1}}} \tanh ^{-1}\left(\sqrt{e^{-i \text{k1}}}\right)\right)-e^{2 i \text{k1}} \left(\log \left(1-e^{-i \text{k1}}\right)+2 \left(1-\sqrt{e^{i \text{k1}}} \tanh ^{-1}\left(\sqrt{e^{i \text{k1}}}\right)\right)\right)\right)}{5 \left(-1+e^{2 i \text{k1}}\right)}$$ Any idea where it came from?

Answer 1

The first should come from $$\sum_{k=1}^\infty \dfrac{\sin(kt)}{k} = -i/2 \left( -\ln \left( 1-{{\rm e}^{it}} \right) +\ln \left( 1-{ {\rm e}^{-it}} \right) \right)$$ which you get by expressing $\sin(kt)$ as a complex exponential and using the Maclaurin series for $\ln$ (with Dirichlet's test showing convergence and Abel's theorem identifying the value)

The second similarly, using $$\sum _{k=0}^{\infty }{\frac {{t}^{k}}{ \left( k+2 \right) \left( k-1/ 2 \right) }}={\frac {2-4\,t}{5t}}-\dfrac{2}{5}\,\sqrt {t} \left( \ln \left( 1-\sqrt {t} \right) -\ln \left( 1+\sqrt {t} \right) \right) +{\frac {2\ln \left( -t+1 \right) }{5{t}^{2}}} $$