A tweak to an integral for $(s-1)\,\Gamma(s)\,\zeta(s)$. What is its value at $s=0$? The following integral:
$$(s-1)\,\Gamma(s)\,\zeta(s)={\int_{0}^{\infty}\!{u}^{s-1} \left( {\frac {u-1}{{{\rm e}^{u}}-1}}+{\frac {u}{ \left( {{\rm e}^{u}}-1 \right) ^{2}}} \right) \,{\rm d}u}\ \qquad \Re(s)>0$$
could be tweaked into:
$$f(s)={\int_{0}^{\infty}\!{u}^{s-1} \left( {\frac {u-1}{{({\rm e}^{u}}-1)^\frac12}}+{\frac {u}{ \left( {{\rm e}^{u}}-1 \right) ^{\frac32}}} \right) \,{\rm d}u}\ \qquad \Re(s)\ge0$$
and yields closed forms expressions for certain values:
\begin{align}
f\left(\frac12\right)&=2 \\
f(1)&= \pi \\
f(2)&= 6\,\pi\,\ln(2) \\
f(3)&= 10\,\pi\,\zeta(2)+20\,\pi\,\ln(2)^2 \\
f(4)&= 84\,\pi\,\zeta(3) +84\,\pi\,\ln(2)\,\zeta(2) +56\,\pi\,\ln(2)^3 \\
f(5)&=...
\end{align}
so, at integer values of $f$, the function can be expressed as a finite series of weighted $\zeta$-values. 
Question: 
f(0) also converges for this integral to: 1.869957636881892752... Curious whether this value could be expressed into other constants? (checked Plouffe's inverter and Mathematica, but no results from those). 
Added:
Made one step forward on $f(0)$, by splitting the associated integral:
$$\int_{0}^{\infty}\! \left( {\frac {1-\frac{1}{u}}{{({\rm e}^{u}}-1)^{\frac12}}}+{\frac {1}{ \left( {{\rm e}^{u}}-1 \right) ^{\frac32}}} \right) \,{\rm d}u$$
into:
$$\overbrace{\int_{0}^{\infty}\! \left( {\frac {1}{{({\rm e}^{u}}-1)^{\frac12}}} \right) \,{\rm d}u}^{\pi} \,\, + \,\, \overbrace{\int_{0}^{\infty}\! \left( {\frac {-\frac{1}{u}}{{({\rm e}^{u}}-1)^{\frac12}}}+{\frac {1}{ \left( {{\rm e}^{u}}-1 \right) ^{\frac32}}} \right) \,{\rm d}u}^{-1.2716350167...}$$
So, the question now boils down to whether a closed form for $-1.2716350167...$ exists.
 A: A partial answer. From the generalized binomial theorem we note that $$\int_{0}^{+\infty}\frac{u^{s-1}}{\left(e^{u}-1\right)^{1/2}}du=\Gamma\left(s\right)\sum_{k\geq0}\dbinom{-1/2}{k}\frac{\left(-1\right)^{k}}{\left(k+1/2\right)^{s}}$$ and using the relation $$2\sum_{k\geq0}\dbinom{-1/2}{k}\left(-1\right)^{k}ke^{-ku}=\frac{e^{-u}}{\left(1-e^{-u}\right)^{3/2}}$$ we note that $$\int_{0}^{+\infty}\frac{u^{s}}{\left(e^{u}-1\right)^{3/2}}du=\int_{0}^{+\infty}e^{-u/2}\frac{e^{-u}u^{s}}{\left(1-e^{-u}\right)^{3/2}}du=2\Gamma\left(s+1\right)\sum_{k\geq0}\dbinom{-1/2}{k}\frac{\left(-1\right)^{k}k}{\left(k+1/2\right)^{s+1}}$$ $$=2\Gamma\left(s+1\right)\left(\sum_{k\geq0}\dbinom{-1/2}{k}\frac{\left(-1\right)^{k}}{\left(k+1/2\right)^{s}}-\frac{1}{2}\sum_{k\geq0}\dbinom{-1/2}{k}\frac{\left(-1\right)^{k}}{\left(k+1/2\right)^{s+1}}\right).$$ So, if we define $$\zeta_{1/2}\left(s\right):=\sum_{k\geq0}\dbinom{-1/2}{k}\frac{\left(-1\right)^{k}}{\left(k+1/2\right)^{s}}$$ we can conclude that $$\int_{0}^{+\infty}u^{s-1}\left(\frac{u-1}{\left(e^{u}-1\right)^{1/2}}+\frac{u}{\left(e^{u}-1\right)^{3/2}}\right)du=2\Gamma\left(s+1\right)\zeta_{1/2}\left(s\right)-\Gamma\left(s\right)\zeta_{1/2}\left(s\right)=\color{red}{\Gamma\left(s\right)\zeta_{1/2}\left(s\right)\left(2s-1\right)}.$$ This is true for $\mathrm{Re}\left(s\right)>0$ and can be easily generalized. Now for $s=0$ numerical experiment shows that $$\int_{0}^{+\infty}u^{-1}\left(\frac{u-1}{\left(e^{u}-1\right)^{1/2}}+\frac{u}{\left(e^{u}-1\right)^{3/2}}\right)du=\lim_{s\rightarrow0^{+}}\left(\Gamma\left(s\right)\zeta_{1/2}\left(s\right)\left(2s-1\right)\right)$$ and I'm quite sure that $\zeta_{1/2}\left(s\right)$ has a closed form and it tends to $0$ as $s\rightarrow0^{+}$ but, at this moment, I'm not able to prove it.
UPDATE: The series like $\zeta_{1/2}\left(s\right)$ are known, in literature, as a generalization of the Hurwitz–Lerch zeta function $$\Phi_{\mu}^{*}\left(z,s,a\right):=\sum_{k\geq0}\frac{\left(\mu\right)_{k}}{k!}\frac{z^{k}}{\left(k+a\right)^{s}},$$ with $\mu\in\mathbb{C},\,a\in\mathbb{\mathbb{C}}\setminus\mathbb{Z}_{0}^{-},\,s\in\mathbb{C}\textrm{ if }\left|z\right|<1\textrm{ and }\mathrm{Re}\left(s-\mu+1\right)>1\textrm{ if }\left|z\right|=1.$ So in our case we have $$\int_{0}^{+\infty}u^{s-1}\left(\frac{u-1}{\left(e^{u}-1\right)^{1/2}}+\frac{u}{\left(e^{u}-1\right)^{3/2}}\right)du=\Gamma\left(s\right)\Phi_{-1/2}^{*}\left(-1,s,\frac{1}{2}\right)\left(2s-1\right)$$ and it is possible to show that $\Phi_{\mu}^{*}\left(z,s,a\right)$ is, in some sense, the fractional derivative of the classical Lerch zeta function. More precisely, we have $$\Phi_{\mu}^{*}\left(z,s,a\right)=\frac{1}{\Gamma\left(\mu\right)}\mathcal{D}_{z}^{\mu-1}\left(z^{\mu-1}\Phi\left(z,s,a\right)\right)$$ (see, for example, [1]) where $$\mathcal{D}_{z}^{-\alpha}\left(f\left(z\right)\right):=\frac{1}{\Gamma\left(\alpha\right)}\int_{0}^{z}\left(z-t\right)^{\alpha-1}f\left(t\right)dt,\,\mathrm{Re}\left(\alpha\right)>0$$ are the classical Riemann-Liouville integrals.
