Finiteness of $\int_0^1 \left(\sum_{n=1}^\infty \frac{n^\alpha e^{- t n^\alpha}}{1 - e^{- t n^\alpha}} \right)^{1/2} \, \mathrm{d}t$ Let $\alpha > 1$.
My question is: is $\int_0^1 \left(\sum_{n=1}^\infty \frac{n^\alpha e^{- t n^\alpha}}{1 - e^{- t n^\alpha}} \right)^{1/2} \, \mathrm{d}t$ finite ?
For $t \in ]0, 1]$, the series $\sum_{n=1}^\infty \frac{n^\alpha e^{- t n^\alpha}}{1 - e^{- t n^\alpha}}$ is convergent (using the ratio test for example). 
I don't think there is a direct way to compute its sum, so I looked for an upper bound, to no avail. The square root also prevents me from switching integral and series.
 A: As you've said, the series converges for $t\in~\!]0,1]$. Thus the remaining issue is to check the asymptotic behavior as $t\to0$ in the series. Note that we have:
$$\sum_{n=1}^\infty\frac{n^\alpha e^{-tn^\alpha}}{1-e^{-tn^\alpha}}\sim\int_1^\infty\frac{x^\alpha e^{-tx^\alpha}}{1-e^{-tx^\alpha}}~\mathrm dx$$
By substituting $tx^\alpha\mapsto x$ we get
$$\sum_{n=1}^\infty\frac{n^\alpha e^{-tn^\alpha}}{1-e^{-tn^\alpha}}\sim\frac1{\alpha t^{(1+\alpha)/\alpha}}\int_t^\infty\frac{\sqrt[\alpha]xe^{-x}}{1-e^{-x}}~\mathrm dx$$
which thus reduces the problem down to testing the convergence of
$$\int_0^1\frac1{t^{(1+\alpha)/2\alpha}}~\mathrm dt$$
A: As Beautiful Art pointed out, it is enough to find the behaviour about $t=0$. Defining $$f(t)=\sum_{n=1}^\infty \frac{n^\alpha}{{\rm e}^{t n^\alpha}-1}$$ you can calculate the Mellin-Transform
$${\cal M}_f(s) = \Gamma(s)\zeta(s)\zeta\left(\alpha(s-1)\right)$$ which converges whenever $\Re(s)>1+\frac{1}{\alpha}$. The inverse then becomes
$$f(t)=\frac{1}{2\pi i} \int_{\sigma - i\infty}^{\sigma + i\infty} \Gamma(s)\zeta(s)\zeta\left(\alpha(s-1)\right) t^{-s} \, {\rm d}s$$
where $\sigma>1+\frac{1}{\alpha}$. This can be used to calculate the series about $t=0$ by shifting the contour to the left. You can assume $\alpha$ not an integer, and then the integrand has poles at $s=1+\frac{1}{\alpha},1,0,-1,-3,-5,-7,...$. So shifting the contour to $\sigma<1+\frac{1}{\alpha}$ we pick up the residue at $s=1+\frac{1}{\alpha}$ and the first terms become
$$f(t)=\frac{\Gamma\left(1+\frac{1}{\alpha}\right)\zeta\left(1+\frac{1}{\alpha}\right)}{\alpha \, t^{1+\frac{1}{\alpha}}} - \frac{1}{2t} - \frac{\zeta(-\alpha)}{2} + \frac{1}{2\pi i} \int_{\sigma - i\infty}^{\sigma + i\infty} \Gamma(s)\zeta(s)\zeta\left(\alpha(s-1)\right) t^{-s} \, {\rm d}s \, ,$$ where now $\sigma<0$.
