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.

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 $\sigma<0$.
