Asymptotic behaviour of sum of decreasing definite integrals I would like to calculate:
\begin{equation*}g(K, T) = \displaystyle \sum_{k=1}^{K} \sum_{t = 1}^{T} \int_{0}^{1} \left(1 - z^k\right)^t \, dz.
\end{equation*}
If no closed form solution exists, I would like to find a tight upper bound. Trivially we have
\begin{equation*}
g(K,T) = O(KT),
\end{equation*} 
but we can do better than this. One can show that 
\begin{equation*}\displaystyle \int_{0}^{1} \left(1 - z^k\right)^t \, dz < \left(1 - \left(\frac{1}{2}\right)^\frac{1}{t}\right)^\frac{1}{k},
\end{equation*}
and further that 
\begin{equation*}
1 - \left(\frac{1}{2}\right)^{\frac{1}{t}}  \le \frac{1}{t},
\end{equation*}
which can be combined and fiddled with to provide the improved bound,
\begin{equation*}
g(K,T) = O(KT^{1-\frac{1}{K}}).
\end{equation*}
Can we do better than this?

The integral over $z$ can be reduced to a Beta function or Euler Integral of the First Kind ${\rm B}\left(x,y\right)$:
$$
\int_{0}^{1}\left(1 - z^{k}\right)^{t}\,{\rm d}z
=
{1 \over k}{\rm B}\left({1 \over k}, t + 1\right)
$$
For the time being, we don't see a further 'nice' reduction.
