Does $\int_0^1 \sum_{n=0}^{\infty}x e^{-nx}\;dx = \sum_{n=0}^{\infty}\int_0^1 x e^{-nx}\;dx$? Does $$\int_0^1 \sum_{n=0}^\infty x e^{-nx}\;dx = \sum_{n=0}^\infty \int_0^1 x e^{-nx}dx$$ ?
This exercise leaves me stumped. On the one hand, it seems the series $\sum_{n=0}^\infty xe^{-nx}$ is not uniformly convergent in $[0,1]$ (it equals $\frac{xe^x}{(e^x-1)}$ in $(0,1]$ and 0 in $x_0=0$, so it cannot be uniformly convergent since it is a series of continuous functions that converges to a non-continuous function). On the other hand, if this is the case, how do I deal with that... thing? 
Perhaps the series is uniformly convergent and I made a mistake?
Thanks!

You can use Fubini's theorem, but it seems overkill. Note that for all integer $N$ we have 
$$\sum_{n=0}^N\int_0^1xe^{-nx}dx=\int_0^1\sum_{n=0}^Nxe^{-nx}dx\leq \int_0^1\sum_{n=0}^{+\infty}xe^{-nx}dx,$$
so 
$$\sum_{n=0}^{+\infty}\int_0^1xe^{-nx}dx\leq \int_0^1\sum_{n=0}^{+\infty}xe^{-nx}dx.$$
For the reversed inequality, fix $\varepsilon>0$. Since $\sum_{n=0}^{+\infty}xe^{-nx}$ is integrable, we can find a $\delta>0$ such that $\int_0^{\delta}\sum_{n=0}^{+\infty}xe^{-nx}\leq \varepsilon$. And the series $\sum_{n=0}^{+\infty}xe^{-nx}$ is normally convergent on $[\delta,1]$. So we have 
\begin{align*}
\int_0^1\sum_{n=0}^{+\infty}xe^{-nx}dx&=\int_0^\delta\sum_{n=0}^{+\infty}xe^{-nx}dx+\int_\delta^1\sum_{n=0}^{+\infty}xe^{-nx}dx\\
&\leq\varepsilon +\int_\delta^1\sum_{n=0}^{+\infty}xe^{-nx}dx\\
&=\varepsilon +\sum_{n=0}^{+\infty}\int_\delta^1xe^{-nx}dx\\
&\leq \varepsilon +\sum_{n=0}^{+\infty}\int_0^1xe^{-nx}dx,
\end{align*}
and since $\varepsilon$ is arbitrary we can conclude the equality.
