# 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!

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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.