Does $\displaystyle \int_{0}^{\infty} \frac{te^{-t}}{1+e^{-t}} \ dt = \int_{0}^{\infty} t \sum_{k=1}^{\infty} (-1)^{k-1} e^{-kt} \ dt $ just because $\displaystyle e^{-t} <1$ for $\displaystyle 0 <t<\infty $?

Or is it also because the geometric series converges uniformly?

And does it matter that $\displaystyle t=0$ is not in the interval?

What about $\displaystyle \int^{\infty}_{0} \sum_{k=1}^{\infty} \frac{(-1)^{k}}{k} e^{-kt} \ dt = \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k} \int_{0}^{\infty} e^{-kt} \ dt $?

Does it suffice to show that $\displaystyle \sum^{\infty}_{k=1} \frac{(-1)^{k-1}}{k} e^{-kt} $ converges uniformly on $\displaystyle 0<t<\infty$? Is there a simpler justification?

I got stuck using the Weierstrass M-test.

Let $\displaystyle f_{n}(x) = \frac{(-1)^{k-1}}{k} \ e^{-kt} $

then $\displaystyle |f_{n}(x)| = \frac{e^{-kt}}{k} < 1/k$ for $\displaystyle 0<t<\infty $

But $\displaystyle \sum \frac{1}{k} $ diverges. So that didn't work.