$\lim_{x\to 0^+}\sum_{n=1}^\infty \frac{\chi(n)}{n} e^{-nx}$

In my answer to a question on this site I googled for a reference and found this. The answers of the user reuns of the latter linked question goes to great length to show that$$\lim_{x\to 0^+}f(x)=L(1,\chi)$$where$$f(x ) =\sum_{n=1}^\infty \frac{\chi(n)}{n} e^{-nx}.$$But since the series of $f$ converges uniformly on $[0,\infty)$ this is trivial. Also the statement $$"\Gamma(s)L(s+1,\chi)-f(0) \Gamma(s) = \int_0^\infty (f(x)-f(0) e^{-x})x^{s-1}dx$$converges and is analytic for $\Re(s) > -1$" needs something even stronger? Something like that the derivative of $f(x)$ is bounded on $(0,\epsilon)$. But to show this one would first show the uniform convergence of $f$... So the problem is reduced to a (slightly) harder problem?

So my question is if I am missing something or if the linked answer is rather confused and bad?

Why the downvote? Please elaborate if I am missing something!