Justification for moving a limit outside an indefinite integral Someone asked on another site about ways to evaluate $\int \ln x \ dx $ without using integration by parts.
My response was the following:
$$ \begin{align} \int\ln x \, dx & =\int\lim_{t\to 0}\frac{x^{t}-1}{t} \, dx \\ &=  \lim_{t\to 0}\frac{1}{t}\int (x^{t}-1) \, dx \\  &=\lim_{t\to 0}\frac{1}{t}\left(\frac{x^{t+1}}{t+1}-x\right)+C \\ &=\lim_{t\to 0}\frac{x^{t+1}-x(t+1)}{t(t+1)}+C \\ &=\lim_{t\to 0}\frac{x^{t+1}\ln x-x}{2t+1}+C  \\ &= x\ln x-x+C \end{align}$$
But I don't know how to justify moving the limit outside the indefinite integral.
 A: First of all, let's fix the interval of integration. We want to show that for any $0<a<b<\infty$, 
$$\int_a^b \ln x\,dx = \lim_{t\to 0}\int_a^b \frac{x^t-1}{t}\,dx \tag1$$
As nbubis said, it suffices to prove that we have uniform convergence. To this end, write $x=e^u$ and use the Taylor approximation 
$$|e^{tu}-1-tu|\le C(tu)^2 \tag2$$
for  sufficiently small $tu$. 
The important point is that $|tu|\le |t|\max(|\log a|, |\log b|)$, which allows us to rewrite (2) as 
$$|e^{tu}-1-tu|\le \widetilde Ct^2 \tag3$$
for sufficiently small $t$, with  $\widetilde C$ independent of $u$. Hence, 
$$
\sup_{x\in [a,b] } \left|\frac{x^{t}-1}{t} -\ln x\right|=
\sup_{u\in [\log a,\log b]} \left|\frac{e^{tu}-1}{t} -u\right| \le \widetilde Ct
$$
as desired.
A: I've seen it said that if the series of functions $f_n$ converges uniformly (not only pointwise!) to the function $f$, then you can write:
$$\lim_{n\to \infty} \int f_n dx =  \int \lim_{n\to \infty}f_n dx = \int fdx$$
I'll try and find the source, and add it here later.
