$ \int_{1}^{n} \frac{x^t}{t} dt \underset{x \to 1^{-}}{\rightarrow} \int_{1}^{n} \frac{1}{t} dt$ $n$ is fixed.

$ x \in ]0,1[$

Coud you help me prove, without theorem of dominated convergence :
$$ \int_{1}^{n} \frac{x^t}{t} dt \underset{x \to 1^{-}}{\rightarrow} \int_{1}^{n} \frac{1}{t} dt$$

My attempt :
$ x^t = e^{ t \ln x}$ and we could use $ e^u \sim 1+u$  for $u \to 0$. If I do that I obtain a $o (t \ln x)$ and I cannot conclude.
 A: Hint:
$$\left|\int_1^ n \frac{x^t}{t} \, dt -  \int_1^ n \frac{1}{t} \, dt\right| \leqslant \int_1^n\left| \frac{x^t}{t}  -  \frac{1}{t}\right| \, dt,$$
and $(x,t) \mapsto x^t/t $ is uniformly continuous on $[1-\delta,1] \times [1,n]$.
A: By the MVT, for any fixed $t\ge 1$
$$\tag 1 x^t-1 = x^t-1^t = tc_x^{t-1}(x-1)$$
for some $c_x\in (x,1).$ The absolute value of the last expression is therefore bounded above by $n\cdot 1 \cdot |x-1|= n|x-1|.$
The difference of the two integrals in question is
$$\tag 2 \int_{1}^{n} \frac{x^t-1}{t} dt.$$
Take the absolute value, move the absolute values inside the integral, and then use the bound we found for $(1).$ This implies the absolute value of $(2)$ is no more than
$$n|x-1|\int_1^n \frac{1}{t}\,dt = n|x-1|\ln n.$$
This $\to 0$ as $x\to 1^-$ as desired.
A: It is enough to show that $x^t\xrightarrow{x\rightarrow1-}1$ uniformly in $t\in[1,n]$.
One my use  the mean value theorem: for all $|x-1|<\frac12$,
$$|x^t-1|=|e^{t\log(x)}-1|=e^u|t\log(x)|<e^{n\log(3/2)}n|\log(x)|$$
where  $u_x$ is some number between $0$ and $t\log(x)$
