For any natural number $a$, determine the value of $\lim_{n\to\infty}\int_1^ex^a(\log x)^ndx.$ Question: For any natural number $a$, determine the value of $$\lim_{n\to\infty}n\int_1^ex^a(\log x)^ndx.$$
My solution: Firstly let $$I_n=\int_1^ex^a(\log x)^ndx, \forall n\in\mathbb{N}.$$ Now substituting $\log x=t$ in $I_n$, we have $$I_n=\int_0^1e^{t(a+1)}t^ndt$$
Now since $$0\le t\le 1\\\implies 0\le t(a+1)\le a+1\\\implies e^0\le e^{t(a+1)}\le e^{a+1}\\\implies 1\le e^{t(a+1)}\le e^{a+1}\\\implies t^n\le e^{t(a+1)}t^n\le e^{a+1}t^n, \forall 0\le t\le 1.$$ 
Thus we have $$e^{t(a+1)}t^n\le e^{a+1}t^n, \forall 0\le t\le 1\implies I_n\le\int_0^1e^{a+1}t^ndt=\frac{e^{a+1}}{n+1}.$$ 
Now let us evaluate $I_n$ by integration by parts. Let $u=e^{t(a+1)}\implies du=(a+1)e^{t(a+1)}dt$ and $dv=t^ndt\implies v=\frac{t^{n+1}}{n+1}.$ 
Thus $$I_n=\frac{e^{a+1}}{n+1}-\frac{a+1}{n+1}I_{n+1}.$$
Now we have $$e^{t{(a+1)}}t^{n+1}\le e^{a+1}t^{n+1}, \forall 0\le t\le 1\\\implies I_{n+1}\le \int_0^1e^{a+1}t^{n+1}dt=\frac{e^{a+1}}{n+2}\\\implies -\frac{a+1}{n+1}I_{n+1}\ge -\frac{(a+1)e^{a+1}}{(n+1)(n+2)}\\\implies I_n=\frac{e^{a+1}}{n+1}-\frac{a+1}{n+1}I_{n+1}\ge \frac{e^{a+1}}{n+1}-\frac{(a+1)e^{a+1}}{(n+1)(n+2)}.$$
Thus we have $$\frac{e^{a+1}}{n+1}-\frac{(a+1)e^{a+1}}{(n+1)(n+2)}\le I_n\le \frac{e^{a+1}}{n+1}\\\implies \frac{n}{n+1}e^{a+1}-\frac{n(a+1)e^{a+1}}{(n+1)(n+2)}\le nI_n\le \frac{n}{n+1}e^{a+1}, \forall n\in\mathbb{N}.$$
Now $$\lim_{n\to\infty} \left(\frac{n}{n+1}e^{a+1}-\frac{n(a+1)e^{a+1}}{(n+1)(n+2)}\right)=e^{a+1}\text{ and }\lim_{n\to\infty}\frac{n}{n+1}e^{a+1}=e^{a+1}.$$ Thus by Sandwich Theorem, we have $$\lim_{n\to\infty}nI_n=e^{a+1}.$$
Is there any other method to solve this problem? 
 A: More generally one can prove that $$n\int_{0}^{1}x^nf(x)\,dx\to f(1)$$ as $n\to\infty $. We assume that $f$ is differentiable with a bounded derivative. There are other ways to prove the result with fewer assumptions on $f$ but we take the simpler approach with more assumptions on $f$.
Since $n/(n+1)$ tends to $1$ we can equivalently deal with $$(n+1)\int_{0}^{1}x^nf(x)\,dx$$ which can be expressed as $$\left. x^{n+1}f(x)\right|_{x=0}^{x=1}-\int_{0}^{1}x^{n+1}f'(x)\,dx$$ via integration by parts. The above equals $$f(1)-\int_{0}^{1}x^{n+1}f'(x)\,dx$$ and integral above does not exceed $$M\int_{0}^{1}x^{n+1}\,dx=\frac{M}{n+2}$$ in absolute value where $M$ is a bound for $|f'| $ on $[0,1]$. Thus the desired limit is $f(1)$ and for current question $f(1)=e^{a+1}$.
A: Indeed $$I_n=\int_0^1 e^{t(a+1)}t^ndt$$
Let $u=t^n$ then
$$ I_n=\frac{1}{n}\int_0^1 e^{(a+1)u^{1/n}}u^{1/n}du $$
$u\mapsto e^{(a+1)u^{1/n}}u^{1/n} $ converges pointwise toward $u\mapsto e^{(a+1)}$ and $\forall u\in[0,1],|e^{(a+1)u^{1/n}}u^{1/n}|\leqslant e^{a+1}$ which is integrable on $[0,1]$. By dominated convergence theorem :
$$ \lim\limits_{n\rightarrow +\infty}nI_n=e^{a+1} $$
