Find $\lim_{n\to\infty} n\int_1^e x^a(\log_ex)^ndx$ Find $\lim_{n\to \infty} n\int_1^e x^a(\log_ex)^ndx$
My idea: The reduction formula by integrating by parts doesn't seem to help so I substituted $\log_ex=t$ which gives
$\int e^{t(a-1)}t^ndt$. I'm not sure how to proceed further.
 A: Substituting $x=e^t$ and integrating by parts give
$$
n\int_1^e {x^a \log ^n xdx}  = n\int_0^1 {e^{(a + 1)t} t^n dt}  \\ =(n+1)\int_0^1 {e^{(a + 1)t} t^n dt}- \int_0^1 {e^{(a + 1)t} t^n dt} \\ = e^{a + 1}  - (a + 1)\int_0^1 {e^{(a + 1)t} t^{n + 1} dt}  - \int_0^1 {e^{(a + 1)t} t^n dt} .
$$
Now
$$
\left| {(a + 1)\int_0^1 {e^{(a + 1)t} t^{n + 1} dt} } \right| \le \left| {a + 1} \right|\max(1,e^{a + 1})  \int_0^1 {t^{n + 1} dt} \\ = \left| {a + 1} \right|\max(1,e^{a + 1})  \frac{1}{{n + 2}}
$$
and
$$
0 < \int_0^1 {e^{(a + 1)t} t^n dt}  \le \max(1,e^{a + 1}) \int_0^1 {t^n dt}  = \max(1,e^{a + 1})  \frac{1}{n + 1}.
$$
Consequently, the limit in question is $e^{a+1}$.
A: Here is an approach that uses only Bernoulli's Inequality and the fact that $1+u\le e^u$.
First notice that for $x\in(0,1)$, $x^{1/(n+1)}\le1$ and
$$
\begin{align}
x^{1/(n+1)}
&=(1+(1-\sqrt{x})/\sqrt{x})^{-2/(n+1)}\tag1\\[6pt]
&\ge1-\frac2{n+1}\frac{1-\sqrt{x}}{\sqrt{x}}\tag2\\
&=1-\frac2{\sqrt{x}(n+1)}+\frac2{n+1}\tag3\\
&\ge1-\frac2{\sqrt{x}(n+1)}\tag4
\end{align}
$$
Explanation:
$(1)$: algebra
$(2)$: Bernoulli's Inequality
$(3)$: algebra
$(4)$: $\frac2{n+1}\gt0$
We will use $[a,b]$ to represent a number between $a$ and $b$.
$$
\begin{align}
\lim_{n\to\infty}n\int_1^ex^a\log(x)^n\,\mathrm{d}x
&=\lim_{n\to\infty}n\int_0^1e^{(a+1)x}x^n\,\mathrm{d}x\tag5\\
&=\lim_{n\to\infty}\frac{n}{n+1}\int_0^1e^{(a+1)x}\,\mathrm{d}x^{n+1}\tag6\\
&=\lim_{n\to\infty}\frac{n}{n+1}\int_0^1e^{(a+1)x^{1/(n+1)}}\,\mathrm{d}x\tag7\\
&=\lim_{n\to\infty}\int_0^1e^{(a+1)\left[1-\frac2{\sqrt{x}(n+1)},1\right]}\,\mathrm{d}x\tag8\\
&=e^{a+1}\lim_{n\to\infty}\int_0^1e^{(a+1)\left[-\frac2{\sqrt{x}(n+1)},0\right]}\,\mathrm{d}x\tag9\\
&=e^{a+1}\lim_{n\to\infty}\int_0^1\left[1-\frac2{\sqrt{x}}\frac{a+1}{n+1},1\right]\,\mathrm{d}x\tag{10}\\
&=e^{a+1}\lim_{n\to\infty}\left[1-4\frac{a+1}{n+1},1\right]\tag{11}\\[9pt]
&=e^{a+1}\tag{12}
\end{align}
$$
Explanation:
$\phantom{1}(5)$: substitute $x\mapsto e^x$
$\phantom{1}(6)$: $x^n\,\mathrm{d}x=\frac1{n+1}\,\mathrm{d}x^{n+1}$
$\phantom{1}(7)$: substitute $x\mapsto x^{1/(n+1)}$
$\phantom{1}(8)$: evaluate the limit of $\frac{n}{n+1}$ and apply $(4)$
$\phantom{1}(9)$: pull the factor of $e^{a+1}$ out front
$(10)$: apply $1+u\le e^u$ to the lower bound
$(11)$: integrate the bounds
$(12)$: evaluate the limit
A: 
I thought it might be instructive to present an approach that appeals to the Dominated Convergence Theorem.  To that end we proceed.


Let $f_n(a)$ be given by
$$f_n(a)=n\int_1^e x^a\log^n(x)\,dx\tag1$$
Enforcing the substitution $x\mapsto e^{1-x/n}$ in $(1)$ reveals
$$\begin{align}
f_n(a)&=e^{a+1}\int_0^\infty \xi_{[0,n]}(x)\,\,e^{-(a+1)x/n}\left(1-\frac xn\right)^n\,dx\tag1
\end{align}$$
where $\xi_{[0,n]}(x)$ is the indicator function.
Inasmuch as $\left|\xi_{[0,n]}(x)\,\,e^{-(a+1)x/n}\left(1-\frac xn\right)^n\right|\le \max(1,e^{-(a+1)})e^{-x}$ and $\int_0^\infty \max(1,e^{-(a+1)})e^{-x}\,dx<\infty$, the Dominated Convergence Theorem guarantees that
$$\begin{align}
\lim_{n\to\infty}f_n(a)&=e^{a+1}\int_0^\infty \lim_{n\to \infty}\left(\xi_{[0,n]}(x)\,\,e^{-(a+1)x/n}\left(1-\frac xn\right)^n\right)\,dx\\\\
&=e^{a+1}\int_0^\infty e^{-x}\,dx\\\\
&=e^{a+1}
\end{align}$$
And we are done!
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{\lim_{n\to \infty}\bracks{n\int_{1}^{\expo{}} \!\! x^{a}\ln^{n}\pars{x}\,\dd x}}:\ {\Large ?}}$.

\begin{align}
&\bbox[5px,#ffd]{\lim_{n\to \infty}\bracks{n\int_{1}^{\expo{}} \!\! x^{a}\ln^{n}\pars{x}\,\dd x}}
\,\,\,\stackrel{x\ =\ \expo{\large t}}{=}\,\,\,
\lim_{n\to \infty}\bracks{%
n\int_{0}^{1}\expo{\pars{a + 1}\, t}\, t^{n}\,\dd t}
\\[5mm] \stackrel{-\pars{a + 1}t\ \mapsto\ t}{=}\,\,\, &
\lim_{n\to \infty}\bracks{%
{n \over \pars{-a - 1}^{n + 1}}\int_{0}^{-a - 1}t^{n}\expo{-t}
\,\dd t}
\\[5mm] = &\
-\lim_{n\to \infty}\braces{%
n\,\pars{-a - 1}^{-n - 1}\bracks{%
\int_{-a - 1}^{\infty}t^{n}\expo{-t}\,\dd t -
\int_{0}^{\infty}t^{n}\expo{-t}\,\dd t}}
\\[5mm] = &\
-\lim_{n\to \infty}\braces{\vphantom{\LARGE A}%
n\,\pars{-a-1}^{-n - 1}\bracks{\vphantom{\Large A}%
\Gamma\pars{n + 1,-a - 1} - \Gamma\pars{n + 1}}}
\label{1}\tag{1}
\end{align}
where $\ds{\Gamma\pars{\ldots,\ldots}}$ and
$\ds{\Gamma\pars{\ldots}}$ are the Incomplete Gamma and
Gamma Functions, respectively.
However, as $\ds{n \to \infty}$:
\begin{equation}
\Gamma\pars{n + 1,-a - 1} - \Gamma\pars{n + 1} \sim
-\,{\pars{-a - 1}^{n + 1}\expo{a + 1} \over n + 1}
\label{2}\tag{2}
\end{equation}
(\ref{1}) and (\ref{2}) lead to
$$
\bbox[5px,#ffd]{\lim_{n\to \infty}\bracks{n\int_{1}^{\expo{}} \!\! x^{a}\ln^{n}\pars{x}\,\dd x}} =
\bbx{\expo{a + 1}}
$$
