$I_n = \int_1^e (\ln x)^n dx$. Compute $\lim_{n\to\infty} nI_n$ If
$$I_n = \int_1^e (\ln x)^n dx$$
Compute $$\lim_{n\to\infty} nI_n$$
One solution would be to determine the recurrence relation of $I_n$ (using integration by parts), namely $$I_{n+1} + (n+1)I_n = e$$
From the recurrence relation and from the fact that $I_n$ is decreasing we can determine that $$e = I_{n+1} + (n+1)I_n \ge I_{n+1} + (n+1)I_{n+1} = (n+2)I_{n+1}$$
which means that $$I_{n+1} \le \frac{e}{n+2}$$ so $$\lim_{n \to \infty} I_n = 0\tag{1}\label{1}$$
The recurrence relation can be rewritten as $$nI_n = e - I_{n+1} - I_{n}$$ and using $\eqref{1}$ we can conclude that $$\lim_{n\to\infty} nI_n = e$$
Is there another "more elegant" solution to this problem? Thank you!
 A: We can consider the exponential generating function
$$\sum_{n \ge 0} I_n \frac{t^n}{n!} = \int_1^e e^{t \ln x} \, dx = \int_1^e x^t \, dx = \frac{e^{t+1} - 1}{t + 1}$$
which gives the exact value
$$\begin{eqnarray*} \frac{I_n}{n!} &=& e \sum_{k=0}^n \frac{(-1)^{n-k}}{k!} - (-1)^n \\
 &=& (-1)^n \left( e \left( e^{-1} - \sum_{k=n+1}^{\infty} \frac{(-1)^k}{k!} \right) - 1 \right) \\
 &=& (-1)^{n+1} e \sum_{k=n+1}^{\infty} \frac{(-1)^k}{k!}. \end{eqnarray*}$$
This gives
$$\begin{eqnarray*} I_n &=& e \left( \frac{1}{n+1} - \frac{1}{(n+1)(n+2)} \pm \dots \right) \\
 &=& \frac{e}{n} + O \left( \frac{1}{n^2} \right) \end{eqnarray*}$$
as desired, and we can extract more terms in the asymptotic expansion too, e.g. the next term is $- \frac{2e}{n^2}$.
A: Here's my less elegant solution.
It turns out that
I recognized the subfactorials
and used one expression for them.
The limit is $e$.
$\begin{array}\\
I_n
&=\int_1^e \ln^n(x) dx\\
&=\int_0^1 y^ne^y dy\\
\end{array}
$
$y=\ln(x),
dy=\dfrac{dx}{x},
dx=x dy = e^y dy
$
$\int u dv = uv-\int v du$,
$u=y^n, dv=e^y,
du=ny^{n-1}, v=e^y,\\
\int y^ne^y dy
=y^ne^y-\int ny^{n-1}e^y dy
$
$\begin{array}\\
I_n
&=(y^ne^y)|_0^1-nI_{n-1}\\
&=e-nI_{n-1}\\
&=a_ne+b_n\\
&=e-n(a_{n-1}e+b_{n-1})\\
&=e(1-na_{n-1})-nb_{n-1}\\
a_n
&=1-na_{n-1}\\
b_n
&=-nb_{n-1}\\
\end{array}
$
$I_0=e-1,
I_1=1,
I_2=e-2$
$a_0=1,
b_0=-1
$
$b_n
=(-1)^{n+1}n!
$
$a_{1, 2, 3, 4, 5}
=0, 1, -2, 9, -44
$
$a_n
=(-1)^{n}!n
$
the subfactorial.
$\begin{array}\\
a_n
&=(-1)^{n}n!\sum_{k=0}^n \dfrac{(-1)^k}{k!}\\
&=(-1)^{n}n!\left(\dfrac1{e}-\sum_{k=n+1}^{\infty} \dfrac{(-1)^k}{k!}\right)\\
&=(-1)^{n}n!\left(\dfrac1{e}-\sum_{k=n+1}^{\infty} \dfrac{(-1)^k}{k!}\right)\\
&=(-1)^{n}\dfrac{n!}{e}-(-1)^{n}n!\sum_{k=n+1}^{\infty} \dfrac{(-1)^k}{k!}\\
&=(-1)^{n}\dfrac{n!}{e}-(-1)^{n}\left(\dfrac{(-1)^{n+1}}{n+1}+O(\dfrac1{n^2})\right)\\
&=(-1)^{n}\dfrac{n!}{e}+\left(\dfrac1{n+1}+O(\dfrac1{n^2})\right)\\
ea_n
&=(-1)^{n}n!+\left(\dfrac{e}{n+1}+O(\dfrac1{n^2})\right)\\
I_n
&=ea_n+b_n\\
&=(-1)^{n}n!+\left(\dfrac{e}{n+1}+O(\dfrac1{n^2})\right)+(-1)^{n+1}n!\\
&=\dfrac{e}{n+1}+O(\dfrac1{n^2})\\
nI_n
&=\dfrac{ne}{n+1}+O(\dfrac1{n})\\
&=e-\dfrac{e}{n+1}+O(\dfrac1{n})\\
&=e+O(\dfrac1{n})\\
\end{array}
$
A: Let $t=\ln(x),$
$$I_n=\int_0^1 t^n e^t dt$$
Upper bound:
$$nI_n\le n\int_0^1 t^n \cdot e\cdot dt=\frac{e\cdot n}{n+1}$$
Lower bound:
$$\text{Take any fixed}~~\delta, ~~\text{where}~~ 0<\delta<1,~~~~ I_n=\int_0^\delta t^n e^t dt+\int_\delta ^1 t^n e^tdt$$
$$\begin{align}
I_n&\ge \int_0^\delta t^n e^0 dt+\int_{\delta} ^1 t^n e^\delta dt\\
\\
I_n&\ge \frac{\delta^{n+1}}{n+1}+e^\delta\cdot \frac{1-\delta^{n+1}}{n+1}\\
\\
nI_n&\ge \frac{n}{n+1}\cdot \delta^{n+1}+e^\delta\cdot \frac{n}{n+1}\cdot(1-\delta^{n+1})\\
\end{align}$$
Combine the lower and upper bound:
$$\frac{n}{n+1}\cdot \delta^{n+1}+e^\delta\cdot \frac{n}{n+1}\cdot(1-\delta^{n+1})\le nI_n\le \frac{e\cdot n}{n+1}$$
Take the limit:
$$\lim_{n\rightarrow \infty}\left(\frac{n}{n+1}\cdot \delta^{n+1}+e^\delta\cdot \frac{n}{n+1}\cdot(1-\delta^{n+1})\right)\le \lim_{n\rightarrow \infty}nI_n\le \lim_{n\rightarrow \infty}\frac{e\cdot n}{n+1}$$
So we get:
$$e^\delta\le \lim_{n\rightarrow \infty}nI_n\le e$$
Since $\delta$ can be chosen arbitrarily close to $1$, so we conclude the limit is $e$.
A: Let $t=\ln(x)$,
$$I_n=\int_0^1 t^n e^t dt$$
Use integration by part:
$$\begin{align}
I_n&=\frac{e}{n+1}-\frac{1}{n+1}\int_0^1 t^{n+1}e^t dt\\
\\
\lim_{n\rightarrow\infty}nI_n&=\lim_{n\rightarrow\infty}\frac{n\cdot e}{n+1}-\lim_{n\rightarrow\infty}\left(\frac{n}{n+1}\int_0^1 t^{n+1}e^t dt\right)\\
\\
&=e-\lim_{n\rightarrow\infty}\int_0^1 t^{n+1}e^t dt
\end{align}$$
Let $f_n(t)=t^{n+1}e^t$, $\Rightarrow|f_n(t)|\le e^t$ for all $n$ and $t\in[0,1]$, and $e^t$ is integrable on $(0,1)$. So by Dominated Convergence Theorem, we can interchange the limit and the integral:
$$\begin{align}
\lim_{n\rightarrow\infty}nI_n&=e-\int_0^1 \left(\lim_{n\rightarrow\infty}t^{n+1}e^t\right) dt\\
\\
&=e-0=e
\end{align}$$
