Finding a closed form for $\int_1^e \frac{(\ln x)^n}{x^2} dx$

Question

I'm trying to find the closed form of the integral $I_n=\int_1^e \frac{(\ln{x})^n}{x^2} dx$.

So I start of with by parts with $u=\frac{1}{x}$ and $v'=\frac{(\ln{x})^n}{x}$

This gives me $I_n=\frac{1}{e(n+1)}+\frac{1}{n+1}I_{n+1}$ which implies that $I_n=-\frac{1}{e}+nI_{n-1}$

From here, I use the recursive $n$ times to get the following expression: $I_n=-\frac{1}{e}-\frac{1}{e}\big(n+n(n-1)+\ldots+n(n-1)\ldots2\big)+n!I_0$ which I re-write (noting that $I_0=1-\frac{1}{e}$ as:

$I_n=n!-\frac{1}{e}\big(1+n+n(n-1)+\ldots+n!\big)$.

Here, I struggle to sum up $1+n+n(n-1)+\ldots+n!$. Looking at the limit as $n$ approaches infinity, the area under $I_n$ should approach zero due to $\ln{x}$ being $<1$ between the bounds so I guess I can deduce that $1+n+n(n-1)+...+n!\longrightarrow en!$ (I'm not sure how rigorous this deduction is though, if it's even true).

So going off this assumption, I guess I could write $1+n+n(n-1)+...+n!=n!\sum_{k=1}^{n}\frac{1}{k!}=n!\sum_{k=1}^{\infty}\frac{1}{k!}-n!\sum_{k=n+1}^{\infty}\frac{1}{k!}=en!-n!\sum_{k=n+1}^{\infty}\frac{1}{k!}$.

But I'm completely stuck here, if I'm even going in the right direction or if this 'simple' looking sum even has a nice closed form. If anyone knows how to move from here, that would be great!

UPDATE: at Expression for $n+n(n-1)+n(n-1)(n-2)+...+n!$, they have provided that $n+n(n-1)+...+n!=\lfloor en! \rfloor$, which gives us $I_n=n!-\frac{\lfloor en! \rfloor}{e}$. I'm not sure how the link deduced that $n!\sum_{k=n+1}^{\infty}\frac{1}{k!}<1$ though.

FURTHER UPDATE: figured out how to deduce that $n!\sum_{k=n+1}^{\infty}\frac{1}{k!}<1$ for $n>1$

We have $n!\sum_{k=n+1}^{\infty}\frac{1}{k!}=\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)(n+3)}+...$

$<\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}+\frac{1}{(n+2)(n+3)}+\frac{1}{(n+3)(n+4)}+...$

$=\frac{1}{n+1}+\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+2}+...$

$=\frac{2}{n+1}$

Answer 1

This may not be the answer you're looking for, since I don't know whether or not you would consider the incomplete gamma function closed form or not. If not, I'll delete.

Perform the substitution $\ln(x)\mapsto x$ on the indefinite integral to get $$\int \frac{(\ln{x})^n}{x^2}\text{ d}x=\int u^n\cdot e^{-u}\text{ d}u=-\Gamma(n+1,u)=-\Gamma(n+1,\ln(x))$$

It is assumed that $n>-1$.

Plug in bounds to get $\Gamma(n+1) - \Gamma(n+1, 1)$ as the answer to that integral.

Answer 2

$$ \begin{aligned} \int_1^e \frac{(\ln x)^n}{x^2} d x &\stackrel{y=\ln x}{=} \int_0^1 \frac{y^n}{e^{2 y}} e^{y }d y \\ &= \int_0^1 y^n e^{-y} d y \\ &= \int_0^{\infty} y^n e^{-y} d y -\int_1^{\infty} y^n e^{-y} d y \\ &=\gamma(n+1,1) \end{aligned} $$