Let $p > -1$ and $r \in \mathbb{N}$, show that $$\lim_n \int_0^n x^p (\ln x)^r \left(1 - \frac{x}{n} \right)^n dx = \int_0^\infty x^p (\ln x)^r e^{-x} dx$$ and that this integral is finite.
To solve that problem, I wanted to use Lebesgue's dominated convergence theorem.
Therefore, I set $f_n(x) = x^p (\ln x )^r \left( 1 - \frac{x}{n} \right)^n$. Then we clearly have that $\lim_n f_n(x) = x^p (\ln x)^r e^{-x}$. Plus $f_n$ are measurable for all $n$. Thus the remaining problem is to find an integrable function $g$ such that $|f_n(x)| \leq g(x)$ almost everywhere (I don't know if it is said like that in english). That is where I'm stuck, I can't find such a function.
At first I wanted to use that $\ln(x) \leq x\ \forall x \geq 0$, thus $f(x) \leq x^{p+r}e^{-x}$ where $f(x) = \lim_n f_n(x)$ but it doesn't seem to lead anywhere.
Any help would be appreciated. Thanks in advance.
Oh I almost forgot about this question. So here's my proof (maybe some parts are a bit messy, so if you want to edit some parts so that it is more rigorous/better, feel free to do so).
As mentioned in the question, we will use Lebesgue's dominated convergence theorem, thus we have two parts to verify:
We set $f_n(x) = x^p (\ln x)^r \left( 1 - \frac{1}{n} \right)^n {\large \chi}_{]0,n[}$. Then $f_n$ is measurable as it is a product of measurable functions and we clearly have $f_n(x) \xrightarrow{n \to \infty} x^p (\ln x)^r e^{-x} {\large \chi}_{]0, \infty[}$.
Now comes the part that is a bit harder. Remember that we have $\ln(1-x) \leq -x$ for all $x < 1$. Now we can use this identity because we consider only $x < n$, thus $\frac{x}{n}<1$ (you will see why this remark is useful). We have $$ \begin{align*} \ln \left( \left( 1 - \frac{x}{n} \right)^n \right) &= n \ln \left( 1 - \frac{x}{n} \right) \leq -n \frac{x}{n} = -x\\ \Rightarrow \underbrace{\exp \left( \ln \left( \left( 1 - \frac{x}{n} \right)^n \right) \right)}_{= \left( 1 - \frac{x}{n} \right)^n} & \leq \exp(-x)\\ |f_n(x)| \leq x^p |\ln x|^r e^{-x} {\large \chi}_{]0, \infty[}=: g(x). \end{align*} $$
Okay, that was the easy part of the hard part. Now that we are warmed up, let's attack the integration part. The first step is to split the integral: $$ \begin{align*} \int_{\mathbb R} g(x)dx &= \int_0^\infty x^p |\ln x|^r e^{-x} dx\\ &= \int_0^1 x^p |\ln x|^r e^{-x}dx + \int_1^\infty x^p (\ln x)^r e^{-x} dx. \end{align*} $$
Once again we will use some property of the natural logarithm, namely $\ln(x) \leq x\ \forall x \geq 1$. We are going to solve the second integral first. $$ \begin{align*} \int_1^\infty x^p |\ln x|^r e^{-x} dx \leq \int_1^\infty x^{p+r} e^{-x} dx \end{align*} $$ (Now comes the first messy part, or not so formal part if you prefer.) We know that if we have a product of a polynomial with an exponential, then the product is asymptotically equivalent to the exponential, thus $x^{p+r}e^{-x} = o(e^{-x})$ on $[1, +\infty[$, therefore $$ \begin{align*} \int_1^\infty x^p |\ln x|^r e^{-x} dx &\leq \int_1^\infty x^{p+r} e^{-x} dx\\ &= \int_1^\infty o(e^{-x})dx\\ &< + \infty \end{align*} $$ because the negative exponential is integrable on $[1, +\infty[$.
Now let's attack the second integral. We will use the following change of variable: $y = \ln x \Rightarrow x = e^y \Rightarrow dx = e^y dy$. We are also going to use that $e^{-x} \leq 1$ on $]0,1]$. $$ \begin{align*} \int_0^1 x^p |\ln x|^r e^{-x}dx &\leq \int_0^1 x^{p+1} \cdot \frac{1}{x} \cdot (-\ln x)^rdx\\ &= (-1)^r \int_0^1 x^{p+1} \cdot \frac{1}{x} \cdot (\ln x)^r dx\\ &= (-1)^r \int_{-\infty}^0 e^{y\overbrace{(p+1)}^{>0}} e^{-y} y^r e^y dy\\ &= (-1)^r \int_{-\infty}^0 e^{y(p+1)}y^r dy \end{align*} $$ and by the same argument we used before (second messy part), this integral is also finite. Thus the sum is finite, therefore $g$ is integrable on $\mathbb R$ as we wanted.
By Lebesgue's dominated convergence theorem, we conclude that $$ \lim_{n \to \infty} \int_0^n x^p (\ln x)^r \left( 1 - \frac{1}{n} \right)^ndx = \int_0^\infty x^p (\ln x)^r e^{-x}dx$$ and that this integral is finite. $\square$