$\int_0^1 (x \ln x)^{2020} \, \mathrm{d}x$ I ran into this integral today
$$\int_0^1 \left( x \ln x \right)^{2020} \,\mathrm{d}x \overset{?}{=} \frac{\Gamma(2021)}{2021^{2021}}$$
My solution goes along these lines. Recalling the identity
$$\ln x = \lim_{n \rightarrow +\infty} n \left( x^{1/n} - 1 \right)$$
one can write the integral as follows:
\begin{align*} 
\int_{0}^{1} \left ( x \ln x \right )^{2020}\, \mathrm{d}x &= \lim_{n \rightarrow +\infty} n^{2020} \int_{0}^{1} \left ( x^{2020} \left ( x^{1/n} -1 \right )^{2020} \right )\, \mathrm{d}x \\ 
&\!\!\!\!\!\!\overset{u=x^{1/n}}{=\! =\! =\! =\!} \lim_{n \rightarrow +\infty} n^{2021} \int_{0}^{1} u^{2021n-1} \left ( 1-u \right )^{2020} \, \mathrm{d}u \\ 
&=\lim_{n \rightarrow +\infty} n^{2021} \int_{0}^{1} u^{2021n-1} \left ( 1-u \right )^{2021-1} \, \mathrm{d}u \\ 
&=\lim_{n \rightarrow +\infty} n^{2021} \mathrm{B} \left ( 2021n, 2021 \right ) \\ 
&=\lim_{n \rightarrow +\infty} n^{2021} \; \frac{\Gamma \left ( 2021 n \right ) \Gamma \left ( 2021 \right )}{\Gamma \left ( 2021 n + 2021 \right )} \\ 
&= \Gamma \left ( 2021 \right ) \lim_{n \rightarrow +\infty} n^{2021} \frac{\Gamma \left ( 2021 n \right )}{\Gamma \left ( 2021 n + 2021 \right )} 
\end{align*}
Using Gautschi's inequality one has that
$$n^{2021}\left ( 2021n -1 \right )^{1-2022}<\frac{ n^{2021} \Gamma\left ( 2021n -1 + 1 \right )}{\Gamma\left ( 2021n-1 + 2022 \right )}< n^{2021}\left ( 2021n\right )^{1-2022}$$
and thus by the squeeze theorem one has the result I stated above.
My doubt however is at the application of the inequality. Is it correct?
Can you suggest other ways of solving the problem?
 A: I don't know if the inequality is correct, but here is another way to solve the integral.
Let
$$I_{n,m} = \int_0^1 x^n \ln^m x \, dx.$$
Notice that the integral we wish to evaluate is $I_{n,n}$. Now, use integration by parts with $u = \ln^m x$ and $dv = x^n \, dx$ to obtain
$$I_{n,m} = \frac{1}{n+1} x^{n+1} \ln^m x \Big|_0^1 - \frac{m}{n+1} \int_0^1 x^n \ln^{m-1} x \, dx$$
$$I_{n,m} = (-1) \frac{m}{n+1} I_{n,m-1}.$$
We can eliminate the natural log term completely by applying this recursive formula $m$ times:
\begin{align}
I_{n,m} &= (-1)^1 \cdot \frac{m}{n+1} \cdot I_{n,m-1} \\
&= (-1)^2 \cdot \frac{m}{n+1} \cdot \frac{m-1}{n+1} \cdot I_{n,m-2} \\
&= (-1)^3 \cdot \frac{m}{n+1} \cdot \frac{m-1}{n+1} \cdot \frac{m-2}{n+1} \cdot I_{n,m-3} \\
&= \cdots \\
&= (-1)^m \cdot \frac{m!}{(n+1)^m} \cdot I_{n,0}.
\end{align}
Since
$$I_{n,0} = \int_0^1 x^n \, dx = \frac{1}{n+1},$$
in totality we have
$$I_{n,m} = (-1)^m \frac{m!}{(n+1)^{m+1}}.$$
A: METHODOLOGY $1$:  Using the Integral Representation of the Gamma Function
Let $I$ be given by the integral 
$$I=\int_0^1 (x\log(x))^{2020}\,dx$$
Now, let $x=e^{-t}$.  Then, we have
$$\begin{align}
I&=\int_0^\infty (e^{-t}t)^{2020}\,e^{-t}\,dt\\\\
&=\int_0^\infty t^{2020}e^{-2021t}\,dt
\end{align}$$
Next, let $t=s/2021$ so that
$$\begin{align}
I&=\frac1{2021}\int_0^\infty \left(\frac{s}{2021}\right)^{2020}e^{-s}\,ds\\\\
&=\frac{\Gamma(2021)}{(2021)^{2021}}
\end{align}$$
And we are done!

METHODOLOGY $2$:  Using Feynman's Trick
Let $I_n(s)$ be given by 
$$I_n(s)=\int_0^1 x^s \log^{n}(x)\,dx$$
Note that we have 
$$\begin{align}
I_n(s)&=\frac{d^n}{ds^n}\int_0^1 x^s\,ds\\\\
&=\frac{d^n}{ds^n}\left(\frac{1}{s+1}\right)\\\\
&=\frac{(-1)^n n!}{(s+1)^{n+1}}
\end{align}$$
Setting $s=n=2020$ yields
$$\begin{align}
\int_0^1 x^{2020}\log^{2020}(x)\,dx&=\frac{(2020)!}{(2021)^{2021}}\\\\
&=\frac{\Gamma(2021)}{(2021)^{2021}}
\end{align}$$
A: Applying Mark Viola's substitution
to Math2718's generalization.
$I_{n,m} 
= \int_0^1 x^n \ln^m x \, dx.
$.
Let $x = e^{-t}$
so $\ln(x) = -t$ and
$dx = -e^{-t}dt$
so
$\begin{array}\\
I_{n,m} 
&= \int_0^1 x^n \ln^m x \, dx.\\
&= -\int_{\infty}^0 e^{-nt} (-t)^me^{-t}dt\\
&= (-1)^m\int_0^{\infty} e^{-(n+1)t} t^mdt\\
&= (-1)^m\int_0^{\infty} e^{-u} (u/(n+1))^mdu/(n+1)
\qquad u=(n+1)t, dt = du/(n+1)\\
&= \dfrac{(-1)^m}{(n+1)^{m+1}}\int_0^{\infty} e^{-u} u^{m}du\\
&= \dfrac{(-1)^mm!}{(n+1)^{m+1}}\\
\end{array}
$
