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?
Best Answer
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
$$I_{n,m} = (-1)^m \frac{m!}{(n+1)^{m+1}}.$$