$\int_0^1 (x \ln x)^{2020} \, \mathrm{d}x$

Question

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?

Answer 1

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

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

Answer 2

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!

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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

$$\begin{align} \int_0^1 x^{2020}\log^{2020}(x)\,dx&=\frac{(2020)!}{(2021)^{2021}}\\\\ &=\frac{\Gamma(2021)}{(2021)^{2021}} \end{align}$$

Answer 3

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} $