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A logarithmic integral

Evaluate the integral

    \[\mathcal{J} = \int_0^1 \left(x \ln x \right)^{2020} \, \mathrm{d}x\]

Solution

Recall the idenity

    \[\mathbf{\ln x = \lim_{n \rightarrow +\infty} n \left ( x^{1/n} - 1 \right )}\]

thus,

    \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 it follows that

\displaystyle 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 hence the integral equals

    \[\mathcal{J} = \frac{\Gamma(2021)}{2021^{2021}}\]

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