Solution: \int_{0}^{1} \frac{x^{2021}-1}{\log x} d x

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Calculate the integral

Solution

Let’s define

\[I(k)=\int_{0}^{1} \frac{x^{k}-1}{\log x} d x\]

\((x, k) \rightarrow \frac{x^{k}-1}{\log x}\) is continuous on \(] 0,1] \times \mathbb{R}_{+}\)

\[\lim _{x \rightarrow 0^{+}} \frac{x^{k}-1}{\log x}=\lim _{x \rightarrow 0^{+}}\left(x^{k}-1\right) \frac{1}{\log x}\]

We know that \[\lim _{x \rightarrow 0^{+}} \log x=-\infty\]

\[\begin{aligned}
&\Rightarrow \lim _{x \rightarrow 0^{+}} \frac{1}{\log x}=0 \\
&\Rightarrow \lim _{x \rightarrow 0^{+}} \frac{x^{k}-1}{\log x}=0
\end{aligned}\]

\(\Rightarrow(x, k) \rightarrow \frac{x^{k}-1}{\log x}\) is continuous on \([0,1] \times \mathbb{R}_{+}\)

\[\begin{aligned} \frac{\partial}{\partial k}\left(\frac{x^{k}-1}{\log x}\right) &=\frac{1}{\log x} \frac{\partial}{\partial k}\left(e^{k \log x}-1\right) \\ &=\frac{1}{\log x}\left((\log x) e^{k \log x}\right) \\ &=x^{k} \end{aligned}\]

\((x, k) \rightarrow \frac{\partial}{\partial k}\left(\frac{x^{k}-1}{\log x}\right)\) is continuous on \([0,1] \times \mathbb{R}_{+}\)

Differentiate both sides with respect to \(k\)

\[I^{\prime}(k)=\frac{d}{d k} \int_{0}^{1} \frac{x^{k}-1}{\log x} d x\]

By Differentiation Under the Integral Sign, we have

\[I^{\prime}\left( k\right) =\int _{0}^{1}\dfrac{\partial }{\partial k}\left( \dfrac{x^{k}-1}{\log x}\right) dx\] \[=\int _{0}^{1}x^{k}dx\]

\[=\left[ \dfrac{x^{k+1}}{k+1}\right] _{0}^{1}=\dfrac{1^{k+1}}{k+1}-\dfrac{0^{k+1}}{k+1}\]

\[=\dfrac{1}{k+1}\]

Integrate with respect to \(k\)

\[I\left( k\right) =\log \left( k+1\right) +c \]

Let \(k=0 \)

\[I\left( 0\right) =\log \left( 0+1\right) +c \]

\[I\left( 0\right) =\int _{0}^{1}\dfrac{x^{0}-1}{logx}dx=0 \]