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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 $$k=0$$
$I\left( 0\right) =\log \left( 0+1\right) +c$
$I\left( 0\right) =\int _{0}^{1}\dfrac{x^{0}-1}{logx}dx=0$
$\Rightarrow c=0$
Hence
$I\left( k\right) =\log \left( k+1\right)$
For $$k=2021$$
\begin{aligned}I\left( 2021\right) =\log \left( 2021+1\right) = \log \left( 2022\right) \end{aligned}
$\Rightarrow \int _{0}^{1}\dfrac{x^{2021}-1}{\log x}dx=\log \left( 2022\right)$
Home -> Solved problems -> Calculate the integral

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