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Calculate the integral

Solution

Let’s define

$$I(k)={\int }_0^1\frac{x^k-1}{\log x}dx$$

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

$$\underset{x\to 0^{+}}{lim}\frac{x^k-1}{\log x}=\underset{x\to 0^{+}}{lim}(x^k-1)\frac{1}{\log x}$$

We know that $$\underset{x\to 0^{+}}{lim}\log x=-\mathrm{\infty }$$

$$\begin{array}{rl}& \Rightarrow \underset{x\to 0^{+}}{lim}\frac{1}{\log x}=0\\ & \Rightarrow \underset{x\to 0^{+}}{lim}\frac{x^k-1}{\log x}=0\end{array}$$

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

$$\begin{array}{rl}\frac{\mathrm{\partial }}{\mathrm{\partial }k}\left(\frac{x^k-1}{\log x}\right)& =\frac{1}{\log x}\frac{\mathrm{\partial }}{\mathrm{\partial }k}(e^{k\log x}-1)\\ & =\frac{1}{\log x}((\log x)e^{k\log x})\\ & =x^k\end{array}$$

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

Differentiate both sides with respect to $k$

$$I^{\mathrm{\prime }}(k)=\frac{d}{dk}{\int }_0^1\frac{x^k-1}{\log x}dx$$

By Differentiation Under the Integral Sign, we have

$$I^{\mathrm{\prime }}\left(k\right)={\int }_0^1{\displaystyle \frac{\mathrm{\partial }}{\mathrm{\partial }k}}\left({\displaystyle \frac{x^k-1}{\log x}}\right)dx$$ $$={\int }_0^1{x}^{k}dx$$

$$={\left[{\displaystyle \frac{x^{k+1}}{k+1}}\right]}_0^1={\displaystyle \frac{1^{k+1}}{k+1}}-{\displaystyle \frac{0^{k+1}}{k+1}}$$

$$={\displaystyle \frac{1}{k+1}}$$

Integrate with respect to $k$

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

Let $k=0$

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

$$I\left(0\right)={\int }_0^1{\displaystyle \frac{x^0-1}{logx}}dx=0$$

$$\Rightarrow c=0$$

Hence

$$I\left(k\right)=\log (k+1)$$

For $k=2021$

$$\begin{array}{r}I\left(2021\right)=\log (2021+1)=\log \left(2022\right)\end{array}$$

$$\Rightarrow {\int }_0^1{\displaystyle \frac{x^{2021}-1}{\log x}}dx=\log \left(2022\right)$$

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